1. ## Integration problem

Question: The area bounded by $y = x^{3} - 3x^{2}$ and the line $y=-4$ is given by the integral ....

I deduced this:

$\int_{-1}^2 (x^{3} - 3x^{2} + 4) dx$

So it's the int from -1 to 2.

$\int_{-1}^2 (4 - x^{3} + 3x^{2}) dx$

Does this have something to do with the graph being below the x-axis?

2. It is $4-x^3-3x^2$, not $x^3-3x^2-4$ because y=4 is above the graph of $x^3-3x^2$. Finding the area of a graph is finding the top part, then subtracting the empty spaces to get your area. In this case y=4 is above the graph of $y=x^3-3x^2$ from -1 to 2.

BTW, you can put a negative limit to the integral by using {-1}.

3. Originally Posted by Lord Darkin
Question: The area bounded by $y = x^{3} - 3x^{2}$ and the line $y=4$ is given by the integral ....

I deduced this:

$\int_1^2 (x^{3} - 3x^{2} + 4) dx$

NOTE: That 1 at the integral should be NEGATIVE 1, the latex wouldn't work when I put in -1.

So it's the int from -1 to 2.

But the correct answer from my answer key seems to be (again, int from NEGATIVE 1 to 2, NOT 1 to 2):

$\int_1^2 (4 - x^{3} + 3x^{2}) dx$

Does this have something to do with the graph being below the x-axis?

I think you mean the area between $3x^2-x^3$ and y=4,

as the other one has no bounded region.

The curve is not below the x-axis from x=-1 to x=2.
You subtract the curve equation from 4 as the curve lies below the line between the points of intersection.
In the bounded region, f(x) is always below y=4.

Integrating the difference between them calculates the area.

The integral is $\int{5-(3x^2-x^3)}dx$

4. Originally Posted by Lord Darkin
Question: The area bounded by $y = x^{3} - 3x^{2}$ and the line $y=4$ is given by the integral ....
what region? did you leave out any additional problem information?

5. Oops, sorry

it was y= NEGATIVE 4

That's why ...

Sorry again, I hate it when I do these typos.

6. No worries,
it's all clear now.
From skeeter's diagram, the curve is now above the line,
hence subtract -4 from the curve equation (ie add 4)

7. ^I see, so why does the right integral turn otu tohave negative x^3 rather than a positive x^3?

8. My earlier integral is incorrect because the line is y=-4, rather than y=4.

If the line was y=4, the curve would need to be $3x^2-x^3$

As the line is y=-4, it's fine to use the equation as you have it... $x^3-3x^2$

Your integral now is $\int{x^3-3x^2-(-4)}dx$

from x=-1 to x=2

Are you sure it's y=-4 and not y=4?

9. Yes, pretty certain.

This means the answer I had in my original post matches yours. My answer key says that it's the second thing I posted, the one where the y=x^3 - 3x^2 function is multiplied by -1. I assumed that had something to do with it being below the x-axis.

10. Yes,

however, the area will give a positive answer if you subtract the line from the curve, since the curve is above the line.

I would say what you had was correct.
the answer key would be correct if the bounds x=-1 and x=2 were reversed above and below the integral sign.

11. Nope, can't find anything like that. I guess that's a mistake.

12. The answer key gives the area between $3x^2-x^3$ and y=4.

13. Ye sit does. This was a multiple choice question and D was the second thing I posted in my original post. Choice A (what I deduced originally and is first in my post) doesn't seem to be right.

14. Originally Posted by Lord Darkin
Question: The area bounded by $y = x^{3} - 3x^{2}$ and the line $y=-4$ is given by the integral ....

I deduced this:

$\int_{-1}^2 (x^{3} - 3x^{2} + 4) dx$ correct for equations given

So it's the int from -1 to 2.

$\int_{-1}^2 (4 - x^{3} + 3x^{2}) dx$ correct for y=4 and $\color{red}f(x)=3x^2-x^3$