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#?

Question

Luke Alvarez · Accepted Answer

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

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#.

Cameron Alexander · Answer

undefined To 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 $, we first need to express the function as a definite integral of a function of $ x $.

Let $ F(x) = \int_{e^x}^{5} 5(\sin(t))^5 dt $.

According to Part 1 of the Fundamental Theorem of Calculus, if $ F(x) $ is defined as above, then $ F'(x) = f(x) $, where $ f(x) $ is the integrand function with the upper limit replaced by $ x $.

In this case, we have:

$ f(x) = 5(\sin(x))^5 $.

Therefore, the derivative of the function $ y = \int_{e^x}^{5} 5(\sin(t))^5 dt $ with respect to $ x $ is:

$ \frac{dy}{dx} = f(x) = 5(\sin(x))^5 $.