##### Integration rules

We already know that finding an integral is the reverse of finding a Derivative.

So firstly you should learn derivates

We talked about two types of integral, but a more scientific definition is:

### Indefinite Integral

$\displaystyle \int{{f(x)dx=F(x)+C}}$ where $\displaystyle {F(x)}$ is an antiderivative of $\displaystyle {f(x)}$.

What is an antiderivative?

An antiderivative of $\displaystyle {f(x)}$ is a function $\displaystyle {F(x)}$, such that $\displaystyle {F}'(x)=f(x)$.

### Definite Integral

Suppose $\displaystyle {f(x)}$ is continuous on $\displaystyle \left[ {a,b} \right]$,divide $\displaystyle \left[ {a,b} \right]$ into n subintervals of width $\displaystyle \Delta x$ and choose $\displaystyle {{x}_{i}}$ from each interval. Then $\displaystyle \int\limits_{a}^{b}{{f(x)dx}}=\underset{{x\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{\infty }{{f({{x}_{i}}}})\Delta x$

### Theorem

If $\displaystyle f(x)$ is continuous on $\displaystyle \left[ {a,b} \right]$ and  $\displaystyle F(x)$  is an antiderivative of $\displaystyle \left[ {a,b} \right]$ then $\displaystyle \int\limits_{a}^{b}{{f(x)dx=}}F(b)-F(a)$

### The Power Rule

The power rule of integration is the inverse of derivates used in differentiation and it gives us the indefinite integral of a variable raised to some power.

$\displaystyle \int{{{{x}^{n}}}}dx=\frac{{{{x}^{{n+1}}}}}{{n+1}}+C$

Example 1: Find the integral of the functions

a) $\displaystyle f(x)={{x}^{5}}$

Applying the power rule of integration we get:

$\displaystyle \int{{{{x}^{5}}}}dx=\frac{{{{x}^{{5+1}}}}}{{5+1}}+C$

$\displaystyle \int{{{{x}^{5}}}}dx=\frac{{{{x}^{6}}}}{6}+C$

b) $\displaystyle f(x)=\sqrt[3]{{{{x}^{2}}}}$

Firstly change the expression into a power term:

$\displaystyle \sqrt[3]{{{{x}^{2}}}}={{x}^{{\frac{2}{3}}}}$

Then we apply the power rule of integration we get:

$\displaystyle \int{{{{x}^{{\frac{2}{3}}}}=\frac{{{{x}^{{1+\frac{2}{3}}}}}}{{1+\frac{2}{3}}}}}+C$

$\displaystyle \int{{{{x}^{{\frac{2}{3}}}}=\frac{{{{x}^{{\frac{5}{3}}}}}}{{\frac{5}{3}}}}}+C$

$\displaystyle \int{{{{x}^{{\frac{2}{3}}}}=\frac{3}{5}\sqrt[3]{{{{x}^{5}}}}}}+C$

### The Constant Coefficient Rule

The constant coefficient rule tells us that the indefinite integral of $\displaystyle c\cdot f(x)$ where $\displaystyle f(x)$ is a function and $\displaystyle c$ is a constant coefficient is equal to the integral of $\displaystyle f(x)$ multiplied by $\displaystyle c$.

$\displaystyle \int{c}\cdot f(x)dx=c\int{{f(x)}}dx$

Example 2: Find the integral of the functions

a) $\displaystyle f(x)=7{{x}^{3}}$

Applying the constant coefficient rule of integration we get:

$\displaystyle \int{7}{{x}^{3}}dx=7\int{{{{x}^{3}}}}dx$

Applying the power rule of integration we get:

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

$\displaystyle 7\int{{{{x}^{3}}}}dx=\frac{7}{4}{{x}^{4}}+C$

b) $\displaystyle f(x)=2\sqrt{x}$

Applying the constant coefficient rule of integration we get:

$\displaystyle \int{{2\sqrt{x}}}dx=2\int{{\sqrt{x}}}dx$

Firstly we need to write the function as a power term and then apply the rule:

$\displaystyle 2\int{{\sqrt{x}}}dx=2\int{{{{x}^{{\frac{1}{2}}}}}}dx$

Applying the power rule of integration we get:

$\displaystyle 2\int{{{{x}^{{\frac{1}{2}}}}}}dx=2\frac{{{{x}^{{\frac{1}{2}}}}}}{{\frac{1}{2}}}+C$

$\displaystyle 2\int{{{{x}^{{\frac{1}{2}}}}}}dx=4\sqrt{x}+C$

### The Sum Rule

The sum rule tells us how we should integrate functions that are the sum of several terms. Just like in the sum rule of derivation we should integrate each term of the sum separately and then add the result.

$\displaystyle \int{{(f(x)+g(x))dx}}=$$\displaystyle \int{{f(x)dx+\int{{g(x)dx}}}} Example 3: Find the integral of the functions a) \displaystyle f(x)={{x}^{4}}+6{{x}^{2}} Using the sum rule of integration we get: \displaystyle {\int{{({{x}^{4}}+6{{x}^{2}})dx}}}$$\displaystyle =\int{{{{x}^{4}}dx}}+\int{{6{{x}^{2}}dx}}$

Using the power rule and costant coefficent rule of integration at each term we get: $\displaystyle \int{{{{x}^{4}}}}dx+\int{{6{{x}^{2}}dx}}$$\displaystyle =\frac{{{{x}^{{4+1}}}}}{{4+1}}+6\frac{{{{x}^{{3+1}}}}}{{3+1}}+C \displaystyle \int{{{{x}^{4}}}}dx+\int{{6{{x}^{2}}dx}}$$\displaystyle =\frac{{{{x}^{5}}}}{5}+\frac{{6{{x}^{4}}}}{4}+C$

b) $\displaystyle f(x)=2{{x}^{2}}+5x+8$

Using the sum rule of integration we get:

$\displaystyle {\int{{(2{{x}^{2}}+5x+8)dx}}}$$\displaystyle =\int{{2{{x}^{2}}dx+\int{{5xdx+\int{{8dx}}}}}} Using the power rule and costant coefficent rule of integration at each term we get: \displaystyle \int{{2{{x}^{2}}dx+\int{{5xdx+\int{{8dx}}}}}}$$\displaystyle =\frac{{2{{x}^{3}}}}{3}+\frac{{5{{x}^{2}}}}{2}+8x+C$

Note! $\displaystyle \int{{8dx=\int{{8\cdot {{x}^{0}}}}}}dx$$\displaystyle =8\int{{{{x}^{0}}}}dx$$\displaystyle =8\frac{{{{x}^{{0+1}}}}}{{0+1}}=8x$ or based on the table of integration of basic functions we will learn that the integral of a constant function is always the constant multiplied by x.

### The difference Rule

The difference rule tells us how we should integrate functions that are the difference of several terms. Just like in the difference rule of derivation we should integrate each term of the difference separately and then subtract the result.

$\displaystyle \int{{(f(x)-g(x))dx=}}$$\displaystyle \int{{f(x)dx-\int{{g(x)dx}}}} Example 4: Find the integral of the functions a) \displaystyle f(x)=3{{x}^{5}}-2x-3 Using the difference rule of integration we get: \displaystyle {\int{{(3{{x}^{5}}-2x-3)dx}}}$$\displaystyle =\int{{3{{x}^{5}}dx-}}\int{{2x}}dx-\int{{3dx}}$

Using the power rule and costant coefficent rule of integration at each term we get:

$\displaystyle \int{{3{{x}^{5}}dx-}}\int{{2xdx-\int{{3dx}}}}$$\displaystyle =3\int{{{{x}^{5}}dx-2\int{{xdx-3\int{{{{x}^{0}}}}}}dx}} \displaystyle 3\int{{{{x}^{5}}}}dx-2\int{{xdx-3\int{{{{x}^{0}}}}}}dx$$\displaystyle =\frac{{3{{x}^{6}}}}{6}-{{x}^{2}}-3x+C$

b) $\displaystyle f(x)=6{{x}^{2}}+x-9$

Using the sum and difference rule of integration we get:

$\displaystyle \int (6{{x}^{2}}+x-9)dx$$\displaystyle =\int{{6{{x}^{2}}dx+\int{x}}}dx-\int{{9dx}} Using the power rule and costant coefficent rule of integration at each term we get: \displaystyle \int{{6{{x}^{2}}dx+\int{x}}}dx-\int{{9dx}}$$\displaystyle =\frac{{6{{x}^{3}}}}{3}+\frac{{{{x}^{2}}}}{2}-9x+C$

$\displaystyle \int{{6{{x}^{2}}dx+\int{x}}}dx-\int{{9dx}}$$\displaystyle =2{{x}^{3}}+\frac{{{{x}^{2}}}}{2}-9x+C Note! All this rules are applied exactly the same in definited integration Example 5: Find the integral of the function a) \displaystyle \int\limits_{2}^{3}{{{{x}^{3}}dx}} Applying the power rule of integration and evaluating based on the values: \displaystyle \int\limits_{2}^{3}{{{{x}^{3}}dx}}=\frac{{{{x}^{4}}}}{4}$$\displaystyle =\frac{{{{{(3)}}^{4}}}}{4}-\frac{{{{{(2)}}^{4}}}}{4}=\frac{{65}}{4}$

b) $\displaystyle \int\limits_{{-1}}^{1}{{(2{{x}^{2}}+1)dx}}$

$\displaystyle \int\limits_{{-1}}^{1}{{(2{{x}^{2}}+1)dx=\int\limits_{{-1}}^{1}{{2{{x}^{2}}dx+\int\limits_{{-1}}^{1}{{dx}}}}}}$

$\displaystyle \int\limits_{{-1}}^{1}{{2{{x}^{2}}}}dx+\int\limits_{{-1}}^{1}{{dx}}$$\displaystyle =\frac{{2{{x}^{3}}}}{3}+x$$\displaystyle =(\frac{2}{3}+1)-(-\frac{2}{3}-1)=\frac{{10}}{3}$

The table of integration of basic and more complicated functions: