absolute integral – Solved Problems

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) $\displaystyle\int_{-1}^1 \! x^2 \, dx=\frac{1}{3}x^3\Bigr|^1_{-1}=\frac{1}{3}(1)^3-\frac{1}{3}(-1)^3=\frac{2}{3}$
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Tag: absolute integral

Problem 2-7: Evaluatin Definite Integrals