I'm taking the AP Calculus BC Exam next week and ran into this problem with no idea how to solve it. Unfortunately, the answer key didn't provide explanations, and I'd really, really appreciate it if someone could explain how to solve this problem - I'm having trouble with the antiderivative of the curve. It's a non-calculator question.
Which of the following is equal to the area of the region bounded by the lines $x = -3$, $x = 1$, $y = 0$, and the curve $y = \dfrac {x+22} {x^{2}+2x-8}$?
a) $\ln 5$
b) $6\ln 5$
c) $7\ln 5$
d) $8\ln 5$
e) $10\ln 5$
Thank you!
Answer
the answer is c. let me explain. the function is negative so we need to find :
$$ -\int f(x)dx $$
between $(-1 , 3)$.
Note that : $x^2 +2x -8 = (x+4)(x-2) $
So, $$\frac{x+22}{x^2 +2x -8} =\frac{x+22}{(x+4)(x-2)} $$
Now $$\frac{x+22}{(x+4)(x-2)} = \frac{A}{x+4} + \frac{B}{x-2}$$
after some calculation you will get that $A = -3$ and $B = 4$.
So we need to find
$$ -\int_{-1}^3 \frac{-3}{x+4} + \frac{4}{x-2}dx $$
and that equals $$ \Big[ -3\ln(|x+4|) + 4\ln(|x-2|)\Big]_{-1}^3 = 7\ln(5) $$
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