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Evaluate an improper integral

mark

Evaluate

\int_0^{\infty} \frac{2}{(x+3)^2}\,dx.
mearing
\int_{0}^{\infty} \frac{2}{(x+3)^2}=\lim\limits_{b \to \infty}\int_{0}^{b} \frac{2}{(x+3)^2}

let u=x+3 so that du=dx

\approx2\int_{3}^{b+3} u^{-2}du\approx2\int_{3}^{b+3} -u^{-1}du\approx\int_{3}^{b+3} -\frac{2}{x+3}=\lim\limits_{b \to \infty}-\frac{2}{x+3} \Big|_3^{b+3}
=-\frac{2}{b+6}+\frac{2}{3+3}=\frac{2}{3}
mark

@mearing I think you got this right but you seem to have dropped your bounds of integration along the way.

chowell1
\int^\infty_0\frac{2}{(x+3)^2} = \lim_{b\rightarrow\infty}\int^b_0\frac{2}{(x+3)^2}

let u = x+3 so that du = dx. Thus,

\approx2\int^b_0-u^{-2}\approx2(-u^{-1}) = \lim_{b\rightarrow\infty}-\frac{2}{x+3} |^b_0
=-\frac{2}{\infty+3}+\frac{2}{0+3} = 0 + \frac{2}{3} = \frac{2}{3}
rstahles
\int_{0}^{\infty} \frac{2}{(x+3)^2}=\lim\limits_{b \to \infty}\int_{0}^{b} \frac{2}{(x+3)^2}

Let u=x+3 so that du = dx

\approx2\int_{0}^{b+3} u^{-2}du\approx2(-u^{-1})\Big|_0^{b+3}\approx -\frac{2}{x+3}\Big|_0^{b}=\lim\limits_{b \to \infty}-\frac{2}{x+3} \Big|_0^{b}
=-\frac{2}{b+3}+\frac{2}{0+3}\to 0 + \frac{2}{3} \to \frac{2}{3}

as b\to \infty.

myost
\int_0^{\infty} \frac{2}{(x+3)^2}\,dx. = \lim\limits_{b \to \infty}\int_{0}^{b} \frac{2}{(x+3)^2}

let u=x+3 so that du=dx

\approx2\int_{0}^{b} u^{-2}du \approx-2u^{-1}\Big|_3^{b+3}\ \approx-\frac{2} {x+3}\Big|_0^{b}\ = \lim\limits_{b \to \infty}-\frac{2}{x+3} \Big|_0^{b}
= -\frac{2}{b+3}+\frac{2}{0+3} = 0 + \frac{2}{3} = \frac{2}{3}

so \lim\limits_{b \to \infty}-\frac{2}{x+3} \Big|_0^{b} = \frac{2}{3}

mark

@mearing @chowell1 @rstahles @myost

I don’t quite see the fully correct response here just yet. I think some combination of mearing’s and rstahles’ might work. I like rstahles’ response because the variables are changed. I like mearing because \infty isn’t plugged in.

We don’t ever plug \infty into an algebraic expression; rather, we take a limit as a variable approaches \infty. Thus,

\frac{1}{\infty} = 0

is complete nonsense because \infty is not even a number. You can’t do algebraic operations with it. On the other hand,

\lim_{x\to0} \frac{1}{x} = 0

is a well defined mathematical equation.

knguyen3
\text{let }u=x+3, \text{ so }du=dx\\ \int_0^\infty\frac{2}{(x+3)^2}dx=\lim\limits_{b\to\infty}\int_0^b\frac{2}{(x+3)^2}dx=\lim\limits_{b\to\infty}\int_3^{b+3}\frac{2}{u^2}du\\=\lim\limits_{b\to\infty}\frac{-2}{u}|_3^{b+3}=\lim\limits_{b\to\infty}\frac{-2}{b+3}+\frac{2}{3}=0+\frac{2}{3}=\frac{2}{3}
agooch
\int_{0}^{\infty} \frac{2}{(x+3)^2}=\lim\limits_{b \to \infty}\int_{0}^{b} \frac{2}{(x+3)^2}

have u=x+3 so then du=dx

\approx2\int_{3}^{b+3} u^{-2}du\approx2\int_{3}^{b+3} -u^{-1}du
\approx\int_{3}^{b+3} -\frac{2}{x+3}=\lim\limits_{b \to \infty}-\frac{2}{x+3} \Big|_3^{b+3}
=\lim\limits_{b \to \infty}-\frac{2}{b+6}+\frac{2}{3+3}=\frac{2}{3}