# Thread: The Area Between Two Curves

1. ## The Area Between Two Curves

Find the area between y= 2-3x^2, y=6, x=1, x=5

2. Originally Posted by McDiesel

Find the area between y= 2-3x^2, y=6, x=1, x=5
Quite unclear on what you just typed. Type it exactly from your homework.

If you type what I think it is, then solve this eqn.
$\int\limits_1^5 {(6 - 2+ 3x^2)} dx$

3. yeah, thats the right equation but shouldn't it be -3x^2? I worked the equation out and I get -140 but the answer is 140...not sure why

4. Originally Posted by McDiesel
yeah, thats the right equation but shouldn't it be -3x^2? I worked the equation out and I get -140 but the answer is 140...not sure why
Nope, because the area under the curve of -3x^2 is negative. And you are trying to find the total absolute area between two curves. Therefore (2-3x^2) becomes (-2+3x^2)

$\int\limits_1^5 {(6 - 2+ 3x^2)} dx = 140$ NOT -140

140 is the total area between the two curves, since the area under the x-axis is negative so you have to take into account for that.

5. Here is my work, where did I go wrong?

$\int\limits_1^5 {(6-2-3x^2)}$

$2x-(3x^3/3)-6x$

$2x-x^3-6x$

$2(5)-5^3-6(5)-[2(1)-1^3-6(1)]$

$=-140$

So where did I go wrong?

6. Originally Posted by McDiesel
Here is my work, where did I go wrong?

$\int\limits_1^5 {(6-2+3x^2)}$

$2x-(3x^3/3)-6x$

$2x-x^3-6x$

$2(5)-5^3-6(5)-[2(1)-1^3-6(1)]$

$=-140$

So where did I go wrong?
You signs are messed up. The first step... 2x should be -2x and -6x and be 6x.

Remember that you have to change the signs of the (2-3x^2) to (-2+3x^2)

7. Here, I just edited your steps

$\int\limits_1^5 {(6-2-3x^2)}dx$

= $\int\limits_1^5 {(4-3x^2)}dx$

$4x+(3x^3/3)$

$4x+x^3$

$[4(5)+5^3]- [4(1)+(1)^3]$

$[145]- [5]$

$=140$

8. Originally Posted by McDiesel
yeah, thats the right equation but shouldn't it be -3x^2? I worked the equation out and I get -140 but the answer is 140...not sure why
Draw a picture, it will make it clear what area the problem is asking for.

CB