Category: Class 12

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.1

Question 1.
Evaluate \(\int \frac{-2}{\sqrt{5 x-4}-\sqrt{5 x-2}} d x\)
Solution:

Question 2.
Evaluate \(\int\left(1+x+\frac{x^{2}}{2 !}\right) d x\)
Solution:

Question 3.
Evaluate \(\int \frac{3 x^{3}-2 \sqrt{x}}{x} d x\)
Solution:

Question 4.
Evaluate ∫(3x² – 5)² dx
Solution:
∫(3x² – 5)² dx
= ∫(9x⁴ – 30x² + 25) dx
= 9∫x⁴ dx – 30∫x² dx + 25∫1 dx
= 9(\(\frac{x^{5}}{5}\)) – 30(\(\frac{x^{3}}{3}\)) + 25x + c
= \(\frac{9x^{5}}{5}\) – 10x³ + 25x + c.

Question 5.
Evaluate \(\int \frac{1}{x(x-1)} d x\)
Solution:

Question 6.
If f'(x) = x² + 5 and f(0) = -1, then find the value of f(x).
Solution:
By the definition of integral
f(x) = ∫f'(x) dx
= ∫(x² + 5) dx
= ∫x² dx + 5∫1 dx
= \(\frac{x^{3}}{3}\) + 5x + c
Now, f(0) = -1 gives
f(0) = 0 + 0 + c = -1
∴ c = -1
∴ from (1), f(x) = \(\frac{x^{3}}{3}\) + 5x – 1.

Question 7.
If f(x) = 4x³ – 3x² + 2x + k, f(0) = -1 and f(1) = 4, find f(x).
Solution:
By the definition of integral
f(x) = ∫f'(x) dx
= ∫(4x³ – 3x² + 2x + k) dx
= 4∫x³ dx – 3∫x² dx + 2∫x dx + k∫1 dx
= 4(\(\frac{x^{4}}{4}\)) – 3(\(\frac{x^{3}}{3}\)) + 2(\(\frac{x^{2}}{2}\)) + kx + c
∴ f(x) = x⁴ – x³ + x² + kx + c
Now, f(0) = 1 gives
f(0) = 0 – 0 + 0 + 0 + c = 1
∴ c = 1
∴ from (1), f(x) = x⁴ – x³ + x² + kx + 1
Further f(1) = 4 gives
f(1) = 1 – 1 + 1 + k + 1 = 4
∴ k = 2
∴ from (2), f(x) = x⁴ – x³ + x² + 2x + 1.

Question 8.
If f(x) = \(\frac{x^{2}}{2}\) – kx + 1, f(0) = 2 and f(3) = 5, find f(x).
Solution:
By the definition of integral

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.10 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.10

Question 1.
Express the truth of each of the following statements by Venn diagrams:
(i) Some hardworking students are obedient.
Solution:
Let U : set of all students
S : set of all hardworking students
O : set of all obedient students.
Then the Venn diagram represents the truth of the given statement is as below:

S ∩ O ≠ φ

(ii) No circles are polygons.
Solution:
Let U : set of closed geometrical figures in the plane
P : set of all polygons
C : set of all circles.
Then the Venn diagram represents the truth of the given statement is as follows:

P ∩ C = φ

(iii) All teachers are scholars and scholars are teachers.
Solution:
Let U : set of all human beings
T : set of all teachers
S : set of all scholars.
Then the Venn diagram represents the truth of the given statement is as below:

(iv) If a quadrilateral is a rhombus, then it is a parallelogram.
Solution:
Let U : set of all quadrilaterals
R : set of all rhombus
P : set of all parallelograms.
Then the Venn diagram represents the truth of the given statement is as below:

R ⊂ P

Question 2.
Draw the Venn diagrams for the truth of the following statements:
(i) Some share brokers are chartered accountants.
Solution:
Let U : set of all human beings
S : set of all share brokers
C : set of all chartered accountants.
Then the Venn diagram represents the truth of the given statement is as below:

S ∩ C ≠ φ

(ii) No wicket-keeper is a bowler in a cricket team.
Solution:
Let U : set of all human beings
W : set of all wicket keepers
B : set of all bowlers.
Then the Venn diagram represents the truth of the given statement is as follows:

W ∩ B = φ

Question 3.
Represent the following statements by Venn diagrams:
(i) Some non-resident Indians are not rich.
Solution:
Let U : set of all human beings
N : set of all non-resident Indians
R : set of all rich people.
Then the Venn diagram represents the truth of the given statement is as below:

N – R ≠ φ

(ii) No circle is a rectangle.
Solution:
Let U : set of all geometrical figures
C : set of all circles
R : set of all rectangles
Then the Venn diagram represents the truth of the given statement is as below:

C ∩ R = φ

(iii) If n is a prime number and n ≠ 2, then it is odd.
Solution:
Let U : set of all real numbers
P : set of all prime numbers n, where n ≠ 2
O : set of all odd numbers.
Then the Venn diagram represents the truth of the given statement is as below:

P ⊂ O

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.5 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.5

Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}+y=e^{-x}\)
Solution:
\(\frac{d y}{d x}+y=e^{-x}\) …….(1)
This is the linear differential equation of the form

This is the general solution.

Question 2.
\(\frac{d y}{d x}\) + y = 3
Solution:
\(\frac{d y}{d x}\) + y = 3
This is the linear differential equation of the form

This is the general solution.

Question 3.
x\(\frac{d y}{d x}\) + 2y = x² . log x.
Solution:
x\(\frac{d y}{d x}\) + 2y = x² . log x
∴ \(\frac{d y}{d x}+\left(\frac{2}{x}\right) \cdot y=x \cdot \log x\) …….(1)
This is the linear differential equation of the form


This is the general solution.

Question 4.
(x + y)\(\frac{d y}{d x}\) = 1
Solution:
(x + y) \(\frac{d y}{d x}\) = 1
∴ \(\frac{d x}{d y}\) = x + y
∴ \(\frac{d x}{d y}\) – x = y
∴ \(\frac{d x}{d y}\) + (-1) x = y ……(1)
This is the linear differential equation of the form

This is the general solution.

Question 5.
y dx + (x – y²) dy = 0
Solution:
y dx + (x – y²) dy = 0
∴ y dx = -(x – y²) dy
∴ \(\frac{d x}{d y}=-\frac{\left(x-y^{2}\right)}{y}=-\frac{x}{y}+y\)
∴ \(\frac{d x}{d y}+\left(\frac{1}{y}\right) \cdot x=y\) ……(1)
This is the linear differential equation of the form

This is the general solution.

Question 6.
\(\frac{d y}{d x}\) + 2xy = x
Solution:
\(\frac{d y}{d x}\) + 2xy = x ………(1)
This is the linear differential equation of the form

This is the general solution.

Question 7.
(x + a) \(\frac{d y}{d x}\) = -y + a
Solution:
(x + a) \(\frac{d y}{d x}\) + y = a
∴ \(\frac{d y}{d x}+\left(\frac{1}{x+a}\right) y=\frac{a}{x+a}\) ……..(1)
This is the linear differential equation of the form

This is the general solution.

Question 8.
dy + (2y) dx = 8 dx
Solution:
dy + (2y) dx = 8 dx
∴ \(\frac{d y}{d x}\) + 2y = 8 …….(1)
This is the linear differential equation of the form

This is the general solution.

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.4

Solve the following differential equations:

Question 1.
x dx + 2y dy = 0
Solution:
x dx + 2y dy = 0
Integrating, we get
∫x dx + 2 ∫y dy = c₁
∴ \(\frac{x^{2}}{2}+2\left(\frac{y^{2}}{2}\right)=c_{1}\)
∴ x² + 2y² = c, where c = 2c₁
This is the general solution.

Question 2.
y² dx + (xy + x²) dy = 0
Solution:
y² dx + (xy + x²) dy = 0
∴ (xy + x²) dy = -y² dx
∴ \(\frac{d y}{d x}=\frac{-y^{2}}{x y+x^{2}}\) ………(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
Substituting these values in (1), we get



This is the general solution.

Question 3.
x²y dx – (x³ + y³) dy = 0
Solution:
x²y dx – (x³ + y³) dy = 0
∴ (x³ + y³) dy = x²y dx
∴ \(\frac{d y}{d x}=\frac{x^{2} y}{x^{3}+y^{3}}\) ……(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)


This is the general solution.

Question 4.
\(\frac{d y}{d x}+\frac{x-2 y}{2 x-y}=0\)
Solution:



This is the general solution.

Question 5.
(x² – y²) dx + 2xy dy = 0
Solution:
(x² – y²) dx + 2xy dy = 0
∴ 2xy dy = -(x² – y²) dx = (y² – x²) dx
∴ \(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\) ………(1)

Question 6.
xy\(\frac{d y}{d x}\) = x² + 2y²
Solution:

Question 7.
x²\(\frac{d y}{d x}\) = x² + xy – y²
Solution:

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.3

Question 1.
Solve the following differential equations:
(i) \(\frac{d y}{d x}\) = x²y + y
Solution:
\(\frac{d y}{d x}\) = x²y + y
∴ \(\frac{d y}{d x}\) = y(x² + 1)
∴ \(\frac{1}{y}\) dy = (x² + 1) dx
Integrating, we get
∫\(\frac{1}{y}\) dy = ∫(x² + 1) dx
∴ log |y|= \(\frac{x^{3}}{3}\) + x + c
This is the general solution.

(ii) \(\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right)\)
Solution:
\(\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right)\)

This is the general solution.

(iii) (x² – yx²) dy + (y² + xy²) dx = 0
Solution:
(x² – yx²) dy + (y² + xy²) dx = 0
∴ x²(1 – y) dy + y²(1 + x) dx = 0
∴ \(\frac{1-y}{y^{2}} d y+\frac{1+x}{x^{2}} d x=0\)
Integrating, we get


This is the general solution.

(iv) \(y^{3}-\frac{d y}{d x}=x \frac{d y}{d x}\)
Solution:
\(y^{3}-\frac{d y}{d x}=x \frac{d y}{d x}\)

∴ 2y² log |x + 1| = 2cy² – 1 is the required solution.

Question 2.
For each of the following differential equations find the particular solution:
(i) (x – y²x) dx – (y + x²y) dy = 0, when x = 2, y = 0.
Solution:
(x – y²x) dx – (y + x²y) dy = 0
∴ x(1 – y²) dx – y(1 + x²) dy = 0

∴ the general solution is
log |1 + x²| + log |1 – y²| = log c, where c₁ = log c
∴ log |(1 + x²)(1 – y²) | = log c
∴ (1 + x²)(1 – y²) = c
When x = 2, y = 0, we have
(1 + 4)(1 – 0) = c
∴ c = 5
∴ the particular solution is (1 + x²)(1 – y²) = 5.

(ii) (x + 1) \(\frac{d y}{d x}\) -1 = 2e^-y, when y = 0, x = 1.
Solution:

∴ log |2 + e^y| = log |c(x + 1)|
∴ 2 + e^y = c(x + 1)
This is the general solution.
Now, y = 0, when x = 1
∴ 2 + e⁰ = c(1 + 1)
∴ 3 = 2c
∴ c = \(\frac{3}{2}\)
∴ the particular solution is
2 + e^y = \(\frac{3}{2}\)(x + 1)
∴ 4 + 2e^y = 3x + 3
∴ 3x – 2e^y – 1 = 0

(iii) y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0, when x = e, y = e².
Solution:

∴ from (1), the general solution is
log |x log x| – log |y| = log c, where c₁ = log c
∴ log |\(\frac{x \log x}{y}\)| = log c
∴ \(\frac{x \log x}{y}\) = c
∴ x log x = cy
This is the general solution.
Now, y = e², when x = e
e log e = ce²
1 = ce ……[∵ log e = 1]
c = \(\frac{1}{e}\)
∴ the particular solution is x log x = (\(\frac{1}{e}\)) y
∴ y = e^x log x

(iv) \(\frac{d y}{d x}\) = 4x + y + 1, when y = 1, x = 0.
Solution:
\(\frac{d y}{d x}\) = 4x + y + 1
Put 4x + y + 1 = v

∴ log |v + 4| = x + c
∴ log |4x + y + 1 + 4| = x + c
i.e. log |4x + y + 5| = x + c
This is the general solution.
Now, y = 1 when x = 0
∴ log|0 + 1 + 5| = 0 + c,
i.e. c = log 6
∴ the particular solution is
log |4x + y + 5| = x + log 6
∴ \(\log \left|\frac{4 x+y+5}{6}\right|\) = x

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2

Question 1.
Obtain the differential equation by eliminating arbitrary constants from the following equations:
(i) y = Ae^3x + Be^-3x
Solution:
y = Ae^3x + Be^-3x ……(1)
Differentiating twice w.r.t. x, we get

This is the required D.E.

(ii) y = \(c_{2}+\frac{c_{1}}{x}\)
Solution:
y = \(c_{2}+\frac{c_{1}}{x}\)
∴ xy = c₂x + c₁
Differentiating w.r.t. x, we get

(iii) y = (c₁ + c₂x) e^x
Solution:
y = (c₁ + c₂x) e^x


This is the required D.E.

(iv) y = c₁ e^3x+ c₂ e^2x
Solution:



This is the required D.E.

(v) y² = (x + c)³
Solution:
y² = (x + c)³
Differentiating w.r.t. x, we get

This is the required D.E.

Question 2.
Find the differential equation by eliminating arbitrary constant from the relation x² + y² = 2ax.
Solution:
x² + y² = 2ax
Differentiating both sides w.r.t. x, we get
2x + 2y\(\frac{d y}{d x}\) = 2a
Substituting value of 2a in equation (1), we get
x² + y² = [2x + 2y \(\frac{d y}{d x}\)]x = 2x² + 2xy \(\frac{d y}{d x}\)
∴ 2xy \(\frac{d y}{d x}\) = y² – x² is the required D.E.

Question 3.
Form the differential equation by eliminating arbitrary constants from the relation bx + ay = ab.
Solution:
bx + ay = ab
∴ ay = -bx + ab
∴ y = \(-\frac{b}{a} x+b\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=-\frac{b}{a} \times 1+0=-\frac{b}{a}\)
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = 0 is the required D.E.

Question 4.
Find the differential equation whose general solution is x³ + y³ = 35ax.
Solution:

Question 5.
Form the differential equation from the relation x² + 4y² = 4b².
Sol ution:
x² + 4y² = 4b²
Differentiating w.r.t. x, we get
2x + 4(2y\(\frac{d y}{d x}\)) = 0
i.e. x + 4y\(\frac{d y}{d x}\) = 0 is the required D.E.

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1

Question 1.
Determine the order and degree of each of the following differential equations:
(i) \(\frac{d^{2} x}{d t^{2}}+\left(\frac{d x}{d t}\right)^{2}+8=0\)
Solution:
The given D.E. is \(\frac{d^{2} x}{d t^{2}}+\left(\frac{d x}{d t}\right)^{2}+8=0\)
This D.E. has highest order derivative \(\frac{d^{2} x}{d t^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

(ii) \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\left(\frac{d y}{d x}\right)^{2}=a^{x}\)
Solution:
The given D.E. is \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\left(\frac{d y}{d x}\right)^{2}=a^{x}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 2.
∴ the given D.E. is of order 2 and degree 2.

(iii) \(\frac{d^{4} y}{d x^{4}}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}\)
Solution:
The given D.E. is \(\frac{d^{4} y}{d x^{4}}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}\)
This D.E. has highest order derivative \(\frac{d^{4} y}{d x^{4}}\) with power 1.
∴ the given D.E. is of order 4 and degree 1.

(iv) (y'”)² + 2(y”)² + 6y’ + 7y = 0
Solution:
The given D.E. is (y”‘)² + 2(y”)² + 6y’ + 7y = 0
This can be written as \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}+2\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+6 \frac{d y}{d x}+7 y=0\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 2.
∴ the given D.E. is of order 3 and degree 2.

(v) \(\sqrt{1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}}=\left(\frac{d y}{d x}\right)^{3 / 2}\)
Solution:
The given D.E. is \(\sqrt{1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}}=\left(\frac{d y}{d x}\right)^{3 / 2}\)
On squaring both sides, we get
\(1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}=\left(\frac{d y}{d x}\right)^{3}\)
∴ \(\left(\frac{d y}{d x}\right)^{2}+1=\left(\frac{d y}{d x}\right)^{5}\)
This D.E. has highest order derivative \(\frac{d y}{d x}\) with power 5.
∴ the given D.E. is of order 1 and degree 5.

(vi) \(\frac{d y}{d x}=7 \frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\frac{d y}{d x}=7 \frac{d^{2} y}{d x^{2}}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

(vii) \(\left(\frac{d^{3} y}{d x^{3}}\right)^{1 / 6}=9\)
Solution:
The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{1 / 6}=9\)
i.e., \(\frac{d^{3} y}{d x^{3}}=9^{6}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 1.
∴ the given D.E. is of order 3 and degree 1.

Question 2.
In each of the following examples, verify that the given function is a solution of the corresponding differential equation:

Solution:
(i) xy = log y + k
Differentiating w.r.t. x, we get

Hence, xy = log y + k is a solution of the D.E. y'(1 – xy) = y².

(ii) y = xⁿ
Differentiating twice w.r.t. x, we get

This shows that y = xⁿ is a solution of the D.E.
\(x^{2} \frac{d^{2} y}{d x^{2}}-n x \frac{d y}{d x}+n y=0\)

(iii) y = e^x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = e^x = y
Hence, y = ex is a solution of the D.E. \(\frac{d y}{d x}\) = y.

(iv) y = 1 – log x
Differentiating w.r.t. x, we get

Hence, y = 1 – log x is a solution of the D.E.
\(x^{2} \frac{d^{2} y}{d x^{2}}=1\)

(v) y = ae^x + be^-x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = a(e^x) + b(-e^-x) = ae^x – be^-x
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = a(e^x) – b(-e^-x)
= ae^x + be^-x
= y
Hence, y = ae^x + be^-x is a solution of the D.E. \(\frac{d^{2} y}{d x^{2}}\) = y.

(vi) ax² + by² = 5
Differentiating w.r.t. x, we get

Hence, ax² + by² = 5 is a solution of the D.E.
\(x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}=y\left(\frac{d y}{d x}\right)\)

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: