MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 18 August 2014

Lecture 22

Integrals

$$f⟶\begin{array}{c} dfdx \end{array}⟶\frac{df}{dx}\frac{df}{dx}⟶\begin{array}{c} ∫dx \end{array}⟶f$$ The integral with respect to $x$ undoes the derivative with respect to $x\text{.}$ $$\int 5x^{4}dx=x^5,\text{since}\frac{d{x}^5}{dx}=5x^4\text{.}$$ ALSO $$\int 5x^{4}dx=x^5+2,\text{since}\frac{d(x^5+2)}{dx}=5x^4\text{.}$$ This is like $\sqrt{9}=3$ and $\sqrt{9}=-3\text{.}$

We usually write $$\int 5x^{4}dx=x^5+c,$$ where $c$ is a constant, so that we can write all possible answers to $\int 5x^{4}dx$ at once. $$\begin{array}{ccc}{\displaystyle \int x^{2}dx}& =& {\displaystyle \frac{x^3}{3}+c,\text{since}\frac{d\left(\frac{x^3}{3}\right)}{dx}=\frac{1}{3}3x^2=x^2\text{.}}\\ {\displaystyle \int x^{3}dx}& =& {\displaystyle \frac{x^4}{4}+c,\text{since}\frac{d\left(\frac{x^4}{4}\right)}{dx}=\frac{1}{4}4x^3=x^3\text{.}}\\ {\displaystyle \int x^{4}dx}& =& {\displaystyle \frac{x^5}{5}+c,\text{since}\frac{d\left(\frac{x^5}{5}\right)}{dx}=\frac{1}{5}5x^4=x^4\text{.}}\\ {\displaystyle \int x^{5}dx}& =& {\displaystyle \frac{x^6}{6}+c,\text{since}\frac{d\left(\frac{x^6}{6}\right)}{dx}=\frac{1}{6}6x^5=x^5\text{.}}\\ {\displaystyle \int x^{642}dx}& =& {\displaystyle \frac{x^{643}}{643}+c,\text{since}\frac{d\left(\frac{x^{643}}{643}\right)}{dx}=\frac{1}{643}643x^{642}=x^{642}\text{.}}\\ {\displaystyle \int x^{-2}dx}& =& {\displaystyle \frac{x^{-1}}{-1}+c,\text{since}\frac{d\left(\frac{x^{-1}}{-1}\right)}{dx}=\frac{1}{-1}(-1)x^{-2}=x^{-2}\text{.}}\\ {\displaystyle \int x^{-3}dx}& =& {\displaystyle \frac{x^{-2}}{-2}+c,\text{since}\frac{d\left(\frac{x^{-2}}{-2}\right)}{dx}=\frac{1}{(-2)}(-2)x^{-3}=x^{-3}\text{.}}\\ {\displaystyle \int x^{-642}dx}& =& {\displaystyle \frac{x^{-641}}{-641}+c,\text{since}\frac{d\left(\frac{x^{-641}}{-641}\right)}{dx}=\frac{1}{-641}(-641)x^{-642}=x^{-642}\text{.}}\\ {\displaystyle \int dx}& =& {\displaystyle \int 1dx=\int x^{0}dx=\frac{x^1}{1}+c,\text{since}\frac{dx}{dx}=1\text{.}}\\ {\displaystyle \int x^{-1}dx}& =& {\displaystyle \int \frac{1}{x}dx=\text{ln}\hspace{0.17em}x+c,\text{since}\frac{d\text{ln}\hspace{0.17em}x}{dx}=\frac{1}{x}\text{.}}\end{array}$$ Note: $\int x^{-1}dx$ is NOT $\frac{x^0}{0}+c\text{;}$ $\frac{x^0}{0}$ DOESN'T MAKE SENSE.

$$\int \sqrt{x}dx=\int x^{\frac{1}{2}}dx=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c=\frac{2}{3}{x}^{\frac{3}{2}}+c,$$ since $$\frac{d\frac{2}{3}{x}^{\frac{3}{2}}}{dx}=\frac{2}{3}\frac{3}{2}{x}^{\frac{1}{2}}=x^{\frac{1}{2}}\text{.}$$

$$\int \sqrt[3]{x}dx=\int x^{\frac{1}{3}}dx=\frac{x^{\frac{4}{3}}}{\frac{4}{3}}+c=\frac{3}{4}{x}^{\frac{4}{3}}+c,$$ since $$\frac{d\frac{3}{4}{x}^{\frac{4}{3}}}{dx}=\frac{3}{4}·\frac{4}{3}{x}^{\frac{4}{3}-1}=x^{\frac{1}{3}}\text{.}$$

$$\int x^{\frac{362}{431}}dx=\frac{x^{\frac{362}{431}+1}}{\frac{362}{431}+1}+c=\frac{x^{\frac{793}{431}}}{\frac{793}{431}}+c=\frac{431}{793}{x}^{\frac{793}{431}}+c,$$ since $$\frac{d\frac{431}{793}{x}^{\frac{793}{431}}}{dx}=\frac{431}{793}\frac{793}{431}{x}^{\frac{793}{431}-1}=x^{\frac{362}{431}}\text{.}$$

$$\begin{array}{ccc}{\displaystyle \int e^{x}dx}& =& {\displaystyle e^x+c,\text{since}\frac{d{e}^x}{dx}=e^x\text{.}}\\ {\displaystyle \int e^{10x}dx}& =& {\displaystyle \frac{e^{10x}}{10}+c,\text{since}\frac{d\left(\frac{e^{10x}}{10}\right)}{dx}=\frac{1}{10}{e}^{10x}·10=e^{10x}\text{.}}\\ {\displaystyle \int e^{-31x}dx}& =& {\displaystyle \frac{e^{-31x}}{-31}+c,\text{since}\frac{d\left(\frac{e^{-31x}}{-31}\right)}{dx}=\frac{1}{-31}{e}^{-31x}(-31)=e^{-31x}\text{.}}\end{array}$$

$$\int e^{x}dx=\int {\left(e^{\text{ln}\hspace{0.17em}2}\right)}^{x}dx=\int e^{x\text{ln}\hspace{0.17em}2}dx=\frac{e^{x\text{ln}\hspace{0.17em}2}}{\text{ln}\hspace{0.17em}2}+c=\frac{2^x}{\text{ln}\hspace{0.17em}2}+c,$$ since $$\frac{d\frac{2^x}{\text{ln}\hspace{0.17em}2}}{dx}=\frac{d\frac{e^{x\text{ln}\hspace{0.17em}2}}{\text{ln}\hspace{0.17em}2}}{dx}=\frac{1}{\text{ln}\hspace{0.17em}2}{e}^{x\text{ln}\hspace{0.17em}2}·\text{ln}\hspace{0.17em}2=e^{x\text{ln}\hspace{0.17em}2}=2^x\text{.}$$

$$\int {38}^{x}dx=\int {\left(e^{\text{ln}\hspace{0.17em}38}\right)}^{x}dx=\int e^{x\text{ln}\hspace{0.17em}38}dx=\frac{e^{x\text{ln}\hspace{0.17em}38}}{\text{ln}\hspace{0.17em}38}+c,$$ since $$\frac{d\frac{e^{x\text{ln}\hspace{0.17em}38}}{\text{ln}\hspace{0.17em}38}}{dx}=\frac{1}{\text{ln}\hspace{0.17em}38}{e}^{x\text{ln}\hspace{0.17em}38}·\text{ln}\hspace{0.17em}38=e^{x\text{ln}\hspace{0.17em}38}={38}^x\text{.}$$

$$\begin{array}{ccc}{\displaystyle \int \text{sin}\hspace{0.17em}xdx}& =& {\displaystyle -\text{cos}\hspace{0.17em}x+c,\text{since}\frac{d(-\text{cos}\hspace{0.17em}x)}{dx}=-(-\text{sin}\hspace{0.17em}x)=\text{sin}\hspace{0.17em}x\text{.}}\\ {\displaystyle \int \text{cos}\hspace{0.17em}xdx}& =& {\displaystyle \text{sin}\hspace{0.17em}x+c,\text{since}\frac{d\text{sin}\hspace{0.17em}x}{dx}=\text{cos}\hspace{0.17em}x\text{.}}\\ {\displaystyle \int {\text{sec}}^{2}xdx}& =& {\displaystyle \text{tan}\hspace{0.17em}x+c,\text{since}\frac{d\text{tan}\hspace{0.17em}x}{dx}={\text{sec}}^{2}x\text{.}}\\ {\displaystyle \int {\text{csc}}^{2}xdx}& =& {\displaystyle -\text{cot}\hspace{0.17em}x+c,\text{since}\frac{d(-\text{cot}\hspace{0.17em}x)}{dx}=-(-{\text{csc}}^{2}x)={\text{csc}}^{2}x\text{.}}\\ {\displaystyle \int \text{tan}\hspace{0.17em}x\hspace{0.17em}\text{sec}\hspace{0.17em}xdx}& =& {\displaystyle \text{sec}\hspace{0.17em}x+c,\text{since}\frac{d\text{sec}\hspace{0.17em}x}{dx}=\text{tan}\hspace{0.17em}x\hspace{0.17em}\text{sec}\hspace{0.17em}x\text{.}}\\ {\displaystyle \int \text{cot}\hspace{0.17em}x\text{csc}\hspace{0.17em}xdx}& =& {\displaystyle -\text{csc}\hspace{0.17em}x+c,\text{since}\frac{d(-\text{csc}\hspace{0.17em}x)}{dx}=-(-\text{cot}\hspace{0.17em}x\hspace{0.17em}\text{csc}\hspace{0.17em}x)=\text{cot}\hspace{0.17em}x\hspace{0.17em}\text{csc}\hspace{0.17em}x\text{.}}\end{array}$$

$$\begin{array}{ccc}{\displaystyle \int \frac{\text{cos}\hspace{0.17em}x}{{\text{sin}}^{2}x}dx}& =& {\displaystyle \int \frac{\text{cos}\hspace{0.17em}x}{\text{sin}\hspace{0.17em}x}\frac{1}{\text{sin}\hspace{0.17em}x}dx}\\ & =& {\displaystyle \int \text{cot}\hspace{0.17em}x\text{csc}\hspace{0.17em}xdx}\\ & =& {\displaystyle -\text{csc}\hspace{0.17em}x+c\text{.}}\end{array}$$

$$\begin{array}{ccc}{\displaystyle \int \frac{1}{1+\text{cos}\hspace{0.17em}x}dx}& =& {\displaystyle \int \frac{1}{(1+\text{cos}\hspace{0.17em}x)}\frac{(1-\text{cos}\hspace{0.17em}x)}{(1-\text{cos}\hspace{0.17em}x)}dx}\\ & =& {\displaystyle \int \frac{1-\text{cos}\hspace{0.17em}x}{1-{\text{cos}}^{2}x}dx}\\ & =& {\displaystyle \int \frac{1-\text{cos}\hspace{0.17em}x}{{\text{sin}}^{2}x}dx}\\ & =& {\displaystyle \int (\frac{1}{{\text{sin}}^{2}x}-\frac{\text{cos}\hspace{0.17em}x}{{\text{sin}}^{2}x})dx}\\ & =& {\displaystyle \int ({\text{csc}}^{2}x-\text{cot}\hspace{0.17em}x\hspace{0.17em}\text{csc}\hspace{0.17em}x)dx}\\ & =& {\displaystyle -\text{cot}\hspace{0.17em}x+\text{csc}\hspace{0.17em}x+c\text{.}}\end{array}$$

Notes and References

These are a typed copy of Lecture 22 from a series of handwritten lecture notes for the class MATH 221 given on October 30, 2000.

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