<s>[INST] Double Integral Confusion A buddy was asking me for help with one of his MV Calc problems, and I ended up getting the same answer as him so I figured I'd ask it here...   
Question Find $$\iint_{R} (x-1) \, dA$$ where $R$ is the region enclosed by $y=x$ and $y=x^3$ in the first quadrant. So naturally I told him to compute $$\int_{0}^{1} \int_{x^3}^{x} (x-1) \, dy \, dx$$ as $x$ varies from $0$ to $1$ and $y$ from $x^3$ to $x$. Using this integral, we got $\frac{-7}{60}$. His textbook gives an answer of $\frac{-1}{2}$.
 [/INST] As mentioned in the comments, your answer is correct.  I'll provide a solution here for future readers.
First, draw a picture.  I'm going to use Wolfram Alpha to show one online--it should be simple enough to graph by hand.  We can see the intersections are at $(0, 0)$ and a $(1, 1)$.  Thus, we have our limits of integration.  Evaluation is as follows:
$$\begin{align}
\int_0^1\int_{x^3}^x(x-1)\,dy\,dx &= \int_0^1y(x-1)\Bigg|_{x^3}^x\,dx\\
&= \int_0^1(x-x^3)(x-1)\,dx\\
&= \int_0^1(-x^4+x^3 + x^2-x)\,dx\\
&= -\frac{x^5}{5} + \frac{x^4}{4} + \frac{x^3}{3} - \frac{x^2}{2}\Bigg|_0^1 \\
&= \frac{-1}{5} + \frac{1}{4} + \frac{1}{3} -\frac{1}{2}\\
&= \frac{-7}{60}
\end{align}$$
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