# Thread: Integral-Area of the region

1. ## Integral-Area of the region

I have a problem that i am completely stuck on, stuck to the point that i don't even know how to approach it. If someone could simply get me started i can do the rest. Any help is appreciated, thanks.

Problem: The area of the region bounded by the graphs of y=x^3 and y=x cannot be found by the single integral ∫(ranging from -1 to 1) [X^3-X]Dx. Explain why this is so. Use symmetry to write a single integral that does represent the area.

2. Originally Posted by Juggalomike
I have a problem that i am completely stuck on, stuck to the point that i don't even know how to approach it. If someone could simply get me started i can do the rest. Any help is appreciated, thanks.

Problem: The area of the region bounded by the graphs of y=x^3 and y=x cannot be found by the single integral ∫(ranging from -1 to 1) [X^3-X]Dx. Explain why this is so. Use symmetry to write a single integral that does represent the area.
$\int_{-1}^{1}(x^3-x) dx = 0$ can easily be seen both from the symmetry of $f(x) = x^3 - x$ and from simply solving the integral, but we can see that the area of the region is definitely not 0!

Let A be the region bounded by the graphs of $g(x) = x^3$ and $h(x) = x$, then, using the same argument as before (symmetry), we can conclude that $A = \int_{0}^{1} (x - x^3) dx + \int_{-1}^{0} (x^3 - x) dx = 2*\int_{0}^{1} (x^3 - x) dx$

3. $\int^{0}_{-1}(x^{3}-x) dx + \int^{1}_{0} (x-x^{3}) \ dx = 2\int^{1}_{0}(x-x^{3}) dx$

too slow

4. thanks a lot for the help, was a lot simpler then i was making it out to be