Finding area in a bounded region

Question

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?

Answer 1

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.

Answer 2

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.