Question

What is the value of `F'(x)` if `F(x) = int_0^sinxsqrt(t)dt` ?

Answer 1

`:. F'(x)=(sqrtsinx)(cosx).`

Explanation:

`F(x)=int_0^sinx sqrttdt`

`because, intsqrttdt=intt^(1/2)dt=t^(1/2+1)/(1/2+1)=2/3t^(3/2)+c,`

`:. F(x)=[2/3t^(3/2)]_0^sinx`

`:. F(x)=2/3sin^(3/2)x`

`:. F'(x)=2/3[{(sinx)}^(3/2)]'`

Using the Chain Rule, `F'(x)=2/3[3/2(sinx)^(3/2-1)]d/dx(sinx)`

`=(sinx)^(1/2)(cosx)`

`:. F'(x)=(sqrtsinx)(cosx).`

Enjoy Maths.!