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.