Improper integrals - evaluate

Question

$$\int_{-2}^2\frac{\sqrt{2-x}-3\sqrt{2+x}}{\sqrt{4-x^2}}\,dx$$

I integrated it to get $$2\sqrt{2+x} + 6\sqrt{2-x}$$ but I dont think it is right. For reference I am doing further maths a level edexcel.

Answer 1

As Fred notes, your integrand $\frac{1}{\sqrt{2+x}}-\frac{3}{\sqrt{2-x}}$ really does have the antiderivative you obtained. Of course, you've been asked to evaluate a definite integral, so you need $[2\sqrt{2+x}+6\sqrt{2-x}]_{-2}^2$. If you want a further spoiler (but I recommend you try the calculation yourself first), hover over this:

$$[2\sqrt{2+x}+6\sqrt{2-x}]_{-2}^2=2\sqrt{4}+6\sqrt{0}-2\sqrt{0}-6\sqrt{4}=-8.$$ This plot provides a nice double-check that we'd expect a negative answer.

Answer 2

$\int_{-2}^2\frac{\sqrt{2-x}-3\sqrt{2+x}}{\sqrt{4-x^2}}$ is an improper integral.

$\int_{-2}^2$ is convergent $ \iff \int_{-2}^0$ and $\int_{0}^2$ are both convergent.

$2\sqrt{2+x} + 6\sqrt{2-x}$ is indeed an anti-derivative of $\frac{\sqrt{2-x}-3\sqrt{2+x}}{\sqrt{4-x^2}}$.

Now investigate the convergence of $ \int_{-2}^0$ and $\int_{0}^2$