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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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    Everything looks right for the first problem. The second one doesn't seem well formulated. Note that, replacing $y=3x\ $ by $y=-3x\ $ in the first problem, you get $\frac {63}2\ $ as wished : [Alpha](http://www.wolframalpha.com/input/?i=int%286*x-x%5E2%2B3*x%2Cx%3D0..3%29)2012-07-04
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    sorry how? $y=-3*x$ why?2012-07-04
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    I merely observed that you got the fine result in this case! Another way to get the $31.5$ would be to compute the area between $y=x^2-6x\ $ and $y=3x\ $. Or to make things short : I think that the second formulation is wrong as provided!2012-07-04
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    but first and second are same ,just different is by $y=0$ this line2012-07-04
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    The second problem is well-formulated. Just draw the graph of the function: the area you need to compute is included between the curves $y=0$ and $y=3x$ ($x \in [0,3]$), and between the curves $y=0$ and $y=6x-x^2$ ($x \in [3,6]$). The answer given is correct.2012-07-04
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    so it means that i have to compute area between [0 3] and [3 6] and add to each other?2012-07-04
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    I think Eugene (+1) got the right answer $\frac {27}2$ for the left part and $18$ for the right part (and I was wrong!).2012-07-04
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    Whoo, I can comment now! Just wanted to add that the second integral should be $\int_0^6 -x^2+6x dx$ minus the result of the integral from the first part. There is a region you may not be accounting for on [0,3].2012-07-04
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    so guys please tell me finally how my integral should look like?2012-07-04
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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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    I contend that it isn't the *only* such region. Observe that the "above the line, below the parabola" slice from the first problem *is* bordered by a (measure zero) portion of the $x$ axis. The problem really should have been phrased better (though of course you've correctly interpreted it).2012-07-04
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    @CameronBuie and in that case, the unbounded region on $[3,\infty)$ works like a charm too!2012-07-04
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    True, though of course that one has infinite area. ^_^2012-07-04