Math 22 Antiderivatives and Indefinite Integrals

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, $f(x)=x^{3}+4x-1$

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