Area between $y=x^4$ and $y=x$ The problem I'm having some trouble solving is this: calculate the area between $y=x^4$ and $y=x$. The points are $a = 0$ and $b = 1$, but the definite integral is negative. What am I doing wrong here?
You have $\displaystyle A=\int_0^1(x-x^4)dx=\left[\frac{x^2}{2}-\frac{x^5}{5}\right]_{0}^1=\frac{3}{10}$ 

or, equivalently,

$\displaystyle A=\int_0^1(y^{\frac{1}{4}}-y)dy=\left[\frac{4}{5}y^{\frac{5}{4}}-\frac{y^2}{2}\right]_0^1=\frac{3}{10}$.
$\int_{0}^{1}\int_{x^4}^{x}dy$ would give you your answer.

Or in simpler form

$\int_{0}^{1}(x-x^4)dy$

I would have even attached the graph of the two functions but I don't know how to.

NOTE: In $[0,1]$, $x\geq x^4$

Always draw the graphs of the corresponding functions before solving such questions. Helps a lot. :)