If f' is continuous on [a,b], show that:

Question

2∫_a ^b f(x) f'(x) dx = [f(b)]² - [f(a)]²

Answer 1

Let u = f(x).  Then du = f'(x)dx.  when x = a, u = f(a) and when x = b,   u = f(b).

 

So, 2∫_{(from a to b)}f(x)f'(x)dx = 2∫_{(from f(a) to f(b))} udu

 

                                              = 2(1/2)u²_{(from f(a) to f(b))}

 

                                      = [f(b)]² - [f(a)]²

 

 

 

   

Answer 2

Note that by the Chain Rule:

 

d/dx f²(x) = 2f(x)f'(x)

 

Thus the identity

 

2∫_a^b f(x) f'(x) dx =f²(b) - f²(a)

 

follows from the Fundamental Theorem of Calculus.