Integration

The opposite of differentiation

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Whereas differentiation is used to find the rate of change of one variable with respect to another, integration does the opposite. When we integrate, we find the area under the graph. That is, we find the product of the function on the y-axis and infinitely thin sections of the x-axis variable.

The anti-derivative

This is the opposite of the derivative of a function. It is not the same as the integral as it does not include the constant of integration. The constant of integration represents the information lost when differentiation occurs.

The integral of $f(x)$, with respect to $x$ is the sum of the anti-derivative, $F(x)$ and the constant of integration, $c$:

$$ \begin{equation}\begin{aligned} \int f(x) dx &= F(x) + c\\ \end{aligned}\end{equation} $$

Power rule

For the power rule, we add $1$ to the power and divide by the new power: $$ \begin{equation}\begin{aligned} \int (x^n) dx&=\frac{x^{n+1}}{n+1}+c\\ \end{aligned}\end{equation} $$

Example

Determine the integral of the function $f(x)=x^2$.

Solution We add $1$ to the power (in this case, $2$) and divide by the new power ($3$): $$ \begin{equation}\begin{aligned} \int x^2 dx&=\frac{x^{2+1}}{2+1}+c\\ &=\frac{x^3}{3}+c\\ \end{aligned}\end{equation} $$

The integral of a scalar multiple

The integral of a scalar multiple of a function is the scalar multiple of the integral of the function: $$ \begin{equation}\begin{aligned} \int kf(x) dx&=k\int f(x) dx\\ \end{aligned}\end{equation} $$

The sums and differences of functions

As with derivatives, the integral of the sum/difference of multiple functions is the sum/difference of the integrals of the individual functions: $$ \begin{equation}\begin{aligned} \int [f(x) + g(x)]dx&=\int f(x) dx+\int g(x) dx\\ \int [f(x) - g(x)]dx&=\int f(x) dx-\int g(x) dx\\ \end{aligned}\end{equation} $$

Reverse chain rule

This only works for $(ax+b)^n$ where $n\neq -1$: $$ \begin{equation}\begin{aligned} \int (ax+b)^n dx&=\frac{(ax+b)^{n+1}}{(n+1)\cdot a}+c\\ \end{aligned}\end{equation} $$

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