Find area of region bounded by g(x) = 4/(2 - x), y = 4, x =

[jlmills5]

Find area of bounded region:
g(x) = 4/(2-x)
y = 4
x = 0

Okay, I have drawn the graph and the region so I believe the integral should be calculated from x = 0 to x = 1.
I've set up the integral as int[4-(4/(2-x))]dx from 0 to 1. I tried substituting u for 2-x but that does not work for me....I'm stuck as to how to integrate. Or I've done something wrong up to this point. Any suggestions?

 

[tkhunny]

Re: Integrate for area

I tried substituting u for 2-x but that does not work for me....

What a strange conclusion.

You have the limits right.

g(0) = 2

\(\displaystyle \frac{4}{2-x} = 4 \implies x = 1\)

You have the integral right.

\(\displaystyle \int_{0}^{1}4-\frac{4}{2-x}\;dx\)

You should be able to do this one in your sleep.

\(\displaystyle \int \frac{1}{x}\;dx = ln|x| + C\)

If we note that the limits are positive and we have a sign problem, we get.

\(\displaystyle \int \frac{4}{2-x}\;dx = -4ln(2-x) + C\)

What does that give for your integral?

** Additional Exploration **

Where might I get this equivalent expression?

\(\displaystyle \int_{2}^{4}2 \cdot \left(\frac{y-2}{y}\right)\;dy\).

Hint: Ponder \(\displaystyle g^{-1}(x)\).