# Math Help - Find the area

1. ## Find the area

Find the exact area of the region between the curve $y = 4x^3 - 2x^2 - 12x$ and the x=axis, over the interval [-1, 2]

My Work:

$\int_{-1}^2\;(4x^3 - 2x^2 - 12x)\;dx = x^4 - \frac{2}{3}\; x^3 - 6x^2$

$Let\; x = 2\;\;\;\;\; 2^4 - \frac{2}{3}\; (2)^3 - 6(2)^2$

$Let\; x = -1\;\;\;\;\; (-1)^4 - \frac{2}{3}\;(-1)^3 - 6(-1)^2$

$\left[2^4 - \frac{2}{3}\;(2)^3 - 6(2)^2\right] - \left[(-1)^4 - \frac{2}{3}\;(-1)^3 - 6(-1)^2\right]$

$= 3.75\; units^2$

2. Originally Posted by R3ap3r
Find the exact area of the region between the curve $y = 4x^3 - 2x^2 - 12x$ and the x=axis, over the interval [-1, 2]

My Work:

$\int_{-1}^2\;(4x^3 - 2x^2 - 12x)\;dx = x^4 - \frac{2}{3}\; x^3 - 6x^2$

$Let\; x = 2\;\;\;\;\; 2^4 - \frac{2}{3}\; (2)^3 - 6(2)^2$

$Let\; x = -1\;\;\;\;\; (-1)^4 - \frac{2}{3}\;(-1)^3 - 6(-1)^2$

$\left[2^4 - \frac{2}{3}\;(2)^3 - 6(2)^2\right] - \left[(-1)^4 - \frac{2}{3}\;(-1)^3 - 6(-1)^2\right]$

$= 3.75\; units^2$

Looks good again. For you, the power rule is obviously not a problem. However, you must be careful that you are answering the question correctly. If there is a root of the equation between -1 and 2, you have a problem.

3. Yeah, the root x = 0 is going to cause you problems. Let $f(x) = 4x^3 - 2x^2 - 12x$. Then you want to integrate $\int_{-1}^{0} f(x)$ and $\int_{0}^{2} f(x)$ separately, and add the absolute values of the answers.

4. so you saying mines wrong then?

5. Originally Posted by R3ap3r
so you saying mines wrong then?
Yes look at the graphs below

we need to break it up into two integrals

$\int_{-1}^{0}f(x)dx-\int_{0}^{2}f(x)dx$

The minus sign is because the area is below the x axis.

I hope this clears it up.

6. I c. how do you show your working for that exactly?

7. State that the curve is less than 0 on [-1,0] and greater than 0 on [0,2] then state that to get the area of a curve when it is below the x-axis you must take the opposite of it...evaluate the integrals..add and poof! you have the answer

8. what I meant to say was what did you get as your answer for it emptyset? does it differ a lot from my own?

9. Originally Posted by R3ap3r
what I meant to say was what did you get as your answer for it emptyset? does it differ a lot from my own?
Taking $\int_{-1}^{0}f(x)dx-\int_0^{2}f(x)dx=\frac{13}{3}+\frac{40}{3}=\frac{5 3}{3}$square units

10. so where exactly did i go wrong in my first post?

11. Originally Posted by R3ap3r
so where exactly did i go wrong in my first post?
Technically the area between any curve and the x-axis is $\int|f(x)|dx$

So when the curve changed signs you had to break the integral at that piont and evaluate them seperately based on whether due to the || if it was positive or negative