Question

Area bounded by curves y=x² -1 and y=1-x² is? A)1/3 B)8/3 C)4/3 D)2/3

Answer 1

Answer:

Answer:–

• Area bounded by the curves=8/3

Given:–

The two curves are,

$y = {x}^{2} - 1$ and

$y = 1 - {x}^{2}$

Equating both the equations,

${x}^{2} - 1 = 1 - {x}^{2}$

$2 {x}^{2} = 2$

${x}^{2} = 1$

$x = ± 1$

Now,

The area between the curves

=$2(^{1} ∫(1 - {x}^{2} - ( {x}^{2} - 1))dx$

$= 2( ^{1} ∫( - 2 {x}^{2} + 2)dx$

$= 2( ^{1} ( \frac{ { - 2x}^{3} }{3} + 2x))$

$= 2( \frac{ - 2}{3} + 2)$

$= 2( \frac{ - 2 + 6}{3} )$

$= 2( \frac{4}{3} )$

$= \frac{8}{3}$

So, The area bounded by the curves is 8/3.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

The area bounded by the curve y = f(x), the lines x = a, x= b and x axis is given by

$\displaystyle \int\limits_{a}^{b} f(x) \, dx = \displaystyle \int\limits_{a}^{b} y \, dx$

EVALUATION

The given curves are

$y = {x}^{2} - 1 \: \: \: \: ....(1)$

$y = 1 - {x}^{2} \: \: ....(2)$

The point of intersection of the above two curves are given by

${x}^{2} - 1 = 1 - {x}^{2}$

$\implies \: 2 {x}^{2} = 2$

$\implies \: {x}^{2} = 1$

$\implies \: {x} = \pm \: 1$

From Equation (1)

Putting x = 1 we get

$y = 0$

Putting x = - 1 we get

$y = 0$

Similarly others points are obtained

So all the point of intersections are

$(1,0), (-1,0),(0,1),(0,-1)$

The shaded region is bounded by given two curves

Now the region is symmetric about both axis

So the required area is

$= \displaystyle4 \times \int\limits_{0}^{1} ( 1- {x}^{2} ) \, dx \: \: \: \: sq \: unit \:$

$\displaystyle \: = 4 \times \bigg(x - \left \frac{x^3}{3} \bigg) \right|_0^1 \: \: \: sq \: unit \:$

$\displaystyle \: = 4 \times \bigg(1 - \frac{1}{3} \bigg) \: \: sq \: unit \:$

$\displaystyle \: = 4 \times \frac{2}{3} \: \: sq \: unit \:$

$\displaystyle \: = \frac{8}{ 3} \: \: \: sq \: unit \:$