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

jlmills5

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Oct 12, 2008
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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?
 
Re: Integrate for area

jlmills5 said:
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

42x=4    x=1\displaystyle \frac{4}{2-x} = 4 \implies x = 1

You have the integral right.

01442x  dx\displaystyle \int_{0}^{1}4-\frac{4}{2-x}\;dx

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

1x  dx=lnx+C\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.

42x  dx=4ln(2x)+C\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?

242(y2y)  dy\displaystyle \int_{2}^{4}2 \cdot \left(\frac{y-2}{y}\right)\;dy.

Hint: Ponder g1(x)\displaystyle g^{-1}(x).
 
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