[tex]\int _ { - 2 } ^ { 2 } | 1 - x ^ { 2 } | d x[/tex]Please Solve It.... Created by abhinavkr01. Math

A fraction (from Latin fractus, "broken") represents a part of a whole or, more generally, any number of equal parts.

tex]Please Solve It......

mathdude500

$\large\underline{\sf{Solution-}}$

Given integral is

$\red{\rm :\longmapsto\:\displaystyle\int_{-2}^{2} |1 - {x}^{2} | \: dx}$

Let we first define the function.

We know,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| = \begin{cases} &\sf{ - x \: \: when \:x \: < \: 0 } \\ &\sf{ \: \: x \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So, using this definition, Let define the given function.

$\red{\rm :\longmapsto\: |1 - {x}^{2} | = |(1 - x)(1 + x)|}$

So, critical points are - 1 and 1.

Thus,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |(1 - x)(1 + x)| = \begin{cases} &\sf{ - (1 - x)(1 + x)\: when \: - 2 \leqslant x \leqslant - 1 } \\ \\ &\sf{ (1 - x)(1 + x) \: when \: - 1 \leqslant x \leqslant 1 } \\ \\ &\sf{ - (1 - x)(1 + x) \: when \: 1 \leqslant x \leqslant 2 } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |1 - {x}^{2} | = \begin{cases} &\sf{ - (1 - {x}^{2}) \: when \: - 2 \leqslant x \leqslant - 1 } \\ \\ &\sf{ 1 - {x}^{2} \: when \: - 1 \leqslant x \leqslant 1 } \\ \\ &\sf{ - (1 - {x}^{2}) \: when \: 1 \leqslant x \leqslant 2 } \end{cases}\end{gathered}\end{gathered}$

Hence,

$\red{\rm :\longmapsto\:\displaystyle\int_{-2}^{2} |1 - {x}^{2} | \: dx}$

can be rewritten as

$\rm \: =\displaystyle\int_{-2}^{ - 1} |1 - {x}^{2} | dx + \displaystyle\int_{-1}^{1} |1 - {x}^{2} | dx + \displaystyle\int_{1}^{2} |1 - {x}^{2} | dx$

$\rm \: = - \displaystyle\int_{-2}^{ - 1} (1 - {x}^{2}) dx + \displaystyle\int_{-1}^{1} (1 - {x}^{2})dx - \displaystyle\int_{1}^{2}(1 - {x}^{2})dx$

We know,

$\boxed{ \bf{ \: \displaystyle\int \: kdx \: = \: kx \: + \: c}}$

and

$\boxed{ \bf{ \: \displaystyle\int \: {x}^{n} dx \: = \: \frac{ {x}^{n + 1} }{n + 1} \: + \: c}}$

So, using this, we get

$\rm \: = - \:\bigg[x - \dfrac{ {x}^{3} }{3} \bigg]^{ - 1}_{ - 2} + \:\bigg[x - \dfrac{ {x}^{3} }{3} \bigg]^{1}_{ - 1} - \:\bigg[x - \dfrac{ {x}^{3} }{3} \bigg]^{2}_{1}$

$\rm \: = - \:\bigg[( - 1 + 2) - \dfrac{ - 1 + 8 }{3} \bigg] + \:\bigg[(1 + 1) - \dfrac{ 1 + 1 }{3} \bigg] - \:\bigg[(2 - 1) - \dfrac{8 - 1 }{3} \bigg]$