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Question
If `f(x) = int_0^pi t sin t dt`, then f' (x) is ______.
Options
cos x + x sin x
x sin x
x cos x
sin x + x cos x
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Solution
If `f(x) = int_0^pi t sin t dt`, then f' (x) is x sin x.
Explanation:
f(x) `= int_0^x t sin t dt`
`= [t * (- cos t)]_0^x - int_0^x 1 * (- cos t)` dt
= - x cos x - 0 cos 0 + `(sin t)_0^x"`
= -x cosx + sin x
Hence, f'(x) = -[cos x - x sin x] + cos x
= -cos x + x sin x + cos x
= x sin x
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