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

$$\int x^2+\int \frac{1}{x^2}+\int 4{\sec }^2(x)dx$$

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

$$\frac{x^2}{3}-\frac{1}{x}+4\tan (x)+c$$

how to find the +c?

f’(x) = 4x+5 f(1) = 4 what is f(x)?

$$\begin{array}{rl}{f}^{\prime }(x)& =4x+5\\ & \\ f(x)& =\int 4x+5dx\\ & \\ & =\frac{4x^{1+1}}{1+1}+5(x^{0+1})\text{ via reverse power rule}\\ & \\ & =\frac{4x^2}{2}+5x^1\\ & \\ & =2x^2+5x+c\\ & \end{array}$$ $$\begin{array}{rl}f(1)& =4\\ & \\ 4& =2(1^2)+5(1)+c\\ & \\ 4& =2+5+c\\ & \\ 4& =7+c\\ & \\ c& =-3\end{array}$$

Exact value

$$f(x)=2x^2+5x+-3$$

demo from lecture 7.2

$$f^{\prime }(x)=x^2+\sin (x)f(0)=3$$

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

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

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 1/2*x^3/3 -> 1/6*x^3 move. Would love more explaination there.

We can figure out the initial values via:

$$\begin{array}{rl}-3& =f^{\prime }(x)\\ & \\ & =\frac{0^2}{2}+e^0+c_1\\ & \\ & =0+1+c_1\\ & \\ c_1& =-4\end{array}$$ $$\begin{array}{rl}2& =f(0)\\ & \\ & =\frac{0^3}{6}+e^0+-4\cdot 0+c_2\\ & \\ & =0+1+0+c_2\\ & \\ c_2& =1\end{array}$$

so..

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

Another go

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

—

$$\begin{array}{rl}{f}^{\prime \prime }(x)& =\sqrt{x}+x\\ & \\ & =x^{\frac{1}{2}}+x^1\\ & \\ & =\frac{1}{2}\cdot x^{0.5+1}+2x^{1+1}+c_1\\ & \\ f^{\prime }(x)& =\frac{1}{2}\cdot x^{1.5}+2x^2+c_1\\ & \\ & =\frac{\frac{1}{2}\cdot x^{1.5+1}}{1.5+1}+\frac{2x^{2+1}}{2+1}+\frac{c_1\cdot x^{0+1}}{1}+c_2\\ & \\ f(x)& =\frac{1}{5.5}{x}^{2.5}+\frac{2}{3}{x}^3+c_{1}x+c_2\end{array}$$ $$\begin{array}{rl}2& =f^{\prime }(1)\\ & \\ & =\frac{1}{2}{1}^{1.5}+2\cdot 1^2+c_1\\ & \\ & =\frac{1}{2}+2+c_1\\ & \\ c_1& =-\frac{1}{2}\end{array}$$ $$\begin{array}{rl}5& =f(1)\\ & \\ & =\frac{1}{5.5}{1}^{2.5}+\frac{2}{3}{1}^3+\frac{1}{2}\cdot 1+c_2\\ & \\ & =\frac{1}{5.5}+\frac{2}{3}+\frac{1}{2}+c_2\\ & \\ & =1+\frac{11}{30}+c_2\\ & \\ c_2& =3+\frac{19}{30}\end{array}$$ $$\begin{array}{rl}f(x)& =\frac{1}{5.5}{x}^{2.5}+\frac{2}{3}{x}^3+-\frac{1}{2}\cdot x+3+\frac{19}{30}\\ & \\ f(4)& =\frac{1}{5.5}{4}^{2.5}+\frac{2}{3}{4}^3+-\frac{1}{2}\cdot 4+3+\frac{19}{30}\\ & \\ & =\frac{1}{5.5}32+\frac{2}{3}64+-2+3+\frac{19}{30}\\ & \\ & =\frac{32}{5.5}+\frac{128}{3}+1+\frac{19}{30}\\ & \\ & =\frac{32\cdot 3\cdot 30}{5.5\cdot 3\cdot 30}+\frac{128\cdot 5.5\cdot 30}{3\cdot 5.5\cdot 30}+\frac{19\cdot 5.5\cdot 3}{30\cdot 3\cdot 5.5}\\ & \\ & =\frac{2880}{495}+\frac{21120}{495}+\frac{313.5}{495}\\ & \\ & =\frac{2880+21120+313.5}{495}\\ & \\ & =\frac{24313.5}{495}\text{(the GCF is 4.5 GCF)}\\ & \\ & =\frac{5403}{110}\end{array}$$

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

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

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

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

37 submit

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

39

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

Techiques of antidifferentiation (ch 6 )

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.

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

• $\int sin(x)cos(x)dx$

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

• $\int x\sqrt{x+3}dx$

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

• $\int \frac{1}{x\ln x}dx$

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

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

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