anti-derivatives / anti-differentiation / integration

Math This is how you reverse derivatives. It seems as though, because we get a + C, these end up as indeterminite results. If you get a few data points like $f(1)=3$, then we can figure out what those c’s are.

“antidifferentiation” is the same as “taking the anti-derivative”

Antidifferentiation is “removing the integral sign”

Reversing differentiation rules (e.g. anti-derivatives)

All of the rules

$\begin{array}{rlr}\int c\cdot f(x)dx& =c\cdot \int f(x)dx& \text{constant multiple}\\ \int (f(x)\pm g(x))dx& =\int f(x)dx\pm \int g(x)dx& \text{sum/difference rule}\\ \int 0dx& =C& \\ \int 1dx& =\int dx=x+C& \\ \int x^{n}dx& =\frac{1}{n+1}{x}^{n+1}+c(ifn\ne -1)& \text{power rule}\end{array}$

$\begin{array}{rl}\int \cos xdx& =\sin x+C\\ \int \sin xdx& =-\cos x+C\\ \int {\sec }^{2}xdx& =\tan x+C\\ \int \csc x\cot xdx& =-\csc x+C\\ \int \sec x\tan xdx& =\sec x+C\\ \int {\csc }^{2}xdx& =-\cot x+C\end{array}$

$\begin{array}{rl}\int e^{x}dx& =e^x+C\\ \int a^{x}dx& =\frac{1}{\ln a}\cdot a^x+C\\ \int \frac{1}{x}dx& =\ln |x|+C\end{array}$

reverse power rule

$\begin{array}{rl}\text{if}{f}^{\prime }(x)& =x^n\text{(for n != -1) then}\\ f(x)& =\frac{x^{n+1}}{n+1}+c\end{array}$

reverse trig

$\begin{array}{rl}\text{if }{f}^{\prime }(x)& =\sin (x)\text{ then }\\ f(x)& =-cos(x)+c\end{array}$

$\begin{array}{rl}\text{if }{f}^{\prime }(x)& =cos(x)\text{ then }\\ f(x)& =cos(x)+c\end{array}$

reverse exponential

$\begin{array}{r}\text{if}{f}^{\prime }(x)=e^x\text{then}\\ f(x)=e^x+c\end{array}$

reverse logarithm

this explains the n=-1 case.

$\begin{array}{rl}\text{if}{f}^{\prime }(x)& =\frac{1}{x}\text{then}\\ f(x)& =\ln (|x|)+c\end{array}$

derivative: $\begin{array}{r}\frac{d}{dx}[\ln (x)]=\frac{1}{x}\end{array}$

summation rule

the antiderivative of (f(x)+g(x)) is the antiderivative of f(x) + g(x)

linearity

the antiderivative of cf(x) is c times antiderivative of f(x)

secant

Derivative of tangent is sec^2. so anti-derivative of sec^2 = tangent.

example of finding an anti-derivative

From lecture 7.1

$\begin{array}{r}\int x^2\frac{1}{x^2}4{\text{sec}}^2(x)dx\end{array}$

linearity says it’s the sum of each one individually

\begin{equation} \int x^{2} ~~ \int\frac{1}{x^{2}} ~~ \int 4\sec^{2}(x) dx \end{equation}

reverse the power rule $\begin{array}{rl}& \int x^2\\ & =\frac{x^{2+1}}{2+1}\\ & =\frac{x^3}{3}\\ & =\frac{1}{3}{x}^3\end{array}$

power rule

$\begin{array}{rl}& \int \frac{1}{x^2}\text{can be rewitten as}\\ & \int x^{-2}\\ & =\frac{x^{-2+1}}{-2+1}\\ & =\frac{x^{-1}}{-1}\end{array}$

$\begin{array}{rl}& \int 4{\sec }^2(x)dx\\ 4\tan (x)& \end{array}$

We end up with

\begin{equation} \frac{x^{2}}{3} - \frac{1}{x} ~~ 4\tan(x) ~~ c \end{equation}

I don’t actually understand how we can simplify the 1/x^2 one

• DONE Completed: 2022-04-05

I’d guess that it would be x^-1 / -1. It simplifies to -1*x^-1 but I don’t get why. —

I had it right. $\frac{x^{-1}}{-1}=\frac{1}{-1}\cdot x^{-1}=-1\cdot x^{-1}$

how to find the +c?

f’(x) = 4x+5 f(1) = 4 what is f(x)?

$\begin{array}{rl}{f}^{\prime }(x)& =4x+5\\ f(x)& =\int 4x+5dx\\ & =\frac{4x^{1+1}}{1+1}+5(x^{0+1})\text{ via reverse power rule}\\ & =\frac{4x^2}{2}+5x^1\\ & =2x^{2}5xc\end{array}$

$\begin{array}{rl}f(1)& =4\\ 4& =2(1^2)5(1)c\\ 4& =25c\\ 4& =7+c\\ c& =-3\end{array}$

Exact value \begin{equation} f(x) = 2x^{2} ~~ 5x ~~ -3 \end{equation}

demo from lecture 7.2

\begin{equation} f’(x) = x^{2}+\sin(x) f(0) = 3 \end{equation}

find f(x) and evaluate f(2) to the nearest hundredth

$\begin{array}{rl}{f}^{\prime }(x)& =x^2+\sin (x)\\ f(x)& =\frac{x^{2+1}}{2+1}c-\cos (x)+\\ f(x)& =\frac{x^3}{3}-\cos (x)+c\\ f(0)& =3\\ 3& =\frac{0^3}{3}-\cos (0)+c\\ 3& =0-1c\\ c& =4\\ f(x)& =\frac{x^3}{3}-\cos (x)+4\end{array}$

Multiple anti-derivatives

First attempt

$\begin{array}{r}{f}^{\prime \prime }(x)=x+e^x\end{array}$

Find f(x)

$\begin{array}{rl}{f}^{\prime }(x)& =\int x+e^{x}dx\\ & =\frac{x^{1+1}}{1+1}{e}^x{c}_1\\ & =\frac{x^2}{2}{e}^x{c}_1\\ f(x)& =\int \frac{1}{2}{x}^2{e}^x{c}_1\\ & =\frac{\frac{1}{2}{x}^3}{3}{e}^x{c}_{1}x+c_2\\ & =\frac{1}{6}{x}^3{e}^x{c}_{1}x+c_2\end{array}$

I’m still really fuzzy on the 1/2x^3/3 → 1/6x^3 move. Would love more explaination there.

We can figure out the initial values via:

$\begin{array}{rl}-3& =f^{\prime }(x)\\ & =\frac{0^2}{2}{e}^0{c}_1\\ & =01c_1\\ c_1& =-4\end{array}$

$\begin{array}{rl}2& =f(0)\\ & =\frac{0^3}{6}{e}^0-4\cdot 0+c_2\\ & =010+c_2\\ c_2& =1\end{array}$

so..

$\begin{array}{r}f(x)=\frac{x^3}{6}{e}^x-4x+1\end{array}$

Another go

$\begin{array}{rl}{f}^{\prime \prime }(x)& =x+\sqrt{x}\\ f(1)& =5\\ f^{\prime }(1)& =2\\ f(4)& =?\end{array}$

—

$\begin{array}{rl}{f}^{\prime \prime }(x)& =\sqrt{x}+x\\ & =x^{\frac{1}{2}}+x^1\\ & =\frac{1}{2}\cdot x^{0.5+1}+2x^{1+1}+c_1\\ f^{\prime }(x)& =\frac{1}{2}\cdot x^{1.5}2x^2{c}_1\\ & =\frac{\frac{1}{2}\cdot x^{1.5+1}}{1.5+1}+\frac{2x^{2+1}}{2+1}+\frac{c_1\cdot x^{0+1}}{1}+c_2\\ f(x)& =\frac{1}{5.5}{x}^{2.5}\frac{2}{3}{x}^3{c}_{1}x+c_2\end{array}$

$\begin{array}{rl}2& =f^{\prime }(1)\\ & =\frac{1}{2}{1}^{1.5}2\cdot 1^2{c}_1\\ & =\frac{1}{2}2c_1\\ c_1& =-\frac{1}{2}\end{array}$

$\begin{array}{rl}5& =f(1)\\ & =\frac{1}{5.5}{1}^{2.5}\frac{2}{3}{1}^3\frac{1}{2}\cdot 1+c_2\\ & =\frac{1}{5.5}\frac{2}{3}\frac{1}{2}+c_2\\ & =1+\frac{11}{30}+c_2\\ c_2& =3+\frac{19}{30}\end{array}$

$\begin{array}{rl}f(x)& =\frac{1}{5.5}{x}^{2.5}\frac{2}{3}{x}^3-\frac{1}{2}\cdot x+3+\frac{19}{30}\\ f(4)& =\frac{1}{5.5}{4}^{2.5}\frac{2}{3}{4}^3-\frac{1}{2}\cdot 4+3+\frac{19}{30}\\ & =\frac{1}{5.5}32\frac{2}{3}64-23\frac{19}{30}\\ & =\frac{32}{5.5}\frac{128}{3}1+\frac{19}{30}\\ & =\frac{32\cdot 3\cdot 30}{5.5\cdot 3\cdot 30}\frac{128\cdot 5.5\cdot 30}{3\cdot 5.5\cdot 30}\frac{19\cdot 5.5\cdot 3}{30\cdot 3\cdot 5.5}\\ & =\frac{2880}{495}\frac{21120}{495}\frac{313.5}{495}\\ & =\frac{2880+21120+313.5}{495}\\ & =\frac{24313.5}{495}\text{(the GCF is 4.5 GCF)}\\ & =\frac{5403}{110}\end{array}$

This was my work, but it was wrong. I mis-factored .5x^0.5+1 and should have had 1.5x^1+0.5 instead.

Practice problems

Section 5.1 (pg 205-206), 9-27, 29-39

9

$\begin{array}{rl}{f}^{\prime }(x)& =3x^{3}dx\\ f(x)& =\frac{3x^{3+1}}{3+1}\\ f(x)& =\frac{3x^4}{4}\\ f(x)& =\frac{3}{4}{x}^4+c\end{array}$

11

$\begin{array}{rl}\int (10x^2-2)dx& \\ f^{\prime }(x)& =10x^2-2x^0\\ f(x)& =\frac{10x^{2+1}}{2+1}-\frac{2x^{0+1}}{0+1}\\ f(x)& =\frac{10}{3}{x}^3-2x\end{array}$

13

$\begin{array}{rl}\int 1ds{f}^{\prime }(x)& =1x^0\\ f(x)& =1\frac{x^{0+1}}{0+1}\\ & =x\end{array}$

15

$\begin{array}{rl}\int \frac{3}{t^2}dt& \\ f^{\prime }(x)& =\frac{3}{x^2}\\ f(x)& =\frac{3}{\frac{x^{2+1}}{2+1}}\\ & =\frac{3}{\frac{1}{3}{x}^3}\\ & =9x^3\end{array}$

17

$\begin{array}{rl}\int {\sec }^2\theta d\theta & \\ f^{\prime }(x)& ={\sec }^{2}x\\ f(x)& =\tan x\end{array}$

19

$\begin{array}{r}\int (\sec x\tan x+\csc x\cot x)dx\end{array}$

21

$\begin{array}{rl}\int 3^{t}dt& \\ f^{\prime }(x)& =3^x\\ f(x)& =3^x+c\text{// exponent rule}\end{array}$

23 submit?

$\begin{array}{rl}\int (2t+3{)}^{2}dt& \\ & =(2t+3)(2t+3)\\ & =4t^2+6t+6t+6\\ f(x)& =\frac{4}{3}{t}^3\frac{12}{2}{t}^{2}6x+c\\ & =\frac{4}{3}{t}^{3}6t^{2}6x+c\end{array}$

25

$\begin{array}{rl}\int x^2{x}^{3}dx& \\ f^{\prime }(x)& =x^2{x}^3\\ f(x)& =\frac{x^{2+1}}{2+1}\frac{x^{3+1}}{3+1}\\ & =\frac{1}{3}{x}^3\frac{1}{4}{x}^4\end{array}$

27 - I’m not entirely sure how to factor this. Is a = c?

• DONE Completed: 2022-03-30

$\begin{array}{rl}\int adx& \\ f^{\prime }(x)& =a\\ & =a^1\\ f(x)& =\frac{1}{\ln a}\cdot a^1+C\end{array}$

Find initial values

29

$\begin{array}{rl}{f}^{\prime }(x)& =\sin x\\ f(0)& =2\end{array}$

31 - submit?

$\begin{array}{rl}{f}^{\prime }(x)& =4x^3-3x^2\\ f(x)& =4\frac{x^{3+1}}{3+1}-3\frac{x^{2+1}}{2+1}\\ & =4\cdot \frac{1}{4}{x}^4-3\cdot \frac{1}{3}{x}^3\\ & =x^4-x^3+c\\ 9& =f(-1)\\ & =-1^4--1^3+c\\ & =-1-1+c\\ c& =7\\ f(x)& =x^4-x^3+7\end{array}$

33

$\begin{array}{rl}{f}^{\prime }(x)& =7^x\\ f(x)& =7^x+c\\ 1& =f(2)\\ & =7^2+c\\ & =49+c\\ c& =-48\end{array}$

35 submit

$\begin{array}{rl}{f}^{\prime \prime }(x)& =7x\\ f^{\prime }(x)& =7\frac{x^{1+1}}{1+1}+c_0\\ & =7=\cdot \frac{1}{2}{x}^2+c_0\\ & =3.5x^2+c_0{x}^0\\ f(x)& =3.5\frac{x^{2+1}}{2+1}+\frac{c_0{x}^{0+1}}{1}+c_1\\ & =3.5\cdot \frac{1}{3}{x}^3{c}_{0}x{c}_1\\ -1& =f^{\prime }(1)\\ & =3.5-1^2+c_0\\ & =-3.5+c_0\\ c_0& =2.5\\ 10& =f(1)\\ & =3.5\cdot \frac{1}{3}{x}^{3}2.5x{c}_1\\ & =\frac{7}{6}2.5c_1\\ & =\frac{7}{6}\frac{15}{6}{c}_1\\ & =\frac{22}{6}+c_1\\ c_1& =\frac{38}{6}\\ f(x)& =\frac{23}{6}{x}^{3}2.5x\frac{38}{6}\end{array}$

37 submit

$\begin{array}{rl}{f}^{\prime \prime }(x)& =\sin x\\ f^{\prime }(x)& =-\cos (x)+c_0\\ f(x)& =-\cos (x)c_{0}x{c}_1\\ 2& =f^{\prime }(\pi )\\ & =-\cos (\pi )+c_0\\ & =--1+c_0\\ c_0& =1\\ 4& =f(\pi )\\ & =-\cos (\pi )\pi c_1\\ & =1\pi c_1\\ c_1& =3-\pi \\ f(x)& =-\cos (x)x3-\pi \end{array}$

39

$\begin{array}{rl}{f}^{\prime \prime }(x)& =0\\ f^{\prime }(x)& =0x+c_0\\ f(x)& =0x^2{c}_{0}x{c}_1\\ 3& =f^{\prime }(1)\\ c_0& =3\\ 1& =f(1)\\ & =03c_1\\ c_1& =-3\\ f(x)& =3x-3\end{array}$

Techiques of antidifferentiation (ch 6 cite:&apex_calc)

integration by substitution

Okay, so if you have a function $f(x)$, you can get the antiderivative ($f^{\prime }(x)$ ) then take the integration of it. That’s the same as $f(x)+C$ the constant. So this section is about substitution.

Given $\int (20x+30)(x^2+3x-5{)}^9$. The second part is complicated, so make it equal to $u$.

$\begin{array}{rl}\int & (20x+30)(x^2+3x-5{)}^9\\ u& =x^2+3x-5\\ \int & (20x+30)u^9\\ du& =(2x+3)dx\end{array}$

In ^^, the $dx$ isn’t “just sitting there”. It gets multiplied by $2x+3$.

$\begin{array}{rl}\int (20x+30)(x^2+3x-5{)}^9& =\int 10(2x+3)(x^2+3x-5{)}^{9}dx\\ & =\int 10u^{9}du\\ & =u^{10}+C\\ \text{anti-derivative of it drops the du?}& \\ & =(x^2+3x-5{)}^{10}+C\end{array}$

Theorum let F and g be differentiable functions where the rnage of g is an intervial i contained in the domain of F. Then

$\begin{array}{r}\int F^{\prime }(g(x))g^{\prime }(x)dx=F(g(x))+C\end{array}$ if $u=g(x)$ then $du=g^{\prime }(x)dx$ and

$\begin{array}{r}\int F^{\prime }(g(x))g^{\prime }(x)dx=\int F^{\prime }(u)du=F(u)C=F(g(x))C\end{array}$

Integration of a linear function

Consider $\int F^{\prime }(ax+b)dx$ where $a\ne 0$ and $b$ are constants. Letting $u=ax+b$ gives $du=a\cdot dx$ leading to the result:

$\begin{array}{r}\int F^{\prime }(ax+b)dx=\frac{1}{a}F(ax+b)+C\end{array}$

Examples

$\int x\sin (x^2+5)dx$ $\begin{array}{rl}u& =x^2+5\\ du& =2xdx\\ \text{same as..}\frac{1}{2}du& =xdx\\ \int xsin(x^2+5)dx& =\int \sin (u)\frac{1}{2}du\\ & =\int \frac{1}{2}sin(u)du\\ & =-\frac{1}{2}cos(u)+C\\ & =-\frac{1}{2}cos(x^2+5)+C\end{array}$

$\int cos(5x)dx$ $\begin{array}{rl}u& =5x\\ du& =5dx\\ \int cos(5x)dx& =\int cos(u)\frac{1}{5}du\\ \int & \frac{1}{5}cos(u)du\\ & =\frac{1}{5}sin(u)+C\\ & =\frac{1}{5}sin(5x)+C\end{array}$

$\int \frac{7}{-3x+1}dx$

In this case, we think of it as the composition of two functions like $f(g(x))$ where $f(x)=7/x$ and $g(x)=-3+1$.

$\begin{array}{rl}u& =-3x+1\\ du& =-3dx\end{array}$

The original function doesn’t have an integrand, so divide $du/-3$ to make balance.

$\begin{array}{rl}\int \frac{7}{-3x+1}dx& =\int \frac{7}{u}\frac{du}{-3}\\ & =\frac{-7}{3}\int \frac{du}{u}\\ & =\frac{-7}{3}ln|u|+C\\ & =-\frac{7}{3}ln|-3x+1|+C\end{array}$

I don’t understand how we made the leap to step 2 above

• DONE Completed: 2022-04-27 They’re solving for $du$, which is a confusing (b/c we’re mixing u’s and x’s), but common way of doing it.

$\int sin(x)cos(x)dx$ $\begin{array}{rl}u& =sin(x)\\ du& =cos(x)dx\\ \int sin(x)cos(x)dx& =\int udu\\ & =\frac{1}{2}{u}^2+C\\ & =\frac{1}{2}si{n}^2(x)+C\end{array}$

$\int x\sqrt{x+3}dx$

$\begin{array}{rl}u& =x+3\\ du& =1dx\\ \int x\sqrt{u}du& =\int (u-3)\cdot u^{\frac{1}{2}}du\\ & =\int u^{\frac{3}{2}}-3u^{\frac{1}{2}}du\\ & =\frac{2}{5}{u}^{\frac{5}{2}}-2u^{\frac{3}{2}}+C\\ & =\frac{2}{5}(x+3{)}^{\frac{5}{2}}-2(x+3{)}^{\frac{3}{2}}+C\end{array}$

$\int \frac{1}{x\ln x}dx$ $\begin{array}{rl}u& =\ln x\\ du& =\frac{1}{x}dx\\ \int \frac{1}{x\ln x}& =\frac{1}{x}\cdot \frac{1}{\ln x}\\ & =\int du\cdot \frac{1}{u}\\ & =\ln |u|+C\\ & =\ln |\ln x|+C\end{array}$

Integral subsitution for things w/ trigonometric functions. (eew) Stoped here on pg 269.

Homework

3 $\begin{array}{rl}& \int 3x^2(x^3-5{)}^{7}dx\\ u& =x^3-5\\ du& =3x^{2}dx\\ \int 3x^2(x^3-5{)}^{7}dx& =\int (u{)}^7\cdot du\\ & =\frac{1}{8}{u}^8+C\\ & =\frac{1}{8}(x^3-5{)}^8+C\end{array}$

6 $\begin{array}{rl}& \int (12x+14)(3x^2+7x-1{)}^5\\ u& =3x^2+7x-1\\ du& =6x+7\\ & =\int 2du(u{)}^5\\ & =\frac{1}{3}{u}^6+C\\ & =\frac{1}{3}(3x^2+7x-1{)}^6+C\end{array}$

9 $\begin{array}{rl}& \int \frac{x}{\sqrt{x+3}}dx\\ u& =x+3\\ du& =dx\\ & =\int (u-3)\cdot \frac{1}{\sqrt{u}}du\\ & =\int u^{.5}-3u^{-.5}du\\ & =\frac{2}{3}{u}^{\frac{3}{2}}-6u^{\frac{1}{2}}\\ & =\frac{2}{3}(x+3{)}^{\frac{3}{2}}-6(x+3{)}^{\frac{1}{2}}\end{array}$

12 $\begin{array}{rl}& \int \frac{x^4}{\sqrt{x^5+1}}dx\\ u& =x^5+1\\ du& =5x^{4}dx\\ & =\int \frac{\frac{du}{5}}{\sqrt{u}}\\ & =\int \frac{du}{5}\cdot u^{0.5}\\ & =\int \frac{1}{5}du\cdot u^{0.5}\\ & =2u^{\frac{3}{2}}+C\\ & =2(x^5+1{)}^{\frac{3}{2}}+C\end{array}$