AP Calculus BC 6.11: Integration by Parts

$$\begin{aligned} u =t+5 d v=\cos \left(\frac{t}{4}\right) d t d u =d t v=4 \sin \left(\frac{t}{4}\right) \end{aligned}$$

u and d v

1 point

$\int(t+5) \cos \left(\frac{t}{4}\right) d t=4(t+5) \sin \left(\frac{t}{4}\right)-\int 4 \sin \left(\frac{t}{4}\right) d t$$u v-\int v d u$

1 point

$=4(t+5) \sin \left(\frac{t}{4}\right)+16 \cos \left(\frac{t}{4}\right)+C$

Answer

1 point

Scoring notes:

- The first and second points are earned with an implied u and d v in the presence of

$4(t+5) \sin \left(\frac{t}{4}\right)-\int 4 \sin \left(\frac{t}{4}\right) d t$ or a mathematically equivalent expression.

- The tabular method may be used to show integration by parts. In this case, the first point is earned

by columns (labeled or unlabeled) that begin with t+5 and $\cos \left(\frac{t}{4}\right)$. The second point is earned

for $4(t+5) \sin \left(\frac{t}{4}\right)-\int 4 \sin \left(\frac{t}{4}\right) d t$ or a mathematically equivalent expression.

- The third point is earned only for an expression mathematically equivalent to

$4(t+5) \sin \left(\frac{t}{4}\right)+16 \cos \left(\frac{t}{4}\right)+C$ (such as $4 t \sin \left(\frac{t}{4}\right)+20 \sin \left(\frac{t}{4}\right)+16 \cos \left(\frac{t}{4}\right)+C$ ) in the presence

of correct supporting work.

- To earn the third point a response must have a final answer that includes a constant of integration.

- Alternate solution:

○ A response can earn the first and second points for correctly applying integration by parts to

$\int t \cos \left(\frac{t}{4}\right) d t$. The tabular method may be used to show integration by parts. The third point is

earned for the correct answer.

Total for part (d)

for question 5

9 points

Part B (BC): Graphing calculator not allowed