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?
Note that on the interval $[0, 1]$, $-3x^2 &gt; (x^2 - 4x)$.  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.
Determine the  point of Intersect bounded by the  curve f(x) = x to the  power 4 and 2x to the power 2 and  g(x)= 2x to the power 2, (( i=(2) , (j) = 3,((( -3 is equal and less than x &lt; 3)))