Find the area between

y= 5x^3 and the liney= 45x, forxhttp://highered.mcgraw-hill.com/tber.../chars/geq.gif 0

Can anybody show me how to go about doing this?

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- May 7th 2008, 02:33 PMzsig013Area between line and curve
Find the area between

*y*= 5*x^*3 and the line*y*= 45*x*, for*x*http://highered.mcgraw-hill.com/tber.../chars/geq.gif 0

Can anybody show me how to go about doing this? - May 7th 2008, 02:48 PMicemanfan
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

$\displaystyle \int_{0}^{3} 45x - 5x^3 dx$. - May 7th 2008, 02:56 PMzsig013
Awesome, thanks!