Indefinite Integration 1 Question 61

62. The value of $\int _{-2}^{2}\left|1-x^{2}\right| d x$ is … .

(1989, 2M)

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Solution:

  1. $\int _{-2}^{2}\left|1-x^{2}\right| d x$

$=\int _{-2}^{-1}\left(x^{2}-1\right) d x+\int _{-1}^{1}\left(1-x^{2}\right) d x+\int _1^{2}\left(x^{2}-1\right) d x$

$=\frac{x^{3}}{3}-x _{-2}^{-1}+x-{\frac{x^{3}}{3}} _{-1}^{1}+\frac{x^{3}}{3}-x _1^{2}$

$=-\frac{1}{3}+1+\frac{8}{3}-2+1-\frac{1}{3}+1-\frac{1}{3}+\frac{8}{3}-2-\frac{1}{3}+1$

$=4$



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