How do you use the second fundamental theorem of Calculus to find the derivative of given #int (cos(t^3))# from #[cosx, 7x]#?

Question

Charles Anctil · Accepted Answer

# 7 cos (7x)^3 + sin x cos (cos^3 x) #

Part 2 of the FTC states that

#color{blue}{d/(du) int_a^u \ f(t) \ dt = f(u)}# where #a = const#

if #u = u(x)#, we can add in the chain rule to conclude that

#d/dx int_a^u \ f(t) \ dt = f(u)* (du)/dx#

Additionally, we can infer from the FTC that

#d/(du) int_u^a \ f(t) \ dt = - d/(du) int_a^u \ f(t) \ dt = - f(u)#

Adding everything up, we arrive at the following from the FTC and the chain rule:

#d/(dx) int_u^v \ f(t) \ dt = f(v)* (dv)/dx - f(u) * (du)/dx#

thus, this is what we have

#f(t) = cos t^3#

# u = cos x, u' = -sin x# #v = 7x, v'= 7#

and so we compare patterns to obtain

# 7 cos (7x)^3 + sin x cos (cos^3 x) #

Luke Alvarez · Answer

undefined To find the derivative of the integral $\int_{\cos(x)}^{7x} \cos(t^3) dt$, using the Second Fundamental Theorem of Calculus, first, find an antiderivative of $\cos(t^3)$, let's call it $F(t)$. Then, evaluate $F(7x) - F(\cos(x))$, and differentiate this expression with respect to $x$. This will give you the derivative of the given integral with respect to $x$.