Calculus Help!

[Guest]

Hi I'm quite confused about this question, idk if I got the right answer of -1?

Find the exact area of the region bounded between the graph of f(x)=2x(1−x^2), the x-axis and the lines x=0 and x=1.

I got -1? Is that correct?

[HallsofIvy]

The area bounded by "[tex]f(x)=2x(1−x^2)[/tex], the x-axis and the lines x=0 and x=1" is given by
[tex]\int_0^1 2x(1- x^2) dx= 2\int_0^1 (x- x^3)dx[/tex]

Can you do that integral?

[Guest]

It should have been immediately obvious that -1 is NOT the "area bounded by [tex]f(x)= 2x(1- x^2)[/tex], the x-axis, and the lines x= 0 and x= 1" because an "area" is not negative!

[Guest]

Hi I'm quite confused about this question, idk if I got the right answer of -1?

Find the exact area of the region bounded between the graph of f(x)=2x(1−x^2), the x-axis and the lines x=0 and x=1.

I got -1? Is that correct?

Since there has been no response by "Guest" for 6 months, the area is given by
[tex]\int_0^1 2x(1- x^2)dx= 2\int_0^1 x- x^2 dx= 2\left[\frac{1}{2}x^2- \frac{1}{3}x^3\right]_0^1= 2\left(\frac{1}{2}-\frac{1}{3}\right)= 2\left(\frac{1}{6}\right)= \frac{1}{3}[/tex].

[Guest]

And that is wrong because of a typo! It should be
[tex]\int_0^1 x(x- x^2)dx= \int_0^1 (x^2- x^3)dx= \left[\frac{1}{3}x^3- \frac{1}{4}x^4\right]_0^1[/tex]
[tex]= \frac{1}{3}- \frac{1}{4}= \frac{4}{12}- \frac{3}{12}= \frac{1}{12}[/tex].