Question

The value of ∫|1- x^2| dx where, x→(- 2, 3) is

Answer 1

Here the interval of integration is $(-2,3)$. Divide it into 3 intervals,

$(-2,3)= (-2,-1)\cup (-1,1) \cup (1,3)$.

In the interval, $(-2,-1) \cup (1,3)$,

$|1-x^2|=x^2-1$

In the interval, $(-1,1)$,

$|1-x^2|=1-x^2$

Thus the integral is,

$\int\limits^3_{-2} {|1-x^2|} \, dx=\int\limits^{-1}_{-2} {(x^2-1)} \, dx+\int\limits^1_{-1} {(1-x^2)} \, dx+\int\limits^3_{1} {(x^2-1)} \, dx\\ \int\limits^3_{-2} {|1-x^2|} \, dx=[\frac{x^3}{3}-x]^ {-1}_{-2}+[x-\frac{x^3}{3}] ^1_{-1}+[\frac{x^3}{3}-x] ^3_{1}\\ \int\limits^3_{-2} {|1-x^2|} \, dx=\frac{4}{3} +\frac{4}{3}+\frac{20}{3}\\ \int\limits^3_{-2} {|1-x^2|} \, dx=\frac{28}{3}$