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.

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

(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:
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q1 (vi)
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
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q1 (vi).1

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

(vii) y2 = 16x, x = 0, x = 4.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q1 (vii)
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
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q1 (vii).1

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)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q2 (i)
For y2 = 16x, y = 4√x
Required area = area of the region OBSAO
= 2 [area of the region OSAO]
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q2 (i).1

(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)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q2 (ii)
Required area = area of the region APBOA
= 2[area of the region OPBO]
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q2 (ii).1

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

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).
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (i)
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 = \(\frac{1}{2}\)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (i).1
Area of the region OCBDO = area under the line y = 2x between x = 0 and x = \(\frac{1}{2}\)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (i).2

(ii) y2 = 4x and y = x.
Solution:
The vertex of the parabola y2 = 4x is at the origin O = (0, 0).
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (ii).jpg
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 = \(\frac{1}{2}\)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (ii).1
Area of the region OCBDO = area under the line y = 2x between x = 0 and x = \(\frac{1}{2}\)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (ii).2

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

(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.
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (iii)
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).
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (iv).jpg
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 = \(\frac{1}{2}\)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (iv).1
Area of the region OCBDO
= area under the line y
= 4ax between x = 0 and x = \(\frac{1}{4 a x}\)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (iv).2

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

(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).
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (v)
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
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (v).1
Area of the region OPCBDO = area under the parabola y = x2 + 3 between x = 0 and x = 1
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (v).2