# Do the two same curves give different area?

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 – Raymond Manzoni Jul 4 '12 at 13:27
sorry how? $y=-3*x$ why? – dato datuashvili Jul 4 '12 at 13:30
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! – Raymond Manzoni Jul 4 '12 at 13:33
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. – D. Thomine Jul 4 '12 at 13:36
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. – Raymond Manzoni Jul 4 '12 at 14:00

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). – Cameron Buie Jul 4 '12 at 14:28
@CameronBuie and in that case, the unbounded region on $[3,\infty)$ works like a charm too! – Eugene Shvarts Jul 4 '12 at 14:31