$$\int_{-1}^0 \frac1{(1+x)^e} \,dx$$
I have tried using integral test and got an answer of converges with answer of $$\frac1{1-e}$$
is the correct?
And also for $$\int_1^∞ e^{-2x} \, dx$$ that i tried using integration test as well with answer of $$\frac {-e^{-2}}2$$ which converges
Everything doesnt look right and please do advice, thank you.
I have also tried comparison test for both question where both are found to be ∞ (therefore diverges). However, when bigger function diverges, smaller function cannot diverge hence inconclusive. Am i doing it right?
They are just straightforward computations: $$\int_{-1}^0 \frac{{\rm d}x}{(1+x)^e} = \lim_{c \to -1}\int_c^0 \frac{{\rm d}x}{(1+x)^e} = \lim_{c \to -1} \frac{(1+x)^{1-e}}{1-e}\Big|_c^0 = +\infty,$$ because of the limit $$\lim_{c \to -1} \frac{1}{(1+c)^{e-1}} = +\infty.$$ For the next one: $$\int_1^{+\infty} e^{-2x}\,{\rm d}x = \lim_{c \to +\infty}\int_1^ce^{-2x}\,{\rm d}x = \lim_{c \to +\infty}-\frac{1}{2}e^{-2x}\Big|_1^c = \frac{1}{2e^2}.$$You just got a sign wrong there in the end, but the integral converges.