How do you use the second fundamental theorem of Calculus to find the derivative of given #int sect tant dt# from #[0, x^3]#?

Question

Isaiah Allen · Accepted Answer

You have two choices for how to do this.

Method 1 #int_0^(x^3) sect tant dt = F(x^3)-F(0)# where #F# is an antiderivative of #sect tant#. Now use the chain rule to differentiate #F(x^3)# with respect to #x# We get #secx^3 tanx^3 (3x^2)# Method 2 (Really assess the integral in definite terms.) Since #d/dt(sect) = sect tant#, we get #sect# is an antiderivative of #sect tant#. Thus, using the calculus second fundamental theorem, we discover #int_0^(x^3) sect tant dt = sect]_0^(x^3) = secx^3-sec0# We now distinguish in order to respond to the query: #d/dx(sec(x^3)) = secx^3 tanx^3 (3x^2)# (rearrange to taste).

Emily Ammerman · Answer

undefined To find the derivative of the integral $ \int_{0}^{x^3} \sec(t) \tan(t) \, dt $ with respect to $ x $, you can use the second fundamental theorem of calculus. According to this theorem, if $ f(x) $ is continuous on the interval $[a, x]$ where $ a $ is a constant, then the derivative of the integral $ \int_{a}^{x} f(t) \, dt $ with respect to $ x $ is $ f(x) $.

In this case, since the integral is from $ 0 $ to $ x^3 $, the lower limit is a constant, $ a = 0 $, and $ f(t) = \sec(t) \tan(t) $. Therefore, the derivative of the given integral with respect to $ x $ is $ \sec(x^3) \tan(x^3) \cdot (3x^2) $.