Math 22 Antiderivatives and Indefinite Integrals

Antiderivatives

 A function ${\displaystyle F}$ is an antiderivative of a function ${\displaystyle f}$ when for every ${\displaystyle x}$ in the domain of ${\displaystyle f}$,
it follows that ${\displaystyle F'(x)=f(x)}$

 The antidifferentiation process is also called integration and is denoted by ${\displaystyle \int }$ (integral sign).
${\displaystyle \int f(x)dx}$ is the indefinite integral of ${\displaystyle f(x)}$

 If ${\displaystyle F'(x)=f(x)}$ for all ${\displaystyle x}$, we can write:
${\displaystyle \int f(x)dx=F(x)+C}$ for ${\displaystyle C}$ is a constant.


Basic Integration Rules

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

${\displaystyle 2.\int kf(x)=k\int f(x)dx}$

${\displaystyle 3.\int [f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx}$

${\displaystyle 4.\int [f(x)-g(x)]dx=\int f(x)dx-\int g(x)dx}$

${\displaystyle 5.\int x^{n}dx={\frac {x^{n+1}}{n+1}}+C}$ for ${\displaystyle n\neq -1}$

Exercises 1 Find the indefinite integral

1) ${\displaystyle \int 7dr}$

Solution:
${\displaystyle \int 7dr=7r+C}$

2) ${\displaystyle \int -4dx}$

Solution:
${\displaystyle \int -4dx=-4x+C}$

3) ${\displaystyle \int 7x^{2}dx}$

Solution:
${\displaystyle \int 7x^{2}dx=7\int x^{2}dx=7{\frac {x^{2+1}}{2+1}}+C={\frac {7}{3}}x^{3}+C}$

4) ${\displaystyle \int 5x^{-3}dx}$

Solution:
${\displaystyle \int 5x^{-3}dx=5\int x^{-3}dx=5{\frac {x^{-3+1}}{-3+1}}+C={\frac {-5}{2}}x^{-2}+C}$

Exercises 2 Solve the initial value problems, given:

5) ${\displaystyle f'(x)={\frac {1}{5}}x-2}$ and ${\displaystyle f(10)=-10}$

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

6) ${\displaystyle f'(x)=3x^{2}+4}$ and ${\displaystyle f(-1)=-6}$

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