Find the area

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April 27th 2008, 04:02 PM
R3ap3r

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:

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

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

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

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

$= 3.75\; units^2$

Hows my answer look here?
April 27th 2008, 04:03 PM
icemanfan

Quote:

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:

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

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

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

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

$= 3.75\; units^2$

Hows my answer look here?

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.
April 27th 2008, 04:09 PM
icemanfan

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.
April 27th 2008, 04:32 PM
R3ap3r

so you saying mines wrong then?
April 27th 2008, 04:40 PM
TheEmptySet

2 Attachment(s)

Quote:

Originally Posted by R3ap3r

so you saying mines wrong then?

Yes look at the graphs below

Attachment 6080

Attachment 6081

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.
April 27th 2008, 06:18 PM
R3ap3r

I c. how do you show your working for that exactly?
April 27th 2008, 06:23 PM
Mathstud28

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
April 27th 2008, 06:50 PM
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?
April 27th 2008, 06:56 PM
Mathstud28

Quote:

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
April 27th 2008, 06:59 PM
R3ap3r

so where exactly did i go wrong in my first post?
April 27th 2008, 07:02 PM
Mathstud28

Quote:

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