Find f(1) if g(x) = the definite integral from 0 to x^2 of f(t) dt = x ln(x)

Question

Here is a link to the picture of the problem. http://imgur.com/0ZiyAzP

Answer 1

Funny - until I saw the picture you posted, I couldn't remember this.  But now that I've seen the picture...

 

 

 

think of it this way - we have a g(x) that is equal to two different things: the integral and x ln x

 

g'(x) = d/dx (x ln x) = ln x + x (1/x) = ln x + 1  (this is the product rule)

 

g'(x) is ALSO = f(x^x) d/dx (x²) = 2x f(x²)   (This is The Fundamental Theorem of Calculus and the chain rule).

 

g'(x) = ln(x) + 1 = 2x f(x²)

 

solve for f(x²)

 

f(x²) = {ln(x) + 1}/2x

f(1) = {ln(1) + 1}/2 = (0+1)/2 = 1/2