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

Question

1634 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-21T17:55:26

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

#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.!