First slide

Evaluation of definite integrals

Question

Let $f$ be an odd function then $\int_{- 1}^{1} (| x | + f (x) \cos x) d x$ is equal to

Easy

A

0
B

1
C

2
D

none of these

Solution

The function $g (x) = | x |$ is an even function and $h (x) = f (x) \cos x$ is an odd function so
$\int_{- 1}^{1} (| x | + f (x) \cos x) d x = 2 \int_{0}^{1} | x | d x = 2 \int_{0}^{1} x d x = 1 .$

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