Determine if $\displaystyle \int_0^6 \frac{1}{x^{10/11}}\ dx$ converges, and if so, evaluate it.
Which part of the integral is improper?
We replace the $0$ by a parameter $R$, and then do the integral, and limit $R \rightarrow 0$
The limit we then need to compute is $\lim_{R \rightarrow 0} 11(6^{1/11}) - 11(R^{1/11})$
The integral converges to $11 6^{1/11}$.
Determine if $\int_{-3}^4 \frac{1}{x^2} \ dx$ converges, and if so, evaluate it.
Why is the integral improper?
There is a vertical asymptote at 0, so we need to split the integral there to solve it.
Once we split the integral, we need to take the limits $\displaystyle \lim_{R \rightarrow 0^-} \frac{1}{R} + \frac{1}{3}$ and $\displaystyle \lim_{R \rightarrow 0^+} \frac{1}{4} - \frac{1}{R}$
The integral diverges.
Determine if $\int_{-\infty}^\infty \frac{x}{1+x^2}\ dx $ converges, and if so, evaluate it.
We have infinity on both endpoints here. What does that mean for evaluating this integral?
We need to split this integral into two pieces, and zero is the best place to do so.
To evaluate $\displaystyle \int_0^\infty \frac{x}{1+x^2}\ dx$, use the $u$-substitution $u = 1+x^2$.
Each of the two integrals becomes $\displaystyle \int_1^\infty \frac{1}{2u}\ du$
The integral diverges.
Determine if $\displaystyle \int_5^\infty \frac{2}{(x-1)(x+3)}\ dx$ converges, and if so, evaluate it.
This integral is improper because of the $\infty$ at the upper endpoint.
How do we evaluate $\displaystyle \int_5^R \frac{2}{(x-1)(x+3)}\ dx$? Partial Fractions.
The decomposition is $\frac{1/2}{x-1} - \frac{1/2}{x+3}$, which integrates to $\frac{1}{2} \ln|x-1| - \frac{1}{2}\ln|x+3|$.
The limit we need to compute is $\displaystyle \lim_{R \rightarrow \infty} \frac{1}{2} \ln\left|\frac{R-1}{R+3}\right| - \frac{1}{2} \ln\left|\frac{5-1}{5+3}\right| $
This integral converges to $-\frac{1}{2}\ln{\frac{1}{2}}$.
Determine if $\int_2^\infty \frac{1}{x^4 + x}\ dx$ converges and explain why.
We don't have a way to evaluate this integral directly, so what else can we do to determine if this integral converges?
We need to use the comparison test. Based on the form of the expression should this converge or diverge?
The $x^4$ tells me that this should converge. Since $x^4 + x \leq x^4$ for all $x \geq 2$, what can we compare to?
Use the function $\frac{1}{x^4}$, whose integral converges, as a comparison.
Since $\frac{1}{x^4 + x} \leq \frac{1}{x^4}$ and $\displaystyle \int _2^\infty \frac{1}{x^4}\ dx$ converges, we know that $\displaystyle \int_2^\infty \frac{1}{x^4 + x}\ dx$ converges.
Determine if $\displaystyle \int_4^\infty \frac{1}{x^6 - x}\ dx$ converges and explain why.
This is another integral that we can not evaluate directly, so we need the comparison test for this.
Referring to the last problem, we could try to use $\frac{1}{x^6}$ as a comparison. However, $x^6 - x < x^6$, not the other way around. What else can we do?
Since we are looking at $x > 4$, we know that $x^5 > 4$ (yes, we could do a lot better, but this is enough), so that $x^6 =x^5 x > 4x$, so that $x < \frac{1}{4}x^6$. This implies that $x^6 - x \geq x^6 - \frac{1}{4}x^6 = \frac{3}{4}x^6$. How does that help?
Do comparison test with the function $\frac{4}{3} \frac{1}{x^6}$
Since $x > 4$, we have that $x^6 > 4x$. Therefore, we know that $\frac{1}{x^6 - x} \leq \frac{4}{3} \frac{1}{x^6}$ and $\displaystyle \int_4^\infty \frac{1}{x^6}\ dx$ converges, we know that $\displaystyle \int_4^\infty \frac{1}{x^6 - x}\ dx$ converges.
Determine if $\int_{-3}^2 \frac{1}{x^{2/3}}\ dx$ converges, and if so, evaluate it.
Determine if $\int_{-\infty}^{\infty} e^{-x^2}\ dx$ converges. Hint: Compare with $e^{x}$ if $x < -1$ and $e^{-x}$ if $x > 1$. Provide the explanation as to why this is the case.