## jocelynevsq 2 years ago idenify the definite integral that represents the area of the region bounded by the graphs of y=x and y=5x-x^3

1. cherio12

try graphing them first to determine which function is the top and which is the bottom

2. jocelynevsq

ok

3. jocelynevsq

still don't get it

4. jocelynevsq

|dw:1359920225466:dw|

5. zepdrix

So ummmm yah that graph looks accurate. I always get confused by these types of problems. When the upper and lower functions switch places. But I think we can figure it out. Hmm.

6. zepdrix

Let's start by finding points of intersection.$\large y=x \qquad y=5x-x^3\qquad \qquad \rightarrow \qquad \qquad x=5x-x^3$

7. zepdrix

Which turn out to be, $$\large x=0, \quad x=-2, \quad x=2$$.

8. zepdrix

Let's look at just the right side for a moment, maybe it will make a little more sense that way.

9. zepdrix

|dw:1359921446750:dw|See how I've drawn a small rectangle? Let's find the area of that rectangle.

10. zepdrix

The area will be base x height. The base has a thickness of $$dx$$. The height is going to be the upper function minus the lower function.

11. zepdrix

$\large A=\left(height\right)(base)$$\large A=\left(5x-x^3-x\right)(dx)$See how the line y=x is BELOW the other one? So we subtract y=x from the other function to get the correct height. So this represents the area of a small rectangular slice. Now we want to Integrate (sum up) all of the rectangles between these two curves.

12. zepdrix

And we'll include our intersecting points.

13. zepdrix

$\large \int\limits_0^2 4x-x^3 \; dx$ This represents the area between the two curves on the RIGHT side. We still need to find the area of the other chunk. Notice how they switch places on the left? The line y=x is now the UPPER function! So we would subtract y=5x-x^3 from the line y=x. And our limits of integration would be a little different.

14. zepdrix

There is a little trick we could pull though c: See how they're both odd functions? They're both symmetric about the origin. So the area on the left will be the same as the area on the right. We could simply multiply this integral by 2.$\huge A_{total}\qquad =\qquad 2\int\limits\limits_0^2 4x-x^3 \; dx$

15. jocelynevsq

oh okay i see how you got that. thanks!