Find the area bounded by y = (x-16) / (x^2 - 3x - 28), x=2, and y=0.

Answer 1

(I'm assuming that the fraction was supposed to be (x²-16)/(x² - 3x - 28).)

The fraction (x²-16)/(x² - 3x - 28) factors into

((x+4)(x-4)) / ((x+4)(x-7))

so we can cancel out the two (x+4)'s and reduce the fraction to (x-4)/(x-7).

Now that it's reduced, we can easily see that function y = (x-4)/(x-7) has a zero at x=4. To the left of this point, the function is always positive. (For example, at x=0 the value of the function is y=(0-4)/(0-7) = 4/7, which is positive.)

So the (vaguely triangular) area is bounded by the function y=(x-4)/(x-7) on top, by the line y=0 below, by the line x=2 on the left, and tapers to a point on the right at the point (4, 0).

To do a definite integral, we can therefore integrate from x=2 on the left, to x=4 on the right:

∫ (from x=2 to x=4) (x-4)/(x-7) dx

Now how to integrate (x-4)/(x-7)?

It's hard because there's an x in the both the numerator and the denominator.

Let's break it into the sum of two fractions:

(x-4)/(x-7) = (x-7)/(x-7) + 3/(x-7)

The first fraction, (x-7)/(x-7), is just a big fat 1, so its antiderivative is plain old x.

The second fraction, 3/(x-7), no longer has an x in both the numerator and the denominator. It has the very simple antiderivative 3·ln |x-7|. (The vertical bars are absolute value signs.)

(Recall that the derivative of ln |x| is 1/x; therefore the derivative of ln |x-7| is 1/(x-7), and the derivative of 3·ln |x-7| is 3/(x-7).)

So the definite integral

∫ (from x=2 to x=4) (x-4)/(x-7) dx

= (x + 3·ln |x-7|) (from x=2 to x=4)

= (4 + 3·ln 3) - (2 + 3·ln 5)

= 2 + 3((ln 3) - (ln 5))

= 2 - 3((ln 5) - (ln 3))

= 2 - 3·ln(5/3)

which can also be written 2 - ln (125/27)