AP Calculus BC 8.4: Areas with Integrals

Area = \int_{0}^{3}(f(x)-g(x)) d x=\int_{0}^{3} f(x) d x-\int_{0}^{3} g(x) d x\)

Integrand

1 point

$=10-\int_{0}^{3} \frac{12}{3+x} d x=10-12[\ln |3+x|]_{0}^{3}$

Antiderivative of
g(x)

1 point

$=10-12(\ln 6-\ln 3)=10-12(\ln 2)$

Answer

1 point

Scoring notes:

- The first point is earned for any of the integrands f(x)-g(x), g(x)-f(x),|f(x)-g(x)|, or

|g(x)-f(x)| in any definite integral. If the limits are incorrect, the response does not earn the

third point.

- The first point is earned with an implied integrand for f and explicit integrand for g, such as

$10-\int_{0}^{3} g(x) d x$.

- The second point is earned for finding $a \int \frac{d x}{3+x}=a \cdot \ln |3+x|$ or $a \cdot \ln (3+x)$.

- A response is eligible for the third point only if it has earned the first 2 points. The third point is

earned only for the correct answer. The answer does not need to be simplified; however, if

simplification is attempted, it must be correct.

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

example, $\int_{0}^{3} \frac{12}{3+x} d x=\int_{0}^{3} \frac{12}{u} d u=12[\ln (x+3)]_{0}^{3}$.

- A response with incorrect communication, such as "Area $=\int_{0}^{3}(g(x)-f(x)) d x=10-12(\ln 2)$,"

does not earn the third point. However, a response of " $\int_{0}^{3}(g(x)-f(x)) d x=12(\ln 2)-10$, so the

area is $10-12(\ln 2)$ " earns all 3 points.

Total for part (a) 3 points

(b) Evaluate the improper integral $\int_{0}^{\infty}(g(x))^{2} d x$, or show that the integral diverges.