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The indefinite and definite are the two types of integrals that make up Integral Calculus. The indefinite integral takes up the form:
\begin{equation*}
{\int {f(x)\normalsubformula{\text{dx}}}}
\end{equation*}
The indefinite integral does not have any upper or lower limits, and can be referred to in a more familiar fashion, the antiderivitive. Since the derivative of a constant is a zero, when taking the antiderivitive of a function, a number known as the constant of integration must be added. For this reason, indefinite integrals are written in the form:
\begin{equation*}
{\int {f(x)\normalsubformula{\text{dx}}}=F(x)+C}
\end{equation*}
The following is a quick reference list of common indefinite integrals (antiderivatives) that in most cases can be referred to when solving a problem:
\begin{equation*}
{\int {x^{{n}}\normalsubformula{\text{dx}}}=\frac{x^{{n+1}}}{n+1}+C}
\end{equation*}
\begin{equation*}
{\int {\frac{1}{x}\normalsubformula{\text{dx}}}=\text{ln}x+C}
\end{equation*}
\begin{equation*}
{\int {a^{{x}}\normalsubformula{\text{dx}}}=\frac{a^{{x}}}{\text{ln}x}+C}
\end{equation*}
\begin{equation*}
{\int {\text{sin}\normalsubformula{\text{xdx}}}=-\text{cos}x+C}
\end{equation*}
\begin{equation*}
{\int {\text{cos}\normalsubformula{\text{xdx}}}=\text{sin}x+C}
\end{equation*}
\begin{equation*}
{\int {\text{tan}\normalsubformula{\text{xdx}}}=\text{ln}(\text{sec}x)+C}
\end{equation*}
\begin{equation*}
{\int {\text{csc}\normalsubformula{\text{xdx}}}=\frac{1}{2}\text{ln}(\frac{1-\text{cos}x}{1+\text{cos}x})+C}
\end{equation*}
\begin{equation*}
{\int {\text{sec}\normalsubformula{\text{xdx}}}=\text{ln}(\text{sec}x+\text{tan}x)+C}
\end{equation*}
\begin{equation*}
{\int {\text{cot}\normalsubformula{\text{xdx}}}=\text{ln}(\text{sin}x)+C}
\end{equation*}
\begin{equation*}
{\int {\text{sin}^{{2}}\normalsubformula{\text{xdx}}}=\frac{1}{2}x-\frac{1}{4}\text{sin}(2x)+C}
\end{equation*}
\begin{equation*}
{\int {\text{cos}^{{2}}\normalsubformula{\text{xdx}}}=\frac{1}{2}x+\frac{1}{4}\text{sin}(2x)+C}
\end{equation*}
\begin{equation*}
{\int {\text{tan}^{{2}}\normalsubformula{\text{xdx}}}=\text{tan}x-x+C}
\end{equation*}
\begin{equation*}
{\int {\text{sec}^{{2}}\normalsubformula{\text{xdx}}}=\text{tan}x+C}
\end{equation*}
\begin{equation*}
{\int {\text{csc}^{{2}}\normalsubformula{\text{xdx}}}=-\text{cot}x+C}
\end{equation*}
\begin{equation*}
{\int {\text{cot}^{{2}}\normalsubformula{\text{xdx}}}=-\text{cot}x-x+C}
\end{equation*}
\begin{equation*}
{\int {\text{sec}x\text{tan}\normalsubformula{\text{xdx}}}=\text{sec}x+C}
\end{equation*}
Example
Solve:
\begin{equation*}
{\int {(5x^{{2}}+\sqrt{x}-\frac{4}{x^{{3}}})}\normalsubformula{\text{dx}}}
\end{equation*}
First things first, put the equation in a form that is better recognized:
\begin{equation*}
{\int {(5x^{{2}}+\sqrt{x}-\frac{4}{x^{{3}}})}\normalsubformula{\text{dx}}=\int
{(5x^{{2}}+x^{{\frac{1}{2}}}-4x^{{-3}})\normalsubformula{\text{dx}}}}
\end{equation*}
Taking the antidirivative of each term we get:
\begin{equation*}
{\frac{5}{3}x^{{3}}+\frac{2}{3}x^{{\frac{3}{2}}}-\frac{4}{-2}x^{{-2}}+C}
\end{equation*}
Simplified:
\begin{equation*}
{\frac{5}{3}x^{{3}}+\frac{2}{3}x^{{\frac{3}{2}}}+2x^{{-2}}+C}
\end{equation*}
Placing upper and lower limits makes the integral definite, such that:
\begin{equation*}
{\int _{{a}}^{{b}}{f(x)\normalsubformula{\text{dx}}}}
\end{equation*}
With the upper and lower limits established, the definite integral can be solved so that:
\begin{equation*}
{\int _{{a}}^{{b}}{f(x)\normalsubformula{\text{dx}}}=F(b)-F(a)}
\end{equation*}
Knowing the following rules will help when solving definite integrals:
\begin{equation*}
{\int _{{a}}^{{b}}{f(x)\normalsubformula{\text{dx}}}=0}
\end{equation*}
And
\begin{equation*}
{\int _{{a}}^{{b}}{f(x)\normalsubformula{\text{dx}}}=-\int _{{b}}^{{a}}{f(x)\normalsubformula{\text{dx}}}}
\end{equation*}
And if ${c\in (a,b)}$ then:
\begin{equation*}
{\int _{{a}}^{{b}}{f(x)\normalsubformula{\text{dx}}}=\int _{{a}}^{{c}}{f(x)\normalsubformula{\text{dx}}}+\int
_{{c}}^{{b}}{f(x)\normalsubformula{\text{dx}}}}
\end{equation*}
Example
Evaluate
\begin{equation*}
{\overset{{5}}{\underset{{1}}{\int }}{x^{{2}}+\sqrt{x}\normalsubformula{\text{dx}}}}
\end{equation*}
The antidirivative is:
\begin{equation*}
{\frac{1}{3}x^{{3}}+\frac{2}{3}x^{{\frac{3}{2}}}}
\end{equation*}
Evaluating at the upper and lower limits:
\begin{equation*}
{\frac{1}{3}x^{{3}}+\frac{2}{3}x^{{\frac{3}{2}}}|_{{1}}^{{5}}}
\end{equation*}
\begin{equation*}
{\left(\frac{1}{3}(5)^{{3}}+\frac{2}{3}(5)^{{\frac{3}{2}}}\right)-\left(\frac{1}{3}(1)^{{3}}+\frac{2}{3}(1)^{{\frac{3}{2}}}\right)}
\end{equation*}
\begin{equation*}
{\left(\text{49}\text{.}\text{12}\right)-\left(1\right)=\text{48}\text{.}\text{12}}
\end{equation*}
Integration by Parts
When an integral of a product of functions is encountered, a process known as integration of parts can be employed to simplify the problem in to one that is more familiar. The integration of parts formula is:
\begin{equation*}
{\int {f(x){g}^{‘}(x)\normalsubformula{\text{dx}}=f(x)g(x)-}\int {{f}^{‘}(x)g(x)\normalsubformula{\text{dx}}}}
\end{equation*}
Substituting the following into the equation:
${u=f(x)}$
${v=g(x)}$
${\normalsubformula{\text{du}}={f}^{‘}(x)\normalsubformula{\text{dx}}}$
${\normalsubformula{\text{dv}}={g}^{‘}(x)\normalsubformula{\text{dx}}}$
We get the more familiar integration by parts formula:
\begin{equation*}
{\int {\normalsubformula{\text{udv}}=\normalsubformula{\text{uv}}-}\int {\normalsubformula{\text{vdu}}}}
\end{equation*}
We can turn the equations above in to solvable definite integrals by adding upper and lower limits to the problem.
Integration and Area
Given a function ${f(x)}$ bound by the x-axis and two parameters, a and b, we are able to find the area under the curve using the integral such that:
\begin{equation*}
{A=\int _{{a}}^{{b}}{f(x)\normalsubformula{\text{dx}}}}
\end{equation*}
Length of a Curve
Given the same parameters as above, ${f(x)}$ and the upper and lower limits of a and b, we are able to compute the length of a curve using the integral such that:
\begin{equation*}
{s=\int
_{{a}}^{{b}}{\sqrt{1+(\frac{\normalsubformula{\text{dy}}}{\normalsubformula{\text{dx}}})^{{2}}}\normalsubformula{\text{dx}}}}
\end{equation*}
Solids of Revolution
When we have a function ${f(x)}$ within the upper and lower limits of a and b, we can rotate it about a given axis to get the surface of the solid of revolution such that the volume is:
\begin{equation*}
{V=\int _{{a}}^{{b}}{A(x)\normalsubformula{\text{dx}}}}
\end{equation*}
Where ${A(x)}$ is the cross sectional area of the solid of revolution. To get ${A(x)}$, we can cut the object perpendicular with the axis of rotation to get either a solid disk or a ring. If the solid of rotation is solid, then the disk area is equal to:
\begin{equation*}
{A=\pi (r)^{{2}}}
\end{equation*}
If the solid of rotation is hollowed out, then the ring area is equal to:
\begin{equation*}
{A=\pi \left((r_{{o}})^{{2}}-(r_{{i}})^{{2}}\right)}
\end{equation*}
Where ${r_{{o}}}$ and ${r_{{i}}}$ is the outer and inner radius. In both instances, the radii will depend on the functions given and the axis of rotation.
Integration by Partial Fractions
It is not easy to evaluate an integral in the form of an algebraic fraction. However, if we express the fraction in partial fractions prior to assessing, we dramatically simplify the problem for solving.
As a recap, given a function in the form ${f(x)=\frac{a(x)}{b(x)}}$, where both ${a(x)}$ and ${b(x)}$ are polynomials and the degree of ${b(x)}$ is greater than that of ${a(x)}$, ${f(x)}$ can be broken down in to partial fractions. For each factor of ${b(x)}$, the following table can be used to determine the new terms from decomposition:
