Category: Class 12

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.5 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5

Question 1.
Apply the given elementary transformation on each of the following matrices:
(i) \(\left[\begin{array}{cc}
3 & -4 \\
2 & 2
\end{array}\right]\), R₁ ↔ R₂
(ii) \(\left[\begin{array}{cc}
2 & 4 \\
1 & -5
\end{array}\right]\), C₁ ↔ C₂
(iii) \(\left[\begin{array}{ccc}
3 & 1 & -1 \\
1 & 3 & 1 \\
-1 & 1 & 3
\end{array}\right]\) 3R₂ and C₂ → C₂ – 4C₁
Solution:

Question 2.
Transform \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
2 & 1 & 3 \\
3 & 2 & 4
\end{array}\right]\) into an upper triangularmatrix by suitable row transformations.
Solution:

Question 3.
Find the cofactor matrix of the following matrices:
(i) \(\left[\begin{array}{cc}
1 & 2 \\
5 & -8
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}
5 & 8 & 7 \\
-1 & -2 & 1 \\
-2 & 1 & 1
\end{array}\right]\)
Solution:

Question 4.
Find the adjoint of the following matrices:
(i) \(\left[\begin{array}{cc}
2 & -3 \\
3 & 5
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
Solution:

Question 5.
Find the inverses of the following matrices by the adjoint mathod:
(i) \(\left[\begin{array}{rr}
3 & -1 \\
2 & -1
\end{array}\right]\)
(ii) \(\left[\begin{array}{cc}
2 & -2 \\
4 & 5
\end{array}\right]\)
(iii) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
Solution:

Question 6.
Find the inverses of the following matrices by the transformation method:
(i) \(\left[\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Solution:

Question 7.
Find the inverse of A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) by elementary column transformations.
Solution:

Question 8.
Find the inverse of \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\) by the elementary row transformations.
Solution:

Question 9.
If A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\), then find matrix X such that XA = B.
Solution:

Question 10.
Find matrix X, if AX = B, where A = \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
-1 & 1 & 2 \\
1 & 2 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
Solution:

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.9 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.9

Question 1.
Without using truth table, show that
(i) p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Solution:
LHS = p ↔ q
≡ (p → q) ∧ (q → p)
≡ (~p ∨ q) ∧ (~(q ∨ p) …..(Conditional Law)
≡ [~p ∧ (~(q ∨ p)] ∨ [q ∧ (~q ∨ p] …..(Distributive Law)
≡ [(~p ∧ ~q) ∨ (~p ∧ p)] ∨ [(q ∧ ~q) ∨ (q ∧ p)] ……(Distributive Law)
≡ [(~p ∧ ~q) ∨ c] ∨ [c ∨ (q ∧ p)] …..(Complement Law)
≡ (~p ∧ ~q) ∨ (q ∧ p) ……(Identity Law)
≡ (~p ∧ ~q) ∨ (p ∧ q) ……(Commutative Law)
≡ (p ∧ q) ∨ (~p∧ q) ……(Commutative Law)
≡ RHS.

(ii) p ∧ [~p ∨ q) ∨ (~q)] ≡ p
Solution:
LHS = p ∧ [(~p ∨ q) ∨ (~q)]
≡ p ∧ [~p ∨ (q ∨ ~q)] ……(Associative Law)
≡ p ∧ [~p ∨ t] …….(Complement Law)
≡ p ∧ t ……(Identity Law)
≡ p ……(Identity Law)
= RHS.

(iii) ~[(p ∧ q) → ~(q)] ≡ p ∧ q
Solution:
LHS = ~[(p ∧ q) → ~(~q)]
≡ (p ∧ q) ∧ ~(~q) ……(Negation of implication)
≡ (p ∧ q) ∧ q …..(Negation of negation)
≡ p ∧ (q ∧ q) …..(Associative Law)
≡ P ∧ q ……(Idempotent Law)
= RHS

(iv) ~r → ~(p ∧ q) ≡ [~(q → r)] → (~p)
Solution:
LHS = ~r → ~(p ∧ q)
≡ ~q → (~p ∨ ~q) ……(De Morgan’s Law)
≡ ~(~r) ∨ (~p ∨~q) …..(Conditional Law)
≡ r ∨ (~p ∨ ~q) …..(Involution Law)
≡ r ∨ ~q ∨ ~p …..(Commutative Law)
≡ (~q ∨ r) ∨ (~p) ……(Commutative Law)
≡ ~(q → r) ∨ (~p) ……(Conditional Law)
≡ ~(q → r) → (~p) ……(Conditional Law)
= RHS.

(v) (p ∨ q) → r ≡ (p → r) ∧ (q → r)
Solution:
LHS = (p ∨ q) → r
≡ ~(p → q) ∨ r ……..(Conditional Law)
≡ (~p ∧ ~q) ∨ r ……….(De Morgan’s Law)
≡ (~p ∨ r) ∧ (~q ∨ r) ………..(Distributive Law)
≡ (p → r) ∧ (q → r) …….(Conditional Law)
= RHS.

Question 2.
Using the algebra of statement, prove that:
(i) [p ∧ (q ∨ r)] ∨ [~r ∧ ~q ∧ p] ≡ p
Solution:
LHS = [p ∧ (q ∨ r)] ∨ [ ~r ∧ ~q ∧ p]
≡ [p ∧ (q ∨ r)] ∨ [(~r ∧ ~q) ∧ p] ……(Associative Law)
≡ [p ∧ (q ∨ r)] ∨ [(~q ∧ ~r) ∧ p] ……(Commutative Law)
≡ [p ∧ (q ∨ r)] ∨ [ ~ (q ∨ r) ∧ p] ……(De Morgan’s Law)
≡ [p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)] ……(Commutative Law)
≡ p ∧ [(q ∨ r) ∨ ~ (q ∨ r) ] …..(Distributive Law)
≡ p ∧ t …….(Complement Law)
≡ p ……(Identity Law)
= RHS.

(ii) (p ∧ q) ∨ (p ∧ ~q) ∨ (~p ∧ ~q) ≡ p ∨ ~q
Solution:
LHS = (p ∧ q) ∨ (p ∧ ~q) ∨ (~p ∧ ~ q)
≡ (p ∧ q) ∨ [(p ∧ ~q) ∨ (~p ∧ ~q)] ……(Associative Law)
≡ (p ∧ q) ∨ [(~q ∧ p) ∨ (~q ∧ ~p)] …..(CommutativeLaw)
≡ (p ∧ q) ∨ [ ~q ∧ (p ∨ ~ p)] …..(Distributive Law)
≡ (p ∧ q) ∨ (~q ∧ t) ……(Complement Law)
≡ (p ∧ q) ∨ (~q) …….(Identity Law)
≡ (p ∨ ~q) ∧ (q ∨ ~q) ……(Distributive Law)
≡ (p ∨ ~q) ∧ t …….(Complement Law)
≡ p ∨ ~q …..(Identity Law)
= RHS.

(iii) (p ∨ q) ∧ (~p ∨ ~q) ≡ (p ∧ ~q) ∨ (~p ∧ q)
Solution:
LHS = (p ∨ q) ∧ (~p ∨ ~q)
≡ [p ∧ (~p ∨ ~q)] ∨ [q ∧ (~p ∨ ~q)] ……(Distributive Law)
≡ [(p ∧ ~p) ∨ (p ∧ ~q)] ∨ [q ∧ ~p) ∨ (q ∧ ~q)] ……(Distributive Law)
≡ [c ∨ (p ∧ ~q)] ∨ [(q ∧ ~p) ∨ c] ……(Complement Law)
≡ (p ∧ ~q) ∨ (q ∧ ~p) ……..(Identity Law)
≡ (p ∧ ~q) ∨ (~p ∧ q) ………(Commutative Law)
= RHS.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.8 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.8

Question 1.
Write the negation of each of the following statements:
(i) All the stars are shining if it is night.
Solution:
The given statement can be written as:
If it is night, then all the stars are shining.
Let p : It is night.
q : All the stars are shining.
Then the symbolic form of the given statement is p → q
Since, ~(p → q) ≡ p ∧ ~q,
the negation of the given statement is ‘It is night and all the stars are not shining.’

(ii) ∀ n ∈ N, n + 1 > 0.
Solution:
The negation of the given statement is
‘∃ n ∈ N, such that n + 1 ≤ 0.’

(iii) ∃ n ∈ N, such that (n² + 2) is odd number.
Solution:
The negation of the given statement is
‘∀ n ∈ N, n² + 2 is not an odd number.’

(iv) Some continuous functions are differentiable.
Solution:
The negation of a given statement is ‘All continuous functions are not differentiable.’

Question 2.
Using the rules of negation, write the negations of the following:
(i) (p → r) ∧ q
Solution:
The negation of (p → r) ∧ q is
~[(p → r) ∧ q] ≡ ~(p → r) ∨ (~q) …..[Negation of conjunction]
≡ (p ∧ ~r) ∨ (~q) ……[Negation of implication]

(ii) ~(p ∨ q) → r
Solution:
The negation of ~(p ∨ q) → r is
~[~(p ∨ q) → r] ≡ ~(p ∨ q) ∧ (~r) …..[Negation of implication]
≡ (~p ∧ ~q) ∧ (~r) ……[Negation of disjunction]

(iii) (~p ∧ q) ∧ (~q ∨ ~r)
Solution:
The negation of (~p ∧ q) ∧ (~q ∨ ~r) is
~[(~p ∧ q) ∧ (~q ∨ ~ r)] ≡ ~(~p ∧ q) ∨ ~(~q ∨ ~r) ……[Negation of conjunction]
≡ [~(~p) ∨ ~q] ∨ [~(~q) ∧ ~(~r)] … [Negation of conjunction and disjunction]
≡ (p ∨ ~q) ∨ (q ∧ r) …..[Negation of negation]

Question 3.
Write the converse, inverse, and contrapositive of the following statements:
(i) If it snows, then they do not drive the car.
Solution:
Let p : It snows.
q : They do not drive the car.
Then the symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q.
i.e. If they do not drive the car, then it snows.
Inverse: ~p → ~q is the inverse of p → q.
i.e. If it does not snow, then they drive the car.
Contrapositive: ~q → ~p is the contrapositive of p → q.
i.e. If they drive the car, then it does not snow.

(ii) If he studies, then he will go to college.
Solution:
Let p : He studies.
q : He will go to college.
Then two symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q.
i.e. If he will go to college, then he studies.
Inverse: ~p → ~q is the inverse of p → q.
i.e. If he does not study, then he will not go to college.
Contrapositive: ~q → ~p is the contrapositive of p → q.
i.e. If he will not go to college, then he does not study.

Question 4.
With proper justification, state the negation of each of the following:
(i) (p → q) ∨ (p → r)
Solution:
The negation of (p → q) ∨ (p → r) is
~[(p → q) ∨ (p → r)] ≡ ~(p → q) ∧ ~(p → r) …..[Negation of disjunction]
≡ (p ∧ ~q) ∧ (p ∧ ~r) …[Negation of implication]

(ii) (p ↔ q) ∨ (~q → ~r)
Solution:
The negation of (p ↔ q) ∨ (~q → ~r) is
~[(p ↔ q) ∨ (~q → ~r)] ≡ ~(p ↔ q) ∧ ~(~q → ~r) …..[Negation of disjunction]
≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ [~q ∧ ~(~r)] ……[Negation of biconditional and implication]
≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ (~q ∧ r) ……[Negation of negation]

(iii) (p → q) ∧ r
Solution:
The negation of (p → q) ∧ r is
~[(p → q) ∧ r] ≡ ~(p → q) ∨ (~r) …..[Negation of conjunction]
≡ (p ∧ ~q) ∨ (~r) …..[Negation of implication]

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.7 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.7

Question 1.
Write the dual of each of the following:
(i) (p ∨ q) ∨ r
Solution:
(p ∧ q) ∧ r

(ii) ~(p ∨ q) ∧ [p ∨ ~(q ∧ ~r)]
Solution:
~(p ∧ q) ∨ [p ∧ ~(q ∨ ~r)]

(iii) p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r
Solution:
p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r

(iv) ~(p ∧ q) ≡ ~p ∨ ~q
Solution:
~(p ∨ q) ≡ ~p ∧ ~q

Question 2.
Write the dual statement of each of the following compound statements:
(i) 13 is prime number and India is a democratic country.
Solution:
13 is prime number or India is a democratic country.

(ii) Karina is very good or everybody likes her.
Solution:
Karina is very good and everybody likes her.

(iii) Radha and Sushmita can not read Urdu.
Solution:
Radha or Sushmita can not read Urdu.

(iv) A number is real number and the square of the number is non-negative.
Solution:
A number is real number or the square of the number is non-negative.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

I. Choose the correct option from the given alternatives:

Question 1.
The area bounded by the region 1 ≤ x ≤ 5 and 2 ≤ y ≤ 5 is given by
(a) 12 sq units
(b) 8 sq units
(c) 25 sq units
(d) 32 sq units
Answer:
(a) 12 sq units

Question 2.
The area of the region enclosed by the curve y = \(\frac{1}{x}\), and the lines x = e, x = e² is given by
(a) 1 sq unit
(b) \(\frac{1}{2}\) sq units
(c) \(\frac{3}{2}\) sq units
(d) \(\frac{5}{2}\) sq units
Answer:
(a) 1 sq unit

Question 3.
The area bounded by the curve y = x³, the X-axis and the lines x = -2 and x = 1 is
(a) -9 sq units
(b) \(-\frac{15}{4}\) sq units
(c) \(\frac{15}{4}\) sq units
(d) \(\frac{17}{4}\) sq units
Answer:
(c) \(\frac{15}{4}\) sq units

Question 4.
The area enclosed between the parabola y² = 4x and line y = 2x is
(a) \(\frac{2}{3}\) sq units
(b) \(\frac{1}{3}\) sq units
(c) \(\frac{1}{4}\) sq units
(d) \(\frac{3}{4}\) sq units
Answer:
(b) \(\frac{1}{3}\) sq units

Question 5.
The area of the region bounded between the line x = 4 and the parabola y² = 16x is
(a) \(\frac{128}{3}\) sq units
(b) \(\frac{108}{3}\) sq units
(c) \(\frac{118}{3}\) sq units
(d) \(\frac{218}{3}\) sq units
Answer:
(a) \(\frac{128}{3}\) sq units

Question 6.
The area of the region bounded by y = cos x, Y-axis and the lines x = 0, x = 2π is
(a) 1 sq unit
(b) 2 sq units
(c) 3 sq units
(d) 4 sq units
Answer:
(d) 4 sq units

Question 7.
The area bounded by the parabola y² = 8x, the X-axis and the latus rectum is
(a) \(\frac{31}{3}\) sq units
(b) \(\frac{32}{3}\) sq units
(c) \(\frac{32 \sqrt{2}}{3}\) sq units
(d) \(\frac{16}{3}\) sq units
Answer:
(b) \(\frac{32}{3}\) sq units

Question 8.
The area under the curve y = 2√x, enclosed between the lines x = 0 and x = 1 is
(a) 4 sq units
(b) \(\frac{3}{4}\) sq units
(c) \(\frac{2}{3}\) sq units
(d) \(\frac{4}{3}\) sq units
Answer:
(d) \(\frac{4}{3}\) sq units

Question 9.
The area of the circle x² + y² = 25 in first quadrant is
(a) \(\frac{25 \pi}{3}\) sq units
(b) 5π sq units
(c) 5 sq units
(d) 3 sq units
Answer:
(a) \(\frac{25 \pi}{3}\) sq units

Question 10.
The area of the region bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is
(a) ab sq units
(b) πab sq units
(c) \(\frac{\pi}{a b}\) sq units ab
(d) πa² sq units
Answer:
(b) πab sq units

Question 11.
The area bounded by the parabola y² = x and the line 2y = x is
(a) \(\frac{4}{3}\) sq units
(b) 1 sq unit
(c) \(\frac{2}{3}\) sq unit
(d) \(\frac{1}{3}\) sq unit
Answer:
(a) \(\frac{4}{3}\) sq units

Question 12.
The area enclosed between the curve y = cos 3x, 0 ≤ x ≤ \(\frac{\pi}{6}\) and the X-axis is
(a) \(\frac{1}{2}\) sq unit
(b) 1 sq unit
(c) \(\frac{2}{3}\) sq unit
(d) \(\frac{1}{3}\) sq unit
Answer:
(d) \(\frac{1}{3}\) sq unit

Question 13.
The area bounded by y = √x and line x = 2y + 3, X-axis in first quadrant is
(a) 2√3 sq units
(b) 9 sq units
(c) \(\frac{34}{3}\) sq units
(d) 18 sq units
Answer:
(b) 9 sq units

Question 14.
The area bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) and the line \(\frac{x}{a}+\frac{y}{b}=1\) is
(a) (πab – 2ab) sq units
(b) \(\frac{\pi a b}{4}-\frac{a b}{2}\) sq units
(c) (πab – ab) sq units
(d) πab sq units
Answer:
(b) \(\frac{\pi a b}{4}-\frac{a b}{2}\) sq units

Question 15.
The area bounded by the parabola y = x² and the line y = x is
(a) \(\frac{1}{2}\) sq unit
(b) \(\frac{1}{3}\) sq unit
(c) \(\frac{1}{6}\) sq unit
(d) \(\frac{1}{12}\) sq unit
Answer:
(c) \(\frac{1}{6}\) sq unit

Question 16.
The area enclosed between the two parabolas y² = 4x and y = x is
(a) \(\frac{8}{3}\) sq units
(b) \(\frac{32}{3}\) sq units
(c) \(\frac{16}{3}\) sq units
(d) \(\frac{4}{3}\) sq units
Answer:
(c) \(\frac{16}{3}\) sq units

Question 17.
The area bounded by the curve y = tan x, X-axis and the line x = \(\frac{\pi}{4}\) is
(a) \(\frac{1}{3}\) log 2 sq units
(b) log 2 sq units
(c) 2 log 2 sq units
(d) 3 log 2 sq units
Answer:
(a) \(\frac{1}{3}\) log 2 sq units

Question 18.
The area of the region bounded by x² = 16y, y = 1, y = 4 and x = 0 in the first quadrant, is
(a) \(\frac{7}{3}\) sq units
(b) \(\frac{8}{3}\) sq units
(c) \(\frac{64}{3}\) sq units
(d) \(\frac{56}{3}\) sq units
Answer:
(d) \(\frac{56}{3}\) sq units

Question 19.
The area of the region included between the parabolas y² = 4ax and x² = 4ay, (a > 0) is given by
(a) \(\frac{16 a^{2}}{3}\) sq units
(b) \(\frac{8 a^{2}}{3}\) sq units
(c) \(\frac{4 a^{2}}{3}\) sq units
(d) \(\frac{32 a^{2}}{3}\) sq units
Answer:
(a) \(\frac{16 a^{2}}{3}\) sq units

Question 20.
The area of the region included between the line x + y = 1 and the circle x² + y² = 1 is
(a) \(\frac{\pi}{2}-1\) sq units
(b) π – 2 sq units
(c) \(\frac{\pi}{4}-\frac{1}{2}\) sq units
(d) π – \(\frac{1}{2}\) sq units
Answer:
(c) \(\frac{\pi}{4}-\frac{1}{2}\) sq units

(II) Solve the following:

Question 1.
Find the area of the region bounded by the following curve, the X-axis and the given lines:
(i) 0 ≤ x ≤ 5, 0 ≤ y ≤ 2
(ii) y = sin x, x = 0, x = π
(iii) y = sin x, x = 0, x = \(\frac{\pi}{3}\)
Solution:
(i) Required area = \(\int_{0}^{5} y d x\), where y = 2
= \(\int_{0}^{5} 2 d x\)
= \([2 x]_{0}^{5}\)
= 2 × 5 – 0
= 10 sq units.

(ii) The curve y = sin x intersects the X-axis at x = 0 and x = π between x = 0 and x = π.

Two bounded regions A₁ and A₂ are obtained. Both the regions have equal areas.
∴ required area = A₁ + A₂ = 2A₁

(iii) Required area = \(\int_{0}^{\pi / 3} y d x\), where y = sin x

Question 2.
Find the area of the circle x² + y² = 9, using integration.
Solution:
By the symmetry of the circle, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 3.

From the equation of the circle, y² = 9 – x².
In the first quadrant, y > 0
∴ y = \(\sqrt{9-x^{2}}\)
∴ area of the circle = 4 (area of the region OABO)

Question 3.
Find the area of the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) using integration.
Solution:

By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 5.
From the equation of the ellipse
\(\frac{y^{2}}{16}=1-\frac{x^{2}}{25}=\frac{25-x^{2}}{25}\)
∴ y² = \(\frac{16}{25}\) (25 – x²)
In the first quadrant y > 0
∴ y = \(\frac{4}{5} \sqrt{25-x^{2}}\)
∴ area of the ellipse = 4(area of the region OABO)

Question 4.
Find the area of the region lying between the parabolas:
(i) y² = 4x and x² = 4y
(ii) 4y² = 9x and 3x² = 16y
(iii) y² = x and x² = y.
Solution:
(i)

For finding the points of intersection of the two parabolas, we equate the values of y² from their equations.
From the equation x² = 4y, y = \(\frac{x^{2}}{4}\)
y = \(\frac{x^{4}}{16}\)
\(\frac{x^{4}}{16}\) = 4x
∴ x⁴ – 64x = 0
∴ x(x³ – 64) = 0
∴ x = 0 or x³ = 64 i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = \(\frac{4^{2}}{4}\) = 4
∴ the points of intersection are 0(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y² = 4x, i.e. y = 2√x between x = 0 and x = 4

(ii)

For finding the points of intersection of the two parabolas, we equate the values of 4y² from their equations.
From the equation 3x² = 16y, y = \(\frac{3 x^{2}}{16}\)
∴ y = \(\frac{3 x^{4}}{256}\)
∴ \(\frac{3 x^{4}}{256}\) = 9x
∴ 3x⁴ – 2304x = 0
∴ x(x³ – 2304) = 0
∴ x = 0 or x³ = 2304 i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = \(\frac{4^{2}}{4}\)
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y² = 4x,
i.e. y = 2√x between x = 0 and x = 4

Area of the region ODABO = area under the rabola x² = 4y,
i.e. y = \(\frac{x^{2}}{4}\) between x = 0 and x = 4

(iii)

For finding the points of intersection of the two parabolas, we equate the values of y² from their equations.
From the equation x² = y, y = \(\frac{x^{2}}{y}\)
∴ y = \(\frac{x^{2}}{y}\)
∴ \(\frac{x^{2}}{y}\) = x
∴ x² – y = 0
∴ x(x³ – y) = 0
∴ x = 0 or x³ = y
i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = \(\frac{4^{2}}{4}\) = 4
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y² = 4x,
i.e. y = 2√x between x = 0 and x = 4

Area ofthe region ODABO = area under the rabola x² = 4y,
i.e. y = \(\frac{x^{2}}{3}\) between x = 0 and x = 4

Question 5.
Find the area of the region in the first quadrant bounded by the circle x² + y² = 4 and the X-axis and the line x = y√3.
Solution:

For finding the points of intersection of the circle and the line, we solve
x² + y² = 4 ………(1)
and x = y√3 ……..(2)
From (2), x² = 3y²
From (1), x² = 4 – y²
3y² = 4 – y²
4y² = 4
y² = 1
y = 1 in the first quadrant.
When y = 1, r = 1 × √3 = √3
∴ the circle and the line intersect at A(√3, 1) in the first quadrant
Required area = area of the region OCAEDO = area of the region OCADO + area of the region DAED
Now, area of the region OCADO = area under the line x = y√3, i.e. y = \(\frac{x}{\sqrt{3}}\) between x = 0
and x = √3

Question 6.
Find the area of the region bounded by the parabola y² = x and the line y = x in the first quadrant.
Solution:
To obtain the points of intersection of the line and the parabola, we equate the values of x from both equations.

∴ y² = y
∴ y² – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = 0
When y = 1, x = 1
∴ the points of intersection are O(0, 0) and A(1, 1).
Required area = area of the region OCABO = area of the region OCADO – area of the region OBADO
Now, area of the region OCADO = area under the parabola y² = x i.e. y = +√x (in the first quadrant) between x = 0 and x = 1

Area of the region OBADO = area under the line y = x between x = 0 and x = 1

Question 7.
Find the area enclosed between the circle x² + y² = 1 and the line x + y = 1, lying in the first quadrant.
Solution:

Required area = area of the region ACBPA = (area of the region OACBO) – (area of the region OADBO)
Now, area of the region OACBO = area under the circle x² + y² = 1 between x = 0 and x = 1

Area of the region OADBO = area under the line x + y = 1 between x = 0 and x = 1

∴ required area = \(\left(\frac{\pi}{4}-\frac{1}{2}\right)\) sq units.

Question 8.
Find the area of the region bounded by the curve (y – 1)² = 4(x + 1) and the line y = (x – 1).
Solution:
The equation of the curve is (y – 1)² = 4(x + 1)
This is a parabola with vertex at A (-1, 1).
To find the points of intersection of the line y = x – 1 and the parabola.
Put y = x – 1 in the equation of the parabola, we get
(x – 1 – 1)² = 4(x + 1)
∴ x² – 4x + 4 = 4x + 4
∴ x² – 8x = 0
∴ x(x – 8) = 0
∴ x = 0, x = 8
When x = 0, y = 0 – 1 = -1
When x = 8, y = 8 – 1 = 7
∴ the points of intersection are B (0, -1) and C (8, 7).

To find the points where the parabola (y – 1)² = 4(x + 1) cuts the Y-axis.
Put x = 0 in the equation of the parabola, we get
(y – 1)² = 4(0 + 1) = 4
∴ y – 1 = ±2
∴ y – 1 = 2 or y – 1 = -2
∴ y = 3 or y = -1
∴ the parabola cuts the Y-axis at the points B(0, -1) and F(0, 3).
To find the point where the line y = x – 1 cuts the X-axis.
Put y = 0 in the equation of the line, we get
x – 1 = 0
∴ x = 1
∴ the line cuts the X-axis at the point G (1, 0).
Required area = area of the region BFAB + area of the region OGDCEFO + area of the region OBGO
Now, area of the region BFAB = area under the parabola (y – 1)² = 4(x + 1), Y-axis from y = -1 to y = 3

Since, the area cannot be negative,
Area of the region BFAB = \(\left|-\frac{8}{3}\right|=\frac{8}{3}\) sq units.
Area of the region OGDCEFO = area of the region OPCEFO – area of the region GPCDG

Since, area cannot be negative,
area of the region = \(\left|-\frac{1}{2}\right|=\frac{1}{2}\) sq units.
∴ required area = \(\frac{8}{3}+\frac{109}{6}+\frac{1}{2}\)
= \(\frac{16+109+3}{6}\)
= \(\frac{128}{6}\)
= \(\frac{64}{3}\) sq units.

Question 9.
Find the area of the region bounded by the straight line 2y = 5x + 7, X-axis and x = 2, x = 5.
Solution:
The equation of the line is
2y = 5x + 7, i.e., y = \(\frac{5}{2} x+\frac{7}{2}\)
Required area = area of the region ABCDA = area under the line y = \(\frac{5}{2} x+\frac{7}{2}\) between x = 2 and x = 5

Question 10.
Find the area of the region bounded by the curve y = 4x², Y-axis and the lines y = 1, y = 4.
Solution:

By symmetry of the parabola, the required area is 2 times the area of the region ABCD.
From the equation of the parabola, x² = \(\frac{y}{4}\)
In the first quadrant, x > 0
∴ x = \(\frac{1}{2} \sqrt{y}\)
∴ required area = \(\int_{1}^{4} x d y\)

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Application of Definite Integration Ex 5.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1

1. Find the area of the region bounded by the following curves, X-axis, and the given lines:

(i) y = 2x, x = 0, x = 5.
Solution:
Required area = \(\int_{0}^{5} y d x\), where y = 2x
= \(\int_{0}^{5} 2x d x\)
= \(\left[\frac{2 x^{2}}{2}\right]_{0}^{5}\)
= 25 – 0
= 25 sq units.

(ii) x = 2y, y = 0, y = 4.
Solution:
Required area = \(\int_{0}^{4} x d y\), where x = 2y
= \(\int_{0}^{4} 2 y d y\)
= \(\left[\frac{2 y^{2}}{2}\right]_{0}^{4}\)
= 16 – 0
= 16 sq units.

(iii) x = 0, x = 5, y = 0, y = 4.
Solution:
Required area = \(\int_{0}^{5} y d x\), where y = 4
= \(\int_{0}^{5} 4 d x\)
= \([4 x]_{0}^{5}\)
= 20 – 0
= 20 sq units.

(iv) y = sin x, x = 0, x = \(\frac{\pi}{2}\)
Solution:
Required area = \(\int_{0}^{\pi / 2} y d x\), where y = sin x
= \(\int_{0}^{\pi / 2} \sin x d x\)
= \([-\cos x]_{0}^{\pi / 2}\)
= -cos \(\frac{\pi}{2}\) + cos 0
= 0 + 1
= 1 sq unit.

(v) xy = 2, x = 1, x = 4.
Solution:
For xy = 2, y = \(\frac{2}{x}\)
Required area = \(\int_{1}^{4} y d x\), where y = \(\frac{2}{x}\)
= \(\int_{1}^{4} \frac{2}{x} d x\)
= \([2 \log |x|]_{1}^{4}\)
= 2 log 4 – 2 log 1
= 2 log 4 – 0
= 2 log 4 sq units.

(vi) y² = x, x = 0, x = 4.
Solution:

The required area consists of two bounded regions A1 and A2 which are equal in areas.
For y² = x, y = √x
Required area = A₁ + A₂ = 2A₁

(vii) y² = 16x, x = 0, x = 4.
Solution:

The required area consists of two bounded regions A1 and A2 which are equal in areas.
For y² = x, y = √x
Required area = A₁ + A₂ = 2A₁

2. Find the area of the region bounded by the parabola:

(i) y² = 16x and its latus rectum.
Solution:
Comparing y² = 16x with y² = 4ax, we get
4a = 16
∴ a = 4
∴ focus is S(a, 0) = (4, 0)

For y² = 16x, y = 4√x
Required area = area of the region OBSAO
= 2 [area of the region OSAO]

(ii) y = 4 – x² and the X-axis.
Solution:
The equation of the parabola is y = 4 – x²
∴ x² = 4 – y
i.e. (x – 0)² = -(y – 4)
It has vertex at P(0, 4)
For points of intersection of the parabola with X-axis,
we put y = 0 in its equation.
∴ 0 = 4 – x²
∴ x² = 4
∴ x = ± 2
∴ the parabola intersect the X-axis at A(-2, 0) and B(2, 0)

Required area = area of the region APBOA
= 2[area of the region OPBO]

3. Find the area of the region included between:

(i) y² = 2x and y = 2x.
Solution:
The vertex of the parabola y² = 2x is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 2x from both the equations we get,
y² = y
∴ y² – y = 0
∴ y = 0 or y = 1
When y = 0, x = \(\frac{0}{2}\) = 0
When y = 1, x = \(\frac{1}{2}\)
∴ the points of intersection are 0(0, 0) and B(\(\frac{1}{2}\), 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO = area under the parabola y² = 2x between x = 0 and x = \(\frac{1}{2}\)

Area of the region OCBDO = area under the line y = 2x between x = 0 and x = \(\frac{1}{2}\)

(ii) y² = 4x and y = x.
Solution:
The vertex of the parabola y² = 4x is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 4x from both the equations we get,
∴ y² = y
∴ y² – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = \(\frac{0}{2}\) = 0
When y = 1, x = \(\frac{1}{2}\)
∴ the points of intersection are O(0, 0) and B(\(\frac{1}{2}\), 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO = area under the parabola y² = 4x between x = 0 and x = \(\frac{1}{2}\)

Area of the region OCBDO = area under the line y = 2x between x = 0 and x = \(\frac{1}{2}\)

(iii) y = x² and the line y = 4x.
Solution:
The vertex of the parabola y = x² is at the origin 0(0, 0)
To find the points of the intersection of a line and the parabola.

Equating the values of y from the two equations, we get
x² = 4x
∴ x² – 4x = 0
∴ x(x – 4) = 0
∴ x = 0, x = 4
When x = 0, y = 4(0) = 0
When x = 4, y = 4(4) = 16
∴ the points of intersection are 0(0, 0) and B(4, 16)
Required area = area of the region OABCO = (area of the region ODBCO) – (area of the region ODBAO)
Now, area of the region ODBCO = area under the line y = 4x between x = 0 and x = 4
= \(\int_{0}^{4} y d x\), where y = 4x
= \(\int_{0}^{4} 4 x d x\)
= 4\(\int_{0}^{4} x d x\)
= 4\([latex]\int_{0}^{4} x d x\)[/latex]
= 2(16 – 0)
= 32
Area of the region ODBAO = area under the parabola y = x² between x = 0 and x = 4
= \(\int_{0}^{4} y d x\), where y = x2
= \(\int_{0}^{4} x^{2} d x\)
= \(\left[\frac{x^{3}}{3}\right]_{0}^{4}\)
= \(\frac{1}{3}\) (64 – 0)
= \(\frac{64}{3}\)
∴ required area = 32 – \(\frac{64}{3}\) = \(\frac{32}{3}\) sq units.

(iv) y² = 4ax and y = x.
Solution:
The vertex of the parabola y² = 4ax is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 4ax from both the equations we get,
∴ y² = y
∴ y² – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = \(\frac{0}{2}\) = 0
When y = 1, x = \(\frac{1}{2}\)
∴ the points of intersection are O(0, 0) and B(\(\frac{1}{2}\), 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO
= area under the parabola y² = 4ax between x = 0 and x = \(\frac{1}{2}\)

Area of the region OCBDO
= area under the line y
= 4ax between x = 0 and x = \(\frac{1}{4 a x}\)

(v) y = x² + 3 and y = x + 3.
Solution:
The given parabola is y = x² + 3, i.e. (x – 0)² = y – 3
∴ its vertex is P(0, 3).

To find the points of intersection of the line and the parabola.
Equating the values of y from both the equations, we get
x² + 3 = x + 3
∴ x² – x = 0
∴ x(x – 1) = 0
∴ x = 0 or x = 1
When x = 0, y = 0 + 3 = 3
When x = 1, y = 1 + 3 = 4
∴ the points of intersection are P(0, 3) and B(1, 4)
Required area = area of the region PABCP = area of the region OPABDO – area of the region OPCBDO
Now, area of the region OPABDO
= area under the line y = x + 3 between x = 0 and x = 1

Area of the region OPCBDO = area under the parabola y = x² + 3 between x = 0 and x = 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Miscellaneous Exercise 4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

I. Choose the correct option from the given alternatives:

Question 1.
\(\int_{2}^{3} \frac{d x}{x\left(x^{3}-1\right)}=\)
(a) \(\frac{1}{3} \log \left(\frac{208}{189}\right)\)
(b) \(\frac{1}{3} \log \left(\frac{189}{208}\right)\)
(c) \(\log \left(\frac{208}{189}\right)\)
(d) \(\log \left(\frac{189}{208}\right)\)
Answer:
(a) \(\frac{1}{3} \log \left(\frac{208}{189}\right)\)

Question 2.
\(\int_{0}^{\pi / 2} \frac{\sin ^{2} x \cdot d x}{(1+\cos x)^{2}}=\)
(a) \(\frac{4-\pi}{2}\)
(b) \(\frac{\pi-4}{2}\)
(c) 4 – \(\frac{\pi}{2}\)
(d) \(\frac{4+\pi}{2}\)
Answer:
(a) \(\frac{4-\pi}{2}\)

Question 3.
\(\int_{0}^{\log 5} \frac{e^{x} \sqrt{e^{x}-1}}{e^{x}+3} \cdot d x=\)
(a) 3 + 2π
(b) 4 – π
(c) 2 + π
(d) 4 + π
Answer:
(b) 4 – π

Question 4.
\(\int_{0}^{\pi / 2} \sin ^{6} x \cos ^{2} x \cdot d x=\)
(a) \(\frac{7 \pi}{256}\)
(b) \(\frac{3 \pi}{256}\)
(c) \(\frac{5 \pi}{256}\)
(d) \(\frac{-5 \pi}{256}\)
Answer:
(c) \(\frac{5 \pi}{256}\)

Question 5.
If \(\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{X}}=\frac{k}{3}\), then k is equal to
(a) √2(2√2 – 2)
(b) \(\frac{\sqrt{2}}{3}\)(2 – 2√2)
(c) \(\frac{2 \sqrt{2}-2}{3}\)
(d) 4√2
Answer:
(d) 4√2

Question 6.
\(\int_{1}^{2} \frac{1}{x^{2}} e^{\frac{1}{x}} \cdot d x=\)
(a) √e + 1
(b) √e − 1
(c) √e(√e − 1)
(d) \(\frac{\sqrt{e}-1}{e}\)
Answer:
(c) √e(√e − 1)

Question 7.
If \(\int_{2}^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right] \cdot d x=a+\frac{b}{\log 2}\), then
(a) a = e, b = -2
(b) a = e, b = 2
(c) a = -e, b = 2
(d) a = -e, b = -2
Answer:
(a) a = e, b = -2

Question 8.
Let \(\mathrm{I}_{1}=\int_{e}^{e^{2}} \frac{d x}{\log x}\) and \(\mathrm{I}_{2}=\int_{1}^{2} \frac{e^{x}}{\boldsymbol{X}} \cdot d x\), then
(a) I₁ = \(\frac{1}{3}\) I₂
(b) I₁ + I₂ = 0
(c) I₁ = 2I₂
(d) I₁ = I₂
Answer:
(d) I₁ = I₂

Question 9.
\(\int_{0}^{9} \frac{\sqrt{X}}{\sqrt{X}+\sqrt{9-X}} \cdot d x=\)
(a) 9
(b) \(\frac{9}{2}\)
(c) 0
(d) 1
Answer:
(b) \(\frac{9}{2}\)

Question 10.
The value of \(\int_{-\pi / 4}^{\pi / 4} \log \left(\frac{2+\sin \theta}{2-\sin \theta}\right) \cdot d \theta\) is
(a) 0
(b) 1
(c) 2
(d) π
Answer:
(a) 0

II. Evaluate the following:

Question 1.
\(\int_{0}^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x\)
Solution:
Let I = \(\int_{0}^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x\)
Put Numerator = A(Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ cos x = A(3 cos x + sin x) + B[\(\frac{d}{d x}\)(3 cos x + sin x)]
= A(3 cos x + sin x) + B(-3 sin x + cos x)
∴ cos x + 0 . sin x = (3A + B) cos x + (A – 3B) sin x
Comparing the coefficients of sinx and cos x on both the sides, we get
3A + B = 1 ………. (1)
A – 3B = 0 ………. (2)
Multiplying equation (1) by 3, we get
9A + 3B = 3 ………(3)
Adding (2) and (3), we get
10A = 3
∴ A = \(\frac{3}{10}\)

Question 2.
\(\int_{\pi / 4}^{\pi / 2} \frac{\cos \theta}{\left[\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right]^{3}} d \theta\)
Solution:

Question 3.
\(\int_{0}^{1} \frac{1}{1+\sqrt{x}} d x\)
Solution:
Let I = \(\int_{0}^{1} \frac{1}{1+\sqrt{x}} d x\)
Put √x = t
∴ x = t² and dx = 2t . dt
When x = 0, t = 0
When x = 1, t = 1

Question 4.
\(\int_{0}^{\pi / 4} \frac{\tan ^{3} x}{1+\cos 2 x} d x\)
Solution:

Question 5.
\(\int_{0}^{1} t^{5} \sqrt{1-t^{2}} d t\)
Solution:

Question 6.
\(\int_{0}^{1}\left(\cos ^{-1} x\right)^{2} d x\)
Solution:

Question 7.
\(\int_{-1}^{1} \frac{1+x^{3}}{9-x^{2}} d x\)
Solution:

Question 8.
\(\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{4} x d x\)
Solution:

Question 9.
\(\int_{0}^{\pi} \frac{x}{1+\sin ^{2} x} d x\)
Solution:

Question 10.
\(\int_{1}^{\infty} \frac{1}{\sqrt{x}(1+x)} d x\)
Solution:

III. Evaluate the following:

Question 1.
\(\int_{0}^{1}\left(\frac{1}{1+x^{2}}\right) \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)
Solution:

Question 2.
\(\int_{0}^{\pi / 2} \frac{1}{6-\cos x} d x\)
Solution:

Question 3.
\(\int_{0}^{a} \frac{1}{a^{2}+a x-x^{2}} d x\)
Solution:

Question 4.
\(\int_{\pi / 5}^{3 \pi / 10} \frac{\sin x}{\sin x+\cos x} d x\)
Solution:

Question 5.
\(\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)
Solution:
Let I = \(\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)

Question 6.
\(\int_{0}^{\pi / 4} \frac{\cos 2 x}{1+\cos 2 x+\sin 2 x} d x\)
Solution:

Question 7.
\(\int_{0}^{\pi / 2}[2 \log (\sin x)-\log (\sin 2 x)] d x\)
Solution:

Question 8.
\(\int_{0}^{\pi}\left(\sin ^{-1} x+\cos ^{-1} x\right)^{3} \sin ^{3} x d x\)
Solution:

Question 9.
\(\int_{0}^{4}\left[\sqrt{x^{2}+2 x+3}\right]^{-1} d x\)
Solution:

Question 10.
\(\int_{-2}^{3}|x-2| d x\)
Solution:
|x – 2|= 2 – x, if x < 2
= x – 2, if x ≥ 2

IV. Evaluate the following:

Question 1.
If \(\int_{a}^{a} \sqrt{x} d x=2 a \int_{0}^{\pi / 2} \sin ^{3} x d x\), find the value of \(\int_{a}^{a+1} x d x\).
Solution:

Question 2.
If \(\int_{0}^{k} \frac{1}{2+8 x^{2}} \cdot d x=\frac{\pi}{16}\), find k.
Solution:

Question 3.
If f(x) = a + bx + cx², show that \(\int_{0}^{1} f(x) d x=\frac{1}{6}\left[f(0)+4 f\left(\frac{1}{2}\right)+f(1)\right]\)
Solution:

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Miscellaneous Exercise 7

(I) Choose the correct alternatives:

Question 1.
Area of the region bounded by the curve x² = y, the X-axis and the lines x = 1 and x = 3 is
(a) \(\frac{26}{3}\) sq units
(b) \(\frac{3}{26}\) sq units
(c) 26 sq units
(d) 3 sq units
Answer:
(a) \(\frac{26}{3}\) sq units

Question 2.
The area of the region bounded by y² = 4x, the X-axis and the lines x = 1 and x = 4 is
(a) 28 sq units
(b) 3 sq unit
(c) \(\frac{28}{3}\) sq units
(d) \(\frac{3}{28}\) sq units
Answer:
(c) \(\frac{28}{3}\) sq units

Question 3.
Area of the region bounded by x² = 16y, y = 1 and y = 4 and the Y-axis, lying in the first quadrant is
(a) 63 sq units
(b) \(\frac{3}{56}\) sq units
(c) \(\frac{56}{3}\) sq units
(d) \(\frac{63}{7}\) sq units
Answer:
(c) \(\frac{56}{3}\) sq units

Question 4.
Area of the region bounded by y = x⁴, x = 1, x = 5 and the X-axis is
(a) \(\frac{3142}{5}\) sq units
(b) \(\frac{3124}{5}\) sq units
(c) \(\frac{3142}{3}\) sq units
(d) \(\frac{3124}{3}\) sq units
Answer:
(b) \(\frac{3124}{5}\) sq units

Question 5.
Using definite integration area of circle x² + y² = 25 is
(a) 5π sq units
(b) 4π sq units
(c) 25π sq units
(d) 25 sq units
Answer:
(c) 25π sq units

(II) Fill in the blanks:

Question 1.
Area of the region bounded by y = x⁴, x = 1, x = 5 and the X-axis is _________
Answer:
\(\frac{3124}{5}\) sq units

Question 2.
Using definite integration area of the circle x² + y² = 49 is ___________
Answer:
49π sq units

Question 3.
Area of the region bounded by x² = 16y, y = 1, y = 4 and the Y-axis lying in the first quadrant is _________
Answer:
\(\frac{56}{3}\) sq units

Question 4.
The area of the region bounded by the curve x² = y, the X-axis and the lines x = 3 and x = 9 is _________
Answer:
234 sq units

Question 5.
The area of the region bounded by y² = 4x, the X-axis and the lines x = 1 and x = 4 is __________
Answer:
\(\frac{28}{3}\) sq units

(III) State whether each of the following is True or False.

Question 1.
The area bounded by the curve x = g(y), Y-axis and bounded between the lines y = c and y = d is given by \(\int_{c}^{d} x d y=\int_{y=c}^{y=d} g(y) d y\)
Answer:
True

Question 2.
The area bounded by two curves y = f(x), y = g(x) and X-axis is \(\left|\int_{a}^{b} f(x) d x-\int_{b}^{a} g(x) d x\right|\)
Answer:
False

Question 3.
The area bounded by the curve y = f(x), X-axis and lines x = a and x = b is \(\left|\int_{a}^{b} f(x) d x\right|\)
Answer:
True

Question 4.
If the curve, under consideration, is below the X-axis, then the area bounded by curve, X-axis, and lines x = a, x = b is positive.
Answer:
False

Question 5.
The area of the portion lying above the X-axis is positive.
Answer:
True

(IV) Solve the following:

Question 1.
Find the area of the region bounded by the curve xy = c², the X-axis, and the lines x = c, x = 2c.
Solution:

= c² log(\(\frac{2 c}{c}\))
= c² . log 2 sq units.

Question 2.
Find the area between the parabolas y² = 7x and x² = 7y.
Solution:

For finding the points of intersection of the two parabolas,
we equate the values of y² from their equations.
From the equation x² = 7y, y² = \(\frac{x^{4}}{49}\)
∴ \(\frac{x^{4}}{49}\) = 7x
∴ x⁴ = 343x
∴ x⁴ – 343x = 0
∴ x(x³ – 343) = 0
∴ x = 0 or x³ = 343, i.e. x = 7
When x = 0, y = 0
When x = 7, 7y = 49
∴ y = 7
∴ the points of intersection are O(0, 0) and A(7, 7)
Required area = area of the region OBACO
= (area of the region ODACO) – (area of the region ODABO)
Now, area of the region ODACO = area under the parabola y² = 7x
i.e. y = √7 √x

Area of the region ODABO = Area under the parabola
x² = 7y
i.e. y = \(\frac{x^{2}}{7}\)

∴ required area = \(\frac{98}{3}-\frac{49}{3}=\frac{49}{3}\) sq units.

Question 3.
Find the area of the region bounded by the curve y = x² and the line y = 10.
Solution:

By the symmetry of the parabola,
the required area is twice the area of the region OABCO
Now, the area of the region OABCO

Question 4.
Find the area of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1.
Solution:
By the symmetry of the ellipse, the required area of the ellipse is 4 times the area of the region OPQO.
For the region OPQO, the limits of integration are x = 0 and x = 4.

Question 5.
Find the area of the region bounded by y = x², the X-axis and x = 1, x = 4.
Solution:
Required area = \(\int_{1}^{4} y d x\), where y = x²
= \(\int_{1}^{4} x^{2} d x\)
= \(\left[\frac{x^{3}}{3}\right]_{1}^{4}=\frac{4^{3}}{3}-\frac{1}{3}=\frac{64-1}{3}\)
= 21 sq units.

Question 6.
Find the area of the region bounded by the curve x² = 25y, y = 1, y = 4, and the Y-axis.
Solution:

Question 7.
Find the area of the region bounded by the parabola y² = 25x and the line x = 5.
Solution:

Given the equation of the parabola is y² = 25x
∴ y = 5√x …… [∵ IIn first quadrant, y > 0]
Required area = area of the region OQRPO
= 2(area of the region ORPO)

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 7 Application of Definite Integration Ex 7.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Application of Definite Integration Ex 7.1

Question 1.
Find the area of the region bounded by the following curves, the X-axis, and the given lines:
(i) y = x⁴, x = 1, x = 5
Solution:

(ii) y = \(\sqrt{6 x+4}\), x = 0, x = 2
Solution:

(iii) \(\sqrt{16-x^{2}}\), x = 0, x = 4
Solution:

(iv) 2y = 5x + 7, x = 2, x = 8
Solution:

(v) 2y + x = 8, x = 2, x = 4
Solution:

(vi) y = x² + 1, x = 0, x = 3
Solution:

(vii) y = 2 – x², x = -1, x = 1
Solution:

Question 2.
Find the area of the region bounded by the parabola y² = 4x and the line x = 3.
Solution:

Required area = area of the region OABO
= 2(area of the region OACO)

Question 3.
Find the area of the circle x² + y² = 25.
Solution:

By the symmetry of the circle, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 5.
From the equation of the circle, y² = 25 – x².
In the first quadrant y > 0
∴ y = \(\sqrt{25-x^{2}}\)
∴ area of the circle = 4(area of region OABO)

Question 4.
Find the area of the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{25}\) = 1
Solution:

By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 2.
From the equation of the ellipse,

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 6 Definite Integration Miscellaneous Exercise 6 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Definite Integration Miscellaneous Exercise 6

(I) Choose the correct alternative:

Question 1.
\(\int_{-9}^{9} \frac{x^{3}}{4-x^{2}} d x\) = ________
(a) 0
(b) 3
(c) 9
(d) -9
Answer:
(a) 0

Question 2.
\(\int_{-2}^{3} \frac{d x}{x+5}\) = _________
(a) -log(\(\frac{8}{3}\))
(b) log(\(\frac{8}{3}\))
(c) log(\(\frac{3}{8}\))
(d) -log(\(\frac{3}{8}\))
Answer:
(b) log(\(\frac{8}{3}\))

Question 3.
\(\int_{2}^{3} \frac{x}{x^{2}-1} d x\) = _________
(a) log(\(\frac{8}{3}\))
(b) -log(\(\frac{8}{3}\))
(c) \(\frac{1}{2}\) log(\(\frac{8}{3}\))
(d) \(\frac{-1}{2}\) log(\(\frac{8}{3}\))
Answer:
(c) \(\frac{1}{2}\) log(\(\frac{8}{3}\))

Question 4.
\(\int_{4}^{9} \frac{d x}{\sqrt{x}}\) = ___________
(a) 9
(b) 4
(c) 2
(d) 0
Answer:
(c) 2

Question 5.
If \(\int_{0}^{a} 3 x^{2} d x=8\), then a = __________
(a) 2
(b) 0
(c) \(\frac{8}{3}\)
(d) a
Answer:
(a) 2

Question 6.
\(\int_{2}^{3} x^{4}\) dx = ________
(a) \(\frac{1}{2}\)
(b) \(\frac{5}{2}\)
(c) \(\frac{5}{211}\)
(d) \(\frac{211}{5}\)
Answer:
(d) \(\frac{211}{5}\)

Question 7.
\(\int_{0}^{2} e^{x}\) dx = _______
(a) e – 1
(b) 1 – e
(c) 1 – e²
(d) e² – 1
Answer:
(d) e² – 1

Question 8.
\(\int_{a}^{b} f(x) d x\) = ________
(a) \(\int_{b}^{a} f(x) d x\)
(b) –\(\int_{a}^{b} f(x) d x\)
(c) –\(\int_{b}^{a} f(x) d x\)
(d) \(\int_{0}^{a} f(x) d x\)
Answer:
(c) –\(\int_{b}^{a} f(x) d x\)

Question 9.
\(\int_{-7}^{7} \frac{x^{3}}{x^{2}+7} d x\) = _________
(a) 7
(b) 49
(c) 0
(d) \(\frac{7}{2}\)
Answer:
(c) 0

Question 10.
\(\int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}} d x\) = _________
(a) \(\frac{7}{2}\)
(b) \(\frac{5}{2}\)
(c) 7
(d) 2
Answer:
(b) \(\frac{5}{2}\)

(II) Fill in the blanks:

Question 1.
\(\int_{0}^{2} e^{x} d x\) = ________
Answer:
e² – 1

Question 2.
\(\int_{2}^{3} x^{4} d x\) = __________
Answer:
\(\frac{211}{5}\)

Question 3.
\(\int_{0}^{1} \frac{d x}{2 x+5}\) = ____________
Answer:
\(\frac{1}{2} \log \left(\frac{7}{5}\right)\)

Question 4.
If \(\int_{0}^{a} 3 x^{2} d x\) = 8, then a = _________
Answer:
2

Question 5.
\(\int_{4}^{9} \frac{1}{\sqrt{x}} d x\) = _________
Answer:
2

Question 6.
\(\int_{2}^{3} \frac{x}{x^{2}-1} d x\) = _________
Answer:
\(\frac{1}{2} \log \left(\frac{8}{3}\right)\)

Question 7.
\(\int_{-2}^{3} \frac{d x}{x+5}\) = _________
Answer:
\(\log \left(\frac{8}{3}\right)\)

Question 8.
\(\int_{-9}^{9} \frac{x^{3}}{4-x^{2}} d x\) = _____________
Answer:
o

(III) State whether each of the following is True or False:

Question 1.
\(\int_{a}^{b} f(x) d x=\int_{-b}^{-a} f(x) d x\)
Answer:
True

Question 2.
\(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(t) d t\)
Answer:
True

Question 3.
\(\int_{0}^{a} f(x) d x=\int_{a}^{0} f(a-x) d x\)
Answer:
False

Question 4.
\(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(x-a-b) d x\)
Answer:
False

Question 5.
\(\int_{-5}^{5} \frac{x^{3}}{x^{2}+7} d x=0\)
Answer:
True

Question 6.
\(\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x=\frac{1}{2}\)
Answer:
True

Question 7.
\(\int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}} d x=\frac{9}{2}\)
Answer:
False

Question 8.
\(\int_{4}^{7} \frac{(11-x)^{2}}{(11-x)^{2}+x^{2}} d x=\frac{3}{2}\)
Answer:
True

(IV) Solve the following:

Question 1.
\(\int_{2}^{3} \frac{x}{(x+2)(x+3)} d x\)
Solution:

Question 2.
\(\int_{1}^{2} \frac{x+3}{x(x+2)} d x\)
Solution:
Let I = \(\int_{1}^{2} \frac{x+3}{x(x+2)} d x\)
Let \(\frac{x+3}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2}\)
∴ x + 3 = A(x + 2) + Bx
Put x = 0, we get
3 = A(2) + B(0)
∴ A = \(\frac{3}{2}\)
Put x + 2 = 0, i.e. x = -2, we get
-2 + 3 = A(0) + B(-2)
∴ 1 = -2B
∴ B = \(-\frac{1}{2}\)

Question 3.
\(\int_{1}^{3} x^{2} \log x d x\)
Solution:

Question 4.
\(\int_{0}^{1} e^{x^{2}} \cdot x^{3} d x\)
Solution:

Question 5.
\(\int_{1}^{2} e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) d x\)
Solution:

Question 6.
\(\int_{4}^{9} \frac{1}{\sqrt{x}} d x\)
Solution:

Question 7.
\(\int_{-2}^{3} \frac{1}{x+5} d x\)
Solution:

Question 8.
\(\int_{2}^{3} \frac{x}{x^{2}-1} d x\)
Solution:

Question 9.
\(\int_{0}^{1} \frac{x^{2}+3 x+2}{\sqrt{x}} d x\)
Solution:

Question 10.
\(\int_{3}^{5} \frac{d x}{\sqrt{x+4}+\sqrt{x-2}}\)
Solution:

Question 11.
\(\int_{2}^{3} \frac{x}{x^{2}+1} d x\)
Solution:

Question 12.
\(\int_{1}^{2} x^{2} d x\)
Solution:

Question 13.
\(\int_{-4}^{-1} \frac{1}{x} d x\)
Solution:

Question 14.
\(\int_{0}^{1} \frac{1}{\sqrt{1+x}+\sqrt{x}} d x\)
Solution:

Question 15.
\(\int_{0}^{4} \frac{1}{\sqrt{x^{2}+2 x+3}} d x\)
Solution:

Question 16.
\(\int_{2}^{4} \frac{x}{x^{2}+1} d x\)
Solution:

Question 17.
\(\int_{0}^{1} \frac{1}{2 x-3} d x\)
Solution:

Question 18.
\(\int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3} d x\)
Solution:

Question 19.
\(\int_{1}^{2} \frac{d x}{x(1+\log x)^{2}}\)
Solution:

Question 20.
\(\int_{0}^{9} \frac{1}{1+\sqrt{x}} d x\)
Solution: