Find the area between
y = 5x^3 and the line y = 45x, for x 0
Can anybody show me how to go about doing this?
First you need to figure out where the curves intersect. So you have the equation $\displaystyle 5x^3 = 45x$ or $\displaystyle 5x^3 - 45x = 0$. Simplifying, expanding and solving for x:
$\displaystyle 5(x^3 - 9x) = 0$
$\displaystyle x^3 - 9x = 0$
$\displaystyle x(x^2 - 9) = 0$
$\displaystyle x(x-3)(x+3) = 0$
So the curve has three roots. But the problem asks us only to consider values of x greater than or equal to 0, so we have the points of intersection at x = 0 and x = 3.
The next question is, which of the two curves is above the other? A quick test reveals $\displaystyle f(1) = -40$ where $\displaystyle f(x) = 5x^3 - 45x$ so it must be that $\displaystyle 45x$ is above $\displaystyle 5x^3$. Hence, you want to use $\displaystyle g(x) = 45x - 5x^3$.
So to compute the area between the curves we integrate
$\displaystyle \int_{0}^{3} 45x - 5x^3 dx$.