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) y^{2} = x, x = 0, x = 4.

Solution:

The required area consists of two bounded regions A1 and A2 which are equal in areas.

For y^{2} = x, y = √x

Required area = A_{1} + A_{2} = 2A_{1}

(vii) y^{2} = 16x, x = 0, x = 4.

Solution:

The required area consists of two bounded regions A1 and A2 which are equal in areas.

For y^{2} = x, y = √x

Required area = A_{1} + A_{2} = 2A_{1}

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

(i) y^{2} = 16x and its latus rectum.

Solution:

Comparing y^{2} = 16x with y^{2} = 4ax, we get

4a = 16

∴ a = 4

∴ focus is S(a, 0) = (4, 0)

For y^{2} = 16x, y = 4√x

Required area = area of the region OBSAO

= 2 [area of the region OSAO]

(ii) y = 4 – x^{2} and the X-axis.

Solution:

The equation of the parabola is y = 4 – x^{2}

∴ x^{2} = 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 – x^{2}

∴ x^{2} = 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) y^{2} = 2x and y = 2x.

Solution:

The vertex of the parabola y^{2} = 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,

y^{2} = y

∴ y^{2} – 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 y^{2} = 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) y^{2} = 4x and y = x.

Solution:

The vertex of the parabola y^{2} = 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,

∴ y^{2} = y

∴ y^{2} – 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 y^{2} = 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 = x^{2} and the line y = 4x.

Solution:

The vertex of the parabola y = x^{2} 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

x^{2} = 4x

∴ x^{2} – 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 = x^{2} 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) y^{2} = 4ax and y = x.

Solution:

The vertex of the parabola y^{2} = 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,

∴ y^{2} = y

∴ y^{2} – 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 y^{2} = 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 = x^{2} + 3 and y = x + 3.

Solution:

The given parabola is y = x^{2} + 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

x^{2} + 3 = x + 3

∴ x^{2} – 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 = x^{2} + 3 between x = 0 and x = 1