How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #y=int_("[e"^x",5]") 5(sin(t))^5 dt#?

1 Answer
Jul 30, 2015

Answer:

If #F(x)=int_(e^x)^5 5sin^5(t)dt#, then #F'(x)=-5sin^5(e^x)e^x#.

Explanation:

Let us first review the first part of the Fundamental Theorem, I will use the formulation used in Analysis I by Terence Tao:

Let #a< b# be real numbers, and let #f:[a,b]toRR# be a Riemann integrable function. Let #F:[a,b]toRR# be the function
#F(x)=int_a^xf(t)dt#.
Then #F# is continuous. Furthermore, if #x_0in[a,b]# and #f# continuous in #x_0#, then #F# is differentiable in #x_0#, and #F'(x_0)=f(x_0)#.

So now we want to know the derivative of
#F(x)=int_(e^x)^5 5sin^5(t)dt#.
We note that if we define #G(x)=int_(x)^5 5sin^5(t)dt#, then #F(x)=G(e^x)#. So in order to find the derivative of #F#, we use the chain rule, so:
#F'(x)=G'(e^x)d/dxe^x=G'(e^x)e^x#.

Now we need to know the derivative of #G#. We note that in the theorem, the variable #x# serves as an upper limit of integration, while in #G# it serves as a lower limit. This changes the way we can find the derivative slightly, as I will show next.

Suppose you have such a function #f# as given in the theorem, then let #H,G:[a,b]toRR# such that:
#H(x)=int_a^xf(t)dt# and #G(x)=int_x^bf(t)dt#.
Now note that #(H+G)(x)=H(x)+G(x)=int_a^bf(t)dt#.
Suppose #f# continuous in #x_0in[a,b]#, then #H'(x_0)=f(x_0)#, using the fundamental theorem. Furthermore #(H+G)(x)# is constant, so #(H+G)'(x)=0#. Since #G(x)=(H+G)(x)-H(x)#:
#G'(x_0)=(H+G)'(x_0)-H'(x_0)=0-f(x_0)=-f(x_0)#.
So #G'(x_0)=-f(x_0)#.

Applying this to #G(x)=int_(x)^5 5sin^5(t)dt#, we note that #5sin^5(t)# is continuous.
Therefore if #x<=5#, it serves as a lower limit, so #G'(x)=-5sin^5(x)#.
If #x>5#, we use the fact that #int_(x)^5 5sin^5(t)dt=-int_5^x 5sin^5(t)dt#. Since #x# now serves as an upper limit, we may use the original theorem and once again see #G'(x)=-5sin^5(x)#.

We already saw #F'(x)=G'(e^x)e^x#, so #F'(x)=-5sin^5(e^x)e^x#.