I was assigned this integral - a fairly simple problem: $$\int^3_0\frac{1}{\sqrt{3-x}}\ dx$$ Becaause $f(x)$ is continuous only on $(0,3]$ we must take a left sided limit in order to properly solve. Doing this gets us: $$\lim_{t\to 3^-}\int^t _0\frac{1}{\sqrt{3-x}}\ dx$$Which is: $$\lim_{t\to3^-}(-2\sqrt{3-x}) |^t_0$$This is: $$-2\sqrt3+\lim_{t\to3^-}(-2\sqrt{3-t})$$ The second part of the problem becomes zero, leaving me with $-2\sqrt3$ as my answer. This looks great to me, but the real answer is $2\sqrt3$.
Why doesn't the answer have a negative? What did I do wrong that gave me that last negative at the end?
As $t$ approaches $3$ from the left we have
\begin{align}\lim_{t\to 3^-}\int^t _0\frac{1}{\sqrt{3-x}}\ dx&=\lim_{t\to 3^-}\Big[-2\sqrt{3-x}\Big]_0^t\\&= \lim_{t\to 3^-}\Big(-2\sqrt{3-t}+2\sqrt{3}\Big)\\&= \lim_{t\to 3^-}\Big(-2\sqrt{3-t}\Big)+2\sqrt{3}\\&= 0+2\sqrt{3}\\&= 2\sqrt{3} \end{align}
so you made the following errors
$$\lim_{t\to 3^-}\Big[-2\sqrt{3-x}\Big]_0^t=-2\sqrt{3}+\lim_{t\to 3^-}\Big(-2\sqrt{3-t}\Big)$$ which is false. You need to carry through the minus sign to the lower integration limit to form $$\lim_{t\to 3^-}\Big[-2\sqrt{3-x}\Big]_0^t=\lim_{t\to 3^-}\Big(-2\sqrt{3-t}\Big)+2\sqrt{3}$$