Cracking The Ap Calculus ab Exam 2018

6.

If we let u = x − 1, then du = dx. Next we can substitute into the integral: ∫ = ∫ u−2

du. Now we can integrate: ∫ u−2 du = + C = − + C. Last, we substitute back and get −

+ C.

7. − (1 − cos 2x)−2 + C

If  we  let u   =   1   −   cos 2x, then    du  =   2   sin 2x  dx. We  need    to  substitute  for sin 2x  dx, so  we  can

divide the du term by 2: = sin 2 x dx. Next we can substitute into the integral: ∫

dx = (^) ∫ u−3 du. Now we can integrate: (^) ∫ u−3 du = + C = −

• C. Last, we substitute back and get − (1 − cos 2 x)−2 + C.

8. 16

Recall   that    the     absolute    value   function    must    be  rewritten   as  a   piecewise   function:   |x|=   

. Thus, we need to split the integral into two separate integrals in order to evaluate

it:  |x|    dx  =       (−  x)  dx  +    x  dx. Now,    according   to  the Fundamental Theorem of  Calculus,

    (−x)    dx  =       =   (0) −       =   8   and     (x) dx  =       =       −   (0) =   8.

Therefore,  the answer  is  8   +   8   =   16.

9. First, let’s draw a picture.