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I have one problem which confuses me, namely I have solve before this problem similar one. Previous problem:

Find the area of the figure bounded by $y=6x-x^2$ and $y=3x$

For solving this problem, I have set to equal these two graph to each other (find intersection points), so $6x-x^2=3x$, from this I have got $3x-x^2=0 \longrightarrow x(x-3)=0$ or $x=0$ and $x=3$, I have used Wolfram|$\alpha$ and then calculate area by this way $\int_0^3(6x-x^2-3x)dx$

When I calculated this one, I have got $4.5$, which is correct answer, because the book has this answer. The next question is similar but I could not solve it:

Calculate the area of the figure bounded by $y=-x^2+6x,y=0,y=3x$

I don't understand. Are they same? What is trick of this problem? The answer is $31.5$, but I could not solve it myself. Please help me.

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    You may compute $\displaystyle \int_0^3 3x\,dx+\int_3^6 -x^2+6x\,dx\ $ or, as proposed by Eugene, $\int_0^6 -x^2+6x\,dx\ $ minus the first area.2012-07-04

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The plot which shows what's going on is :

Rendered With Mathematica

The problem as phrased the first way asks you to calculate the area above the line, below the parabola on [0,3]. The way the question is phrased the second time, requires you to find a region bordered by all three of the x-axis, the line, and the parabola. The only such region lies on [0,6], visible in the plot.

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    True, though of course that one has infinite area. ^_^2012-07-04