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Question
`int_(-2)^2 |x cos pix| "d"x` is equal to ______.
Options
`8/pi`
`4/pi`
`2/pi`
`1/pi`
Solution
`int_(-2)^2 |x cos pix| "d"x` is equal to `8/pi`.
Explanation:
Since I = `int_(-2)^2 |x cos pix| "d"x`
= `2 int_0^2 |x cos pix| "d"x`
= `2 {int_0^(1/2) |x cos pix|"d"x + int_(1/2)^(3/2) |x cos pix| "d"x + int_(3/2)^2 |x cos pix| "d"x}`
= `8/pi`
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