sketch the region enclosed by y = 3x and y = 5x^2. Decide wether to integrate with respect to x or y. What is the area of the region?
Your limits will be where the two graphs intersect to find there solve $\displaystyle 3x=5x^2$
(I get $\displaystyle x=0$ and $\displaystyle x=0.6$).
You'll nearly always integrate with respect to x and this is no exception
If you sketch a graph it will help give you an idea of the area. From what I can gather between 0 and 0.6: 3x will increase faster than 5x^2 so the integral becomes
$\displaystyle \int^{0.6}_0 (3x-5x^2) \, .dx$
I was trying to solve this problem too and I'm just wondering if this would be right:
$\displaystyle 5x^2 = 3x$
$\displaystyle 5x^2 - 3x = 0$
$\displaystyle x(5x - 3) = 0; x = 0; x = \frac{3}{5}$
$\displaystyle \int_0^\frac{3}{5} 5x^2 - 3x \,dx = \frac{5}{3}x^3 - \frac{3}{2}x^2$
$\displaystyle \frac{5}{3}(\frac{3}{5})^3 - \frac{3}{2}(\frac{3}{5})^2 = \frac{9}{25} - \frac{27}{50} = -\frac{9}{50}$
so the area is:
$\displaystyle \frac{9}{50}$
Is that right?
Everything is correct except your use of integration. You cannot subtract right minus left unless you're going to use horizontal strips. You must use vertical strips therefore you take the top function minus the bottom function.
Therefore your integral should be
$\displaystyle \int_0^\frac{3}{5} 3x - 5x^2 dx$
This gives the answer $\displaystyle \frac{9}{50} units^2$