How do you use the Fundamental Theorem of Calculus to find the derivative of #int xsin(πx) # from 1 to x^2?

1 Answer
Lovecraft
Dec 9, 2016

#d/dxI = 2x^3*sin(pi*x^2)#

Explanation:

Strictly speaking you don't, because you can't have the variable you're integrating show up in the limits of integration. Assuming you meant something like

#I = int_1^(x^2)tsin(pit)dt#

We see that the fundamental theorem of calculus states that

# int_a^bf(x)dx = F(b) - F(a)#

Where #f(x) = d/dxF(x)# so using the Fundamental Theorem of Calculus we have

#I = F(x^2) - F(1)#

Now, we just differentiate both sides

#d/dxI = d/dxF(x^2) - d/dxF(1)#

#F(1)# is a constant so its derivative is #0#

#d/dxI = d/dxF(x^2)#

Using the chain rule:

#u = x^2# and #(du)/dx = 2x#

#d/dxI = d/(du)F(u)*(du)/dx = 2x*d/(du)F(u)#

However we know that #F(u)# is a function such that its derivative is #u*sin(pi*u)#

#d/dxI = 2x*u*sin(pi*u)#

But since #u = x^2#

#d/dxI = 2x*x^2*sin(pi*x^2) = 2x^3*sin(pi*x^2)#

Related questions
See all questions in The Fundamental Theorem of Calculus
Impact of this question
1280 views around the world
You can reuse this answer
Creative Commons License