How do you use the second fundamental theorem of Calculus to find the derivative of given #int 1/(t^5) dt# from #[1,x]#?

Question

Scarlett Allison · Accepted Answer

You could either evaluate the integral and then differentiate or reason as follows:

The Second Fundamental Theorem of Calculus says that If #f# is continuous on #[a,b]#, then #int_a^b f(x) dx = F(b)-F(a)# where #F# is a function for which #F'(x) = f(x)# for all #x# in #[a,b]#. In this case we are using the variable #t# in the integrand and the variable #x# as the upper limit of integration. We want the derivative of #int_1^x 1/t^5 dt#. Note that since we are asked about the interval #[1,x]#, we must have #x > 1# (otherwise the interval is either empty or undefined). So, #1/t^5# is continuous on #[1,x]#, and #int_1^x 1/t^5 dt= F(x) - F(1)# where #F# is a function such that #F'(x) = 1/x^5#. And there is our answer. The derivative of #int_1^x 1/t^5 dt# is #1/x^5#.

Paisley Johnson · Answer

undefined To find the derivative of the integral $ \int_{1}^{x} \frac{1}{t^5} \, dt $, you use the second fundamental theorem of calculus, which states that if $ f(x) = \int_{a}^{x} g(t) \, dt $, then $ f'(x) = g(x) $. Applying this theorem, the derivative of the given integral is $ \frac{d}{dx} \left( \int_{1}^{x} \frac{1}{t^5} \, dt \right) = \frac{1}{x^5} $.