AP Calculus BC 6.13: Improper Integrals

$\int_{0}^{\infty}(g(x))^{2} d x=\lim _{b \rightarrow \infty} \int_{0}^{b} \frac{144}{(3+x)^{2}} d x$

Limit notation

1 point

$=\lim _{b \rightarrow \infty}\left(-\left.\frac{144}{(3+x)}\right|_{0} ^{b}\right)$

Antiderivative

1 point

$=\lim _{b \rightarrow \infty}\left(-\frac{144}{3+b}+\frac{144}{3}\right)=48$

Answer

1 point

Scoring notes:

- To earn the first point a response must correctly use limit notation throughout the problem and not

include arithmetic with infinity, for example, $\left[-\frac{144}{3+x}\right]_{0}^{\infty}$ or $-\frac{144}{3+\infty}+48$.

- The second point can be earned by finding an antiderivative of the form $-\frac{a}{(3+x)}$ for a>0, from

an indefinite or improper integral, with or without correct limit notation. If $a \neq 144$, the response

does not earn the third point.

- The third point is earned only for an answer of 48 (or equivalent).

- A response is not eligible for the third point with incorrect limits of integration for u-substitution, for

example, $\lim _{b \rightarrow \infty} \int_{0}^{b} \frac{144}{u^{2}} d u=\lim _{b \rightarrow \infty}\left[-\frac{144}{3+x}\right]_{0}^{b}$.

Total for part (b) 3 points

(c) Let h be the function defined by $h(x)=x \cdot f^{\prime}(x)$. Find the value of $\int_{0}^{3} h(x) d x$.