3.1: Improper Integrals
- Page ID
- 494
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)An improper integral is the limit of a definite integral as an endpoint of the interval(s) of integration approaches either a specified real number or \({\displaystyle \infty }\) or \({\displaystyle -\infty }\) or, in some cases, as both endpoints approach limits. Such an integral is often written symbolically just like a standard definite integral, in some cases with infinity as a limit of integration.
If \(f(x)\) is continuous on \(\left(a,b\right]\) and not continuous at \(x = a\), then we define the improper integral:
\[ \int _a^b f(x)\, dx = \lim_{m \rightarrow a^+} \int_m^b f(x)\, dx \nonumber \]
\[\begin{align} \int_0^1\dfrac{1}{\sqrt{1-x^2}}dx &= \lim_{m\to{1^-}} \int_0^m \dfrac{1}{\sqrt{1-x^2}}dx \\ &= \left( \lim_{m\to1^-} \sin^{-1}x\right]_0^m \\ &= \lim_{m{\to}1^-} (\sin^{-1}m-\sin^{-1}0) \\&=\dfrac{\pi}{2}. \end{align} \nonumber \]
Improper Integrals Involving Infinity
We define the improper integral with an infinity as a one of the limits.
If \(f(x)\) is continuous on \(\left(a,b\right]\) and not continuous at \(x = a\), then we define
\[ \int _a^\infty f(x)\, dx = \lim_{m \rightarrow \infty} \int_a^m f(x)\, dx. \nonumber \]
\[\begin{align} \int_{1}^{\infty} \dfrac{1}{x^2} dx &= \lim_{m\to\infty} \int_{1}^{m} \dfrac{1}{x^2} dx \\ &= \lim_{m\to\infty}\left( -\dfrac{1}{x}\right]_1^m \\ &= \lim_{m\to\infty}\big[-\dfrac{1}{m}+1\big] \\ &=1. \end{align} \nonumber \]
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\[\int_{0}^{\infty}xe^{-x} dx \nonumber \]
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\[\int_{-\infty}^{\infty} \dfrac{1}{1+x^2} dx \nonumber \]
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\[\int_{0}^{\infty} \sin x dx \nonumber \]
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Determine for which values of \(p\) the integral \( \int _1^m \dfrac{1}{x^p}dx\) is defined.
- Use the formula for arc length to show that the circumference of the semi-circle, \( y=\sqrt{r^2-x^2} \) is \( \pi \, r \).
Vertical Asymptotes
Let \(f(x)\) have a vertical asymptote at \(x = c\). Then
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\[ \int_{a}^{b} f(x) dx = \lim_{m\to{c^-}}\int_{a}^{m} f(x) dx \nonumber \]
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\[ \int_{c}^{a} f(x) dx =\lim_{m\to{c^+}}\int_{m}^{a} f(x) dx \nonumber \]
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\[\int_{a}^{b} f(x) dx = \lim_{m\to{c^-}} \int_{a}^{m} f(x)dx = \lim_{M\to{c^+}} \int_{M}^{b}f(x) dx. \nonumber \]
Evaluate
\[ \int _0^1 \dfrac{1}{\sqrt{x}} \, dx. \nonumber \]
Solution
\[\begin{align} \int_{0}^{1}\dfrac{1}{\sqrt{x}} dx &= \lim_{m\to{0^-}}\int_{m}^{1}\dfrac{1}{\sqrt{x}} dx \\ &= \lim_{m\to{0^-}}\left(2\sqrt{x}\right]_m^1 \\ &=\lim_{m\to{0^-}}(2-2\sqrt{m}) \\ &= 2. \end{align} \nonumber \]
Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.