# area between curves

• Dec 2nd 2007, 10:09 PM
MathNeedy18
area between curves
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
http://webwork1.math.utah.edu/math12...995593img1.gif

From (5/3) to (25/3) 3x^2-5x dx

I keep getting 407.40737 and it's not right! I dunno what I am doing wrong! BAH!
• Dec 2nd 2007, 11:28 PM
earboth
Quote:

Originally Posted by MathNeedy18
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
http://webwork1.math.utah.edu/math12...995593img1.gif

From (5/3) to (25/3) 3x^2-5x dx

I keep getting 407.40737 and it's not right! I dunno what I am doing wrong! BAH!

Hello,

first calculate the intercepts of the two functions: (0, 0), (5/3, 25/3)

The enclosed area is calculated by the difference of the two functions:

$\displaystyle d(x)=5x-3x^2$

$\displaystyle A=\int_0^{\frac53}(5x-3x^2)dx$

I'm sure that you can do the rest.

(For confirmation only: $\displaystyle A=\frac{125}{54}\approx 2.315$ )
• Dec 2nd 2007, 11:31 PM
tbyou87
So first you need to find the points of intersection: (0,0) and (5/3,25/3).
Then you can integrate with respect to x. Make sure you take the function on top 5x and subtract the lower function 3x^2.
$\displaystyle \int_0^\frac{5}{3}(5x-3x^2)dx$
I get 125/54.
• Dec 3rd 2007, 05:50 PM
MathNeedy18
omg! Thanks! I made the dumbest mistakes! I was finding it from 25/3 to 5/3 instead of 0 to 5/3!!!(Tmi)