How do you use part I of the Fundamental Theorem of Calculus to find the derivative of #h(x) = int (cos(t^4) + t) dt# from -4 to sinx? Can someone walk me through this? I'm having a lot of issues getting a grasp on how to do this.?

1 Answer
Jul 30, 2015

Answer:

The answer is #h'(x)=(cos(sin^{4}(x))+sin(x))*cos(x)#.

Explanation:

If you define a function #g# by the formula #g(x)=int_{-4}^{x} (cos(t^{4})+t)\ dt#, then the Fundamental Theorem of Calculus says that its derivative is #g'(x)=cos(x^{4})+x# (get rid of the integral sign and the #dt#, and replace the #t# in the integrand with #x#...the #-4# in the lower limit of the integral is irrelevant (it could be any number and the answer would be the same), but the #x# in the upper limit of the integral is essential)

Now notice that #h(x)=int_{-4}^{sin(x)}(cos(t^{4})+t)\ dt=g(sin(x))# (#h# is a composition of #g# with the sine function).

You can now apply the Chain Rule to say that

#h'(x)=g'(sin(x)) * d/dx(sin(x))#

#=(cos(sin^{4}(x))+sin(x))*cos(x)#

Is this helpful?

Perhaps there is still confusion about what #g# and #h# are. In other words, do they have "ordinary formulas" that don't involve integral signs. The answer, in this case, is "no". The integral #int cos(t^4)\ dt# cannot be evaluated in terms of "elementary functions" (functions that you are "used to").

The definite integral symbol #h(x)=int_{-4}^{sin(x)}(cos(t^{4})+t)\ dt# most certainly defies a function because the integrand is continuous. For any #x#, you can always approximate the value of #h(x)# by numerical integration (like Simpson's Rule).