Indefinite Integration 1 Question 61
62. The value of $\int _{-2}^{2}\left|1-x^{2}\right| d x$ is … .
(1989, 2M)
Show Answer
Solution:
- $\int _{-2}^{2}\left|1-x^{2}\right| d x$
$=\int _{-2}^{-1}\left(x^{2}-1\right) d x+\int _{-1}^{1}\left(1-x^{2}\right) d x+\int _1^{2}\left(x^{2}-1\right) d x$
$=\frac{x^{3}}{3}-x _{-2}^{-1}+x-{\frac{x^{3}}{3}} _{-1}^{1}+\frac{x^{3}}{3}-x _1^{2}$
$=-\frac{1}{3}+1+\frac{8}{3}-2+1-\frac{1}{3}+1-\frac{1}{3}+\frac{8}{3}-2-\frac{1}{3}+1$
$=4$
