Rd Sharma XII Vol 2 2021 for Class 12 Commerce Maths Chapter 2 - Areas Of Bounded Regions

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Page No 20.14:

Question 1:

Using integration, find the area of the region bounded between the line x = 2 and the parabola y² = 8x.

Answer:

$y^{2} = 8 x represents a parabola with vertex at origin and axis of symmetry along the + ve direction of x - axis x = 2 is line parallel to y - axis Let (x, y) be a given point on the parabola, y^{2} = 8 x Since parabola y^{2} = 8 x is symmetric about x - axis, ∴ Required area = 2 (area OCAO) On slicing the area above x - axis into vertical strips of length = |y| and width = d x \Rightarrow area of rectangular strip = |y| d x The approximating rectangle moves between x = 0 and x = 2 . So, area A = 2 (area OCAO) \Rightarrow A = 2 \int_{0}^{2} |y| d x = 2 \int_{0}^{2} y d x as y > 0 \Rightarrow A = 2 \int_{0}^{2} \sqrt{8 x} d x \Rightarrow A = 2 \times 2 \int_{0}^{2} \sqrt{2 x} d x = 4 \sqrt{2} \int_{0}^{2} \sqrt{x} d x \Rightarrow A = 4 \sqrt{2} {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{2} = \frac{8}{3} \sqrt{2} [2^{\frac{3}{2}} - 0] = \frac{8}{3} \times 2^{2} = \frac{32}{3} sq . units ∴ Area A = \frac{32}{3} sq . units$

x	0	$\frac{π}{6}$	$\frac{π}{2}$	$\frac{5 π}{6}$	$π$
sin x	0	$\frac{1}{2}$	1	$\frac{1}{2}$	0
$y = \frac{π}{2} + 2 \sin^{2} x$	1.57	2.07	3.57	2.07	1.57

y = \frac{π}{2} + 2 \sin^{2} x is an arc cutting y - axis at (1.57, 0) and x = π at (π, 1.57) x = π is a line parallel to y - axis Consider, a vertical strip of length = |y| and width = d x in the first quadrant ∴ Area of the approximating rectangle = |y| d x The approximating rectangle moves from x = 0 to x = π \Rightarrow Area of the shaded region = \int_{0}^{π} |y| d x \Rightarrow A = \int_{0}^{π} y d x \Rightarrow A = \int_{0}^{π} (\frac{π}{2} + 2 \sin^{2} x) d x \Rightarrow A = \int_{0}^{π} (\frac{π}{2} + 2 (\frac{1 - \cos 2 x}{2})) d x \Rightarrow A = \frac{π}{2} \int_{0}^{π} d x + \int_{0}^{π} (1 - \cos 2 x) d x \Rightarrow A = \frac{π}{2} {[x]}_{0}^{π} + {[x - \frac{\sin 2 x}{2}]}_{0}^{π} \Rightarrow A = \frac{π}{2} (π) + [π - \frac{\sin 2 π}{2} - 0] \Rightarrow A = π (\frac{π}{2} + 1) \Rightarrow A = π (\frac{π + 2}{2}) \Rightarrow A = \frac{π}{2} (π + 2) sq . units ∴ Area of curve bound by x = 0 and x = π is \frac{π}{2} (π + 2) sq . units

Page No 20.15:

Question 22:

Draw a rough sketch of the curve $y = \frac{x}{π} + 2 \sin^{2} x$ and find the area between the x-axis, the curve and the ordinates x = 0 and x = π.

Answer:

The table for different values of x and y is

x	$\frac{π}{6}$	$\frac{π}{2}$	$\frac{5 π}{6}$	$π$
sinx	$\frac{1}{2}$	1	$\frac{1}{2}$	0
$y = \frac{x}{π} + 2 \sin^{2} x$	$\frac{2}{3}$	$\frac{5}{2}$	$\frac{4}{3}$	1

y = \frac{x}{π} + 2 \sin^{2} x, is an arc cutting y - axis at O (0, 0) and cutting x = π at (π, 1) Consider a vertical strip of length = |y| and width = d x in the first quadrant ∴ Area of approximating rectangle = |y| d x The approximating rectangle moves from x = 0 to x = π \Rightarrow Area of the shaded area = \int_{0}^{π} |y| d x \Rightarrow A = \int_{0}^{π} y d x \{As, y > 0 \Rightarrow |y| = y\} \Rightarrow A = \int_{0}^{π} (\frac{x}{π} + 2 \sin^{2} x) d x \Rightarrow A = \frac{1}{π} \int_{0}^{π} x d x + 2 \int_{0}^{π} \sin^{2} x d x \Rightarrow A = \frac{1}{π} {[\frac{x^{2}}{2}]}_{0}^{π} + 2 {[\frac{x}{2} - \frac{1}{2} \sin x \cos x]}_{0}^{π} \Rightarrow A = \frac{π^{2}}{2 π} + \frac{2}{2} [π - \frac{1}{2} \sin π \cos π - 0] \Rightarrow A = \frac{π}{2} + π \Rightarrow A = \frac{3 π}{2} sq . units ∴ Area of the curve enclosed between x = 0 and x = π is \frac{3 π}{2} sq . units

x	0	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$\frac{5 π}{6}$	$π$
$y = \cos^{2} x$	1	0.5	0.25	0	0.25	0.75	1
$y = \sin^{2} x$	0	0.5	0.75	1	0.75	0.25	0

Let A₁ be the area of curve

y = \cos^{2} x between x = 0 and x = π

Let A₂ be the area of curve

y = \sin^{2} x between x = 0 and x = π

Consider, a vertical strip of length

= |y|

and width

= d x

in the shaded region of both the curves

The area of approximating rectangle

= |y| d x

The given region is intersection of
$y^{2} \leq 5 x and 5 x^{2} + 5 y^{2} \leq 36$
Clearly, $y^{2} \leq 5 x$ is a parabola with vertex at origin and the axis is along the x-axis opening in the positive direction.Also $5 x^{2} + 5 y^{2} \leq 36$ is a circle with centre at the origin and has a radius $\sqrt{\frac{36}{5}} or \frac{6}{\sqrt{5}}$ .
Corresponding equations of given inequations are
$y^{2} = 5 x . . . . . (1) 5 x^{2} + 5 y^{2} = 36 . . . . . (2)$
Substituting the value of y² from (1) into (2), we get

5 x^{2} + 25 x = 36

\Rightarrow 5 x^{2} + 25 x - 36 = 0

\Rightarrow x = \frac{- 25 \pm \sqrt{625 + 720}}{10}

\Rightarrow x = \frac{- 25 \pm \sqrt{1345}}{10}

From the figure we see that x-coordinate of intersecting point can not be negative.

∴ x = \frac{- 25 + \sqrt{1345}}{10}

Now assume that x-coordinate of intersecting point,

a = \frac{- 25 + \sqrt{1345}}{10}

The Required area,
A = 2(Area of OACO + Area of CABC)
Approximating the area of OACO the length =

|y_{1}|

and a width = dx=|y1=dx

Area of O A C O = \int_{0}^{a} |y_{1}| d x

= \int_{0}^{a} y_{1} d x

= \int_{0}^{a} \sqrt{5 x} d x (∵ {y_{1}}^{2} = 5 x \Rightarrow y_{1} = \sqrt{5 x})

\sqrt{5} {[\frac{2 x^{\frac{3}{2}}}{3}]}^{a}_{0}

Therefore, Area of OACO

= \frac{2 \sqrt{5} a^{\frac{3}{2}}}{3}

Similarly approximating the area of CABC the length

= |y_{2}|

and the width = dx

Area of C A B C = \int_{a}^{\frac{6}{\sqrt{5}}} |y_{2}| d x

= \int_{a}^{\frac{6}{\sqrt{5}}} y_{2} d x

= \int_{a}^{\frac{6}{\sqrt{5}}} \sqrt{\frac{36}{5} - x^{2}} d x

= {[\frac{x}{2} \sqrt{\frac{36}{5} - x^{2}} + \frac{18}{5} \sin^{- 1} (\frac{x \sqrt{5}}{6})]}_{a}^{\frac{6}{\sqrt{5}}}

= \frac{6}{2 \sqrt{5}} \sqrt{\frac{36}{5} - \frac{36}{5}} + \frac{18}{5} \sin^{- 1} (1) - \frac{a}{2} \sqrt{\frac{36}{5} - a^{2}} - \frac{18}{5} \sin^{- 1} (\frac{a \sqrt{5}}{6})

= 0 + \frac{18}{5} \sin^{- 1} (1) - \frac{a}{2} \sqrt{\frac{36}{5} - a^{2}} - \frac{18}{5} \sin^{- 1} (\frac{a \sqrt{5}}{6})

= \frac{18}{5} \times \frac{π}{2} - \frac{a}{2} \sqrt{\frac{36}{5} - a^{2}} - \frac{18}{5} \sin^{- 1} (\frac{a \sqrt{5}}{6})

Area of CABC

= \frac{9 π}{5} - \frac{a}{2} \sqrt{\frac{36}{5} - a^{2}} - \frac{18}{5} \sin^{- 1} (\frac{a \sqrt{5}}{6})

Thus the Required area, A = 2(Area of OACO + Area of CABC)

A = 2 [\frac{2 \sqrt{5} a^{\frac{3}{2}}}{3} + \frac{9 π}{5} - \frac{a}{2} \sqrt{\frac{36}{5} - a^{2}} - \frac{18}{5} \sin^{- 1} (\frac{a \sqrt{5}}{6})]

= \frac{4 \sqrt{5} a^{\frac{3}{2}}}{3} + \frac{18 π}{5} - a \sqrt{\frac{36}{5} - a^{2}} - \frac{36}{5} \sin^{- 1} (\frac{a \sqrt{5}}{6})

Where, a = \frac{- 25 + \sqrt{1345}}{10}

Page No 20.51:

Question 15:

Using integration, find the area of the region enclosed by the parabola $y = 3 x^{2}$ and the line $3 x - y + 6 = 0$ .

Answer:

Curves $y = 3 x^{2}$ and $y = 3 x^{2} + 6$ .