If f(x)= 1/sqrt(1+t^5) dt, with bounds 0 to g(x). Where g(x) = [1+sin(t^3)]dt with bounds 0 to cosx, Find f'(pi/2)
Yeah where do I even begin with this... please help. And if someone could solve so I could see the steps involved more carefully I'd appreciate it!!
Hello,
Let P(t) be an antiderivative of 1/sqrt(1+t^5).
The definition of the integral is that :
(because g(x) and 0 are the boundaries of the integral :-))
P(0) is a number, independent of x.
Hence
Use the chain rule :
But we know that (because P is an antiderivative of this function).
--->
~~~~~~~~~~~~~~
Wait !
What is ?
We know that
-->
But .
Therefore, the boundaries of the integral are the same. So the integral is 0 !
--->
~~~~~~~~~~~~~~
Can you do it the same way I've done for calculating ? It may not look evident, but it has been awful for me to type all of this... I hope it will really help