1. ## Integrals. Area.

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?

2. Originally Posted by B-lap
Sketch the region enclosed by y = 3x and y = 5x^2. Decide whether to integrate with respect to x or y. The find 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

$\displaystyle \int^{0.6}_0 (3x-5x^2) \, .dx$

3. You learned how to sketch back in algebra, and also how to solve systems of equations. So you should have a nice picture with labelled intersection points.

Where are you stuck after that?

4. 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?

5. looks fine at a glance Korupt

6. Originally Posted by Korupt
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.

$\displaystyle \int_0^\frac{3}{5} 3x - 5x^2 dx$

This gives the answer $\displaystyle \frac{9}{50} units^2$

7. Originally Posted by millerst
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.

$\displaystyle \int_0^\frac{3}{5} 3x - 5x^2 dx$
This gives the answer $\displaystyle \frac{9}{50} units^2$
Well, he didn't "subtract right minus left", he subtracted "bottom minus top", probably because he did not graph this, as stapel originally suggested, and did not realize that $\displaystyle 3x\ge 5x^2$ on this interval.