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

Answer 1

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