Tag Archives: definite integral

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}
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