Find the Area (HELP--QUICKLY)!

[OhhCalculus]

So, I'm doing this quiz thing, and I have to find the area of this equation and graph it.

The first problem has y=x^3-x and y=3x as the equations.

I set them equal to each other to end up getting the numbers -2, 0, and 2.

I attempted to graph it in my graphing calculator at this point and got a graph with two symmetrical shapes. So, I figured that adding them together would be the correct answer. Yea. I don't really know. The correct answer is supposed to be 8 (I think). Any work would help.

----------------------------

The second one is x=8-y^2 and x=y^2

I did the same process, and ended up with an answer of -64/3. That seems ridiculous. :/

 

[galactus]

So, I'm doing this quiz thing, and I have to find the area of this equation and graph it.

The first problem has y=x^3-x and y=3x as the equations.

I set them equal to each other to end up getting the numbers -2, 0, and 2.

I attempted to graph it in my graphing calculator at this point and got a graph with two symmetrical shapes. So, I figured that adding them together would be the correct answer. Yea. I don't really know. The correct answer is supposed to be 8 (I think). Any work would help.

It would appear they are looking for the signed area under the curves.

\(\displaystyle \int_{0}^{2}(x^{3}-x-3x)dx=-4\)

\(\displaystyle \int_{-2}^{0}(x^{3}-x-3x)dx=4\)

Just integrating from -2 to 2 will give 0 because they cancel one another. But, each separately, gives the area.

One is below the x-axis and one above. 4+4=8. Think of the negative area as an absolute value.

Or, do it this way:

\(\displaystyle \int_{-2}^{0}(x^{3}-x-3x)dx=4\)

\(\displaystyle \int_{0}^{2}(3x-(x^{3}-x))dx=4\)

See. From 0 to 2, y=3x bounds above. From -2 to 0, it bounds below.

----------------------------

 

[Pklarreich]

So, I'm doing this quiz thing, and I have to find the area of this equation and graph it.

The first problem has y=x^3-x and y=3x as the equations.

I set them equal to each other to end up getting the numbers -2, 0, and 2.

I attempted to graph it in my graphing calculator at this point and got a graph with two symmetrical shapes. So, I figured that adding them together would be the correct answer. Yea. I don't really know. The correct answer is supposed to be 8 (I think). Any work would help.

It would appear they are looking for the signed area under the curves.

\(\displaystyle \int_{0}^{2}(x^{3}-x-3x)dx=-4\)

\(\displaystyle \int_{-2}^{0}(x^{3}-x-3x)dx=4\)

==============================================
Worth noting:

x^3 - 4x is an Odd function. Useful observations about even and odd functions:

\(\displaystyle \int_{-a}^{a}Odd(x)dx = 0\)

and

\(\displaystyle \int_{-a}^{a}Even(x)dx = 2\int_{0}^{a}Even(x)dx\)