# Math 22 Antiderivatives and Indefinite Integrals

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## Antiderivatives

 A function $F$ is an antiderivative of a function $f$ when for every $x$ in the domain of $f$ ,
it follows that $F'(x)=f(x)$ The antidifferentiation process is also called integration and is denoted by $\int$ (integral sign).
$\int f(x)dx$ is the indefinite integral of $f(x)$ If $F'(x)=f(x)$ for all $x$ , we can write:
$\int f(x)dx=F(x)+C$ for $C$ is a constant.


## Basic Integration Rules

$1.\int kdx=kx+C$ for $k$ is a constant.

$2.\int kf(x)=k\int f(x)dx$ $3.\int [f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$ $4.\int [f(x)-g(x)]dx=\int f(x)dx-\int g(x)dx$ $5.\int x^{n}dx={\frac {x^{n+1}}{n+1}}+C$ for $n\neq -1$ Exercises 1 Find the indefinite integral

1) $\int 7dr$ Solution:
$\int 7dr=7r+C$ 2) $\int -4dx$ Solution:
$\int -4dx=-4x+C$ 3) $\int 7x^{2}dx$ Solution:
$\int 7x^{2}dx=7\int x^{2}dx=7{\frac {x^{2+1}}{2+1}}+C={\frac {7}{3}}x^{3}+C$ 4) $\int 5x^{-3}dx$ Solution:
$\int 5x^{-3}dx=5\int x^{-3}dx=5{\frac {x^{-3+1}}{-3+1}}+C={\frac {-5}{2}}x^{-2}+C$ Exercises 2 Solve the initial value problems, given:

5) $f'(x)={\frac {1}{5}}x-2$ and $f(10)=-10$ Solution:
Notice $f(x)=\int f'(x)dx=\int ({\frac {1}{5}}x-2)dx={\frac {1}{5}}{\frac {x^{2}}{2}}-2x+C={\frac {1}{10}}x^{2}-2x+C$ So, $f(x)={\frac {1}{10}}x^{2}-2x+C$ we are given $f(10)=-10$ , so ${\frac {1}{10}}(10)^{2}-2(10)+C=10$ Hence, $C=20$ Therefore, $f(x)={\frac {1}{10}}x^{2}-2x+20$ 6) $f'(x)=3x^{2}+4$ and $f(-1)=-6$ Solution:
Notice $f(x)=\int f'(x)dx=\int (3x^{2}+4)dx=x^{3}+4x+C$ So, $f(x)=x^{3}+4x+C$ we are given $f(-1)=-6$ , so $(-1)^{3}+4(-1)+C=-6$ Hence, $C=-1$ Therefore, $f(x)=x^{3}+4x-1$ 