# Thread: area between curves

1. ## 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. 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!

2. 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. 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: $\displaystyle A=\frac{125}{54}\approx 2.315$ )

3. 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.

4. omg! Thanks! I made the dumbest mistakes! I was finding it from 25/3 to 5/3 instead of 0 to 5/3!!! #### Search Tags

area, curves 