# Help me! test tomorrow!

#### ssolaming

##### New member

I tried using integration by parts, but i can't get the answer. The answer is A.

#### HallsofIvy

##### Elite Member
Yes, integration by parts should do it. Let u= f(x) and dv= g'' dx so that du= f' dx and v= g'. $$\displaystyle \int_2^4 fg''dx= \left[f(x)g'(x)\right]_2^4- \int_2^4 f'g' dx$$.

The first part is $$\displaystyle \left[f(x)g'(x)\right]_2^4= f(4)g'(4)- f(2)g'(2)= (13)(7)- (7)(1)= 84$$. Now we need to calculate $$\displaystyle \int_2^4 f'g'dx$$. But we are told that "f'(x)= 3 for all x". Since f' is a constant, we can take it outside the integral: $$\displaystyle 3\int_2^4 g'(x)dx$$.