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January 22nd, 2014, 02:28 AM   #1
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Finding area by integration

How to find the area of the shaded region? I can't find the answer.
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 January 22nd, 2014, 04:19 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Finding area by integration What "region" do you mean? Your graph doesn't appear to make a lot of sense. What is that left boundary? You have the place where the line y= 8- 2x crosses it labeled "4". Is that the y value? 8- 2x= 4 when 2x= 4 or x= 2. Is the left boundary x= 2? And what is the bottom boundary? I see a "-2". Is y= -2 the lower boundary? If all of that is true, rotate it 90 degrees and integrate with respect to y rather than x. From y= -2 to y= 2, the left boundary is x= 2 and the right boundary is the parabola, $y^2= 2x- 2$ or $x= \frac{y^2}{2}+ 1$. From y= 2 to y= 4, the left boundary is x= 2 and the right boundary is y= 8x- 4 or $x= \frac{y}{8}- \frac{1}{2}$. So the area is given by $\int_{-2}^2 \left(\frac{y^2}{2}+ 1- 2\right) dy+ \int_2^4 \left(\frac{y}{8}- \frac{1}{2}- 2\right) dy$

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