Question

need help on improper integrals
a) ∫ -2 to 2 dx/(x^2-4)
b) ∫0 to 2pi/3 tanxdx

Answer 1

That is an improper integral, hence you need put it into limit

Answer 2

\[\int_{-2}^2 \dfrac{dx}{x^2-4}=\int_{-2}^0 \dfrac{dx}{x^2-4}+\int_{0}^{2}\dfrac{dx}{x^2-4}\]
so far so good?

Answer 3

Now, put both into limit

Answer 4

\[lim_{t\rightarrow -2}\int_{t}^0 \dfrac{dx}{x^2-4}+\lim_{t\rightarrow 2}\int_{0}^{t}\dfrac{dx}{x^2-4}\]

Answer 5

Take integral, then limit of each term, then + them together. Dat sit.