# Finding area in a bounded region

How would I solve the following problem?

Sketch the region bounded by the graphs of $f(x)=x^2-4x$ and $-3x^2$ and then find the area of the region.

I have made a sketch for this question but what I am not sure how to do is find the area.

I have found the intersections $x=0$ and $x=1$

$x^2-4x=-3x^2$

$4x^2-4x=0$

$x=0,x=1$

$$f(x)=\int_0^1 x^2-4x-(-3x^2)(dx)$$

$$f(x)=\int_0^1 4x^2-4x(dx)$$

$$G(X)=\frac{4x^3}{3}-2x^2$$

So I did

$$A=\frac{4}{3}(1)-2(1)^2-0$$

$$A=\frac{-2}{3}$$

But how can an Area be negative?

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Note that on the interval $[0, 1]$, $-3x^2 > (x^2 - 4x)$. So the "upper curve" we are interested in is $-3x^2$ and the lower curve $(x^2 - 4x)$.

Always subtract the curve that is UNDER, from the curve that is OVER.

So you need for your integrand to be $$\int_0^1 -3x^2 - (x^2 - 4x) \,dx$$

Otherwise, your work was fine! You found your bounds correctly, and you integrated and evaluated correctly, just wrong choice of the order of subtraction.

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I see that makes sense but to tell which one is over and which one is under should I plug in a test point like f(1/4) and the greater one would the OVER one? – Fernando Martinez Apr 13 '13 at 20:53
Test points are good. But one of the reasons your exercise asked you to sketch the graphs of the functions and area bounded by them was to help you visualize which is on top. Sometimes, two graphs might intersect at three points, where between the first two, one function is greater than the other, and between the second and third point, that reverses. (Think of $\sin x$ and $\cos x$ over the interval $[0, \pi]$. So you might need to check a test point on each side of the points of intersection. – amWhy Apr 13 '13 at 20:57
Yes, in this example, 1/4 would be a fine text point: the function that is OVER the other will evaluate greater than the other. – amWhy Apr 13 '13 at 20:59
Ok that makes sense thanks. – Fernando Martinez Apr 13 '13 at 21:02
@amWhy: Always helpful and instructive! +1 – Amzoti Apr 14 '13 at 1:07

When finding the area bounded by the two curves, make sure you're always integrating top minus bottom. In this case, the $-3x^2$ is actually greater than $x^2 - 4x$. Thus, your integral should be: $$\int_0^1-3x^2-\left(x^2 - 4x\right)\;dx$$ That was the only error you made.

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I see thanks for the help. – Fernando Martinez Apr 13 '13 at 20:46