Find the derivative of #int_0^(x^3e^x) \(t^3+3)^17 \ dt#?

2 Answers
Jul 28, 2017

Answer:

Because we need the chain rule.

Explanation:

#y = g(u) = int_0^u (t^3+3)^17 dt# has

#dy/(du) = (u^3+3)^17#

and

#dy/dx = dy/(du) * (du)/dx#

so

#g'(x) = dy/dx = (u^3+3)^17 * d/dx(x^3e^3)#

Jul 28, 2017

Answer:

# d/dx int_0^(x^3e^x) \ (t^3+3)^17 \ dt = (x^3e^x + 3x^2e^x)((x^3e^x)^3+3)^17#

Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

# d/dx \ int_a^x \ f(t) \ dt = f(x) # for any constant #a#

(ie the derivative of an integral gives us the original function back).

We are asked to find:

# d/dx int_0^(x^3e^x) \(t^3+3)^17 \ dt# ..... [A}

(notice the upper bounds of the first integral are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral using a substitution the product rule, and the chain rule. Let:

# u = x^3e^x => (du)/dx = (x^3)(d/de^x) + (d/dxx^3)(e^x) #
# :. (du)/dx = x^3e^x + 3x^2e^x #

The substituting into the integral [A], and applying the chain rule, we get:

# d/dx int_0^(x^3e^x) \ (t^3+3)^17 \ dt = (du)/dx*d/(du) \ int_0^u \ (t^3+3)^17 \ dt #

And now the derivative of the integral is in the correct form for the FTOC to be applied, giving:

# d/dx int_0^(x^3e^x) \ (t^3+3)^17 \ dt = (du)/(dx)(u^3+3)^17#

And restoring the initial substitution we get:

# d/dx int_0^(x^3e^x) \ (t^3+3)^17 \ dt = (x^3e^x + 3x^2e^x)((x^3e^x)^3+3)^17#

Which is option (2) in the initial question.