# Problem 2-7: Evaluatin Definite Integrals

Evaluate

a) $\displaystyle\int_{-1}^1 \! x^2 \, dx$
b) $\displaystyle\int_{-2}^1 \! \sin(x) \, dx$
c) $\displaystyle\int_{-1}^2 \! 2x^3+x-1 \, dx$
d) $\displaystyle\int_{-2}^2 \! |x| \, dx$

Solution
a) $\displaystyle\int_{-1}^1 \! x^2 \, dx=\frac{1}{3}x^3\Bigr|^1_{-1}=\frac{1}{3}(1)^3-\frac{1}{3}(-1)^3=\frac{2}{3}$

b) $\displaystyle\int_{-2}^1 \! \sin(x) \, dx=-\cos(x)\Bigr|^1_{-2}=-\cos(1)-(-\cos(-2))=-0.9564$

c) $\displaystyle\int_{-1}^2 \! (2x^3+x-1) \, dx=\displaystyle\int_{-1}^2 \! 2x^3 \, dx +\displaystyle\int_{-1}^2 \! x \, dx -\displaystyle\int_{-1}^2 \! 1 \, dx$
$= \frac{2}{4}x^4\Bigr|^{-1}_{2}+\frac{1}{2}x^2\Bigr|^{-1}_{2}-x\Bigr|^{-1}_{2}= (\frac{1}{2}2^4- \frac{1}{2}(-1)^4)+(\frac{1}{2}2^2-\frac{1}{2}(-1)^2)-(2-(-1))$

$= 8-\frac{1}{2}+2-\frac{1}{2}-2-1=6$

d) $\displaystyle\int_{-2}^2 \! |x| \, dx=\displaystyle\int_{-2}^0 \! |x| \, dx+\displaystyle\int_{0}^2 \! |x| \, dx=\displaystyle\int_{-2}^0 \! (-x) \, dx+\displaystyle\int_{0}^2 \! x \, dx$
$= -\frac{1}{2}{x^2}\Bigr|_{-2}^0+\frac{1}{2}{x^2}\Bigr|_{0}^2=-\frac{1}{2}{(0)^2}-\frac{1}{2}(-{(-2)^2})+\frac{1}{2}{2^2}-\frac{1}{2}{0^2}=4$

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1 Comment

1. antony says:

wow,finally