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Also known as Simpson’s \(\textstyle\frac{1}{3}\) Rule is a numerical integration technique that improves upon the Trapezoidal Rule by utilizing the geometry of parabolic arcs. The number of partitions \(n\) must be even.

\(\displaystyle S_n=\frac{b-a}{3n}\big( f(x_0)+4f(x_1)+2f(x_2)+\cdots+4f(x_{n-1})+2f(x_n) \big)\)

\(\displaystyle \Delta x=\frac{b-a}{n}\quad \displaystyle x_k=a+k\Delta x\)

Virtual Lessons

Need some additional help understanding how to apply this integration technique? Click Here to visit the virtual lesson section.

Practice Problems

Approximate the given interval using Simpson Rule using the indicated partitions \(n\). 

  1. \(\displaystyle \int (2x+1)^2dx\)
  2. \(\displaystyle \int x\sqrt{2x+3}dx\)
  3. \(\displaystyle \int \frac{x}{2x+1}dx\)
  4. \(\displaystyle \int \frac{x^3}{x^2-4}dx \)
  5. \(\displaystyle \int \texttt{e}^{x^2+1} dx \)
  1. \(\displaystyle \int {\frac{\texttt{e}^{2x}}{1+\texttt{e}^{2x}}} dx \)
  2. \(\displaystyle \int_1^\texttt{e} {\frac{\texttt{ln}(x)}{x} dx} \)
  3. \(\displaystyle \int {\frac{\texttt{sin}(x)}{1+\texttt{cos}(x)}} dx \)
  4. \(\displaystyle \int {\frac{\texttt{e}^x}{1+\texttt{e}^{2x}}} dx \)
  5. \(\displaystyle \int {\frac{1}{\texttt{e}^x + \texttt{e}^{-x}}} dx \)
  1. \(\displaystyle \int {\frac{1}{\sqrt{x}(x+1)}} dx \)
  2. \(\displaystyle \int {\frac{x}{\sqrt{1-x^4}}} dx \)
  3. \(\displaystyle \int {\frac{x^2}{x^2+1}} dx \)

Suggestive Solution Guide

By setting \(u=2x+1\) gives \(du=2dx\) and substituting these values into the integral leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int (2x+1)^2 dx
&= \int ( \color{red}{\underbrace{\color{black}{2x+1}}_{\large u}} )^2 \color{red}{\underbrace{\color{black}{dx}}_{\large \frac{du}{2}}} \\[5pt]
&= \frac{1}{2}\int u^2
&\color{gray}{\text{[Apply the Substitution]}} \\[5pt]
&= \frac{1}{2}\cdot\frac{u^3}{3} + C
&\color{gray}{\text{[Apply the Power Rule]}}\\[5pt]
&= \color{green}{\frac{(2x+1)^3}{6} + C}
&\color{gray}{\text{[Convert back to } x \text{]}}
\end{align*}
\)

Let \(u=2x+3\), \(x = \frac{u-3}{2}\), and \(du=2dx\). Substituting these values into the integral leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int x\sqrt{2x+3} dx
&= \int \color{red}{\underbrace{\color{black}{x}}_{\large \frac{u-3}{2}}} \color{red}{\underbrace{\color{black}{\sqrt{2x+3}}}_{\large u}}
\color{red}{\underbrace{\color{black}{dx}}_{\large \frac{du}{2}}} \\[5pt]
&= \frac{1}{4}\int (u-3)\sqrt{u} du
&\color{gray}{\text{[Apply the Substitution]}} \\[5pt]
&= \frac{1}{4} \left(\int{u\sqrt{u}\: du}\: -\: 3\int \sqrt{u} \: du\right)
&\color{gray}{\text{[Distribute the Integrals]}} \\[5pt]
&= \frac{1}{4} \left(\int{u^\frac{3}{2}\: du}\: -\: 3\int u^\frac{1}{2} \: du\right)
&\color{gray}{\text{[Use power rules to simplify]}} \\[5pt]
&= \frac{1}{4}\left(\frac{u^\frac{5}{2}}{\frac{5}{2}} – \frac{u^\frac{3}{2}}{\frac{3}{2}} \right) + C
&\color{gray}{\text{[Apply the Power Rule]}} \\[5pt]
&= \frac{1}{4}\left[\left(\frac{2}{5}\right)u^\frac{5}{2} – \left(\frac{2}{3}\right)u^\frac{3}{2} \right] + C
&\color{gray}{\text{[Simplify the fractions]}} \\[5pt]
& = \frac{1}{10}u^\frac{5}{2}\: – \frac{1}{6}u^\frac{3}{2} + C
&\color{gray}{\text{[Distribute the outside fraction]}} \\[5pt]
&= \color{green}{\frac{1}{10}(2x+3)^\frac{5}{2}\: – \frac{1}{6}(2x+3)^\frac{3}{2} + C}
&\color{gray}{\text{[Substitute back to x]}} \\[5pt]
\end{align*}
\)

Let \(u=2x+1\), \(x = \frac{u-1}{2}\), and \(du=2dx\). Substituting these values into the integral leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \frac{x}{2x+1} dx
&= \int \left( \frac{\color{red}{\frac{u-1}{2}}}{\color{red}{u}}\right) \left( \color{red}{\frac{du}{2}}\right)
&\color{gray}{\text{[Apply the Substitution]}} \\[5pt]
&= \frac{1}{4} \int\frac{u-1}{u} du
&\color{gray}{\text{[Rearrange the intergand]}} \\[5pt]
&= \frac{1}{4} \left( \int \frac{u}{u} du \: -\int \frac{du}{u} \right)
&\color{gray}{\text{[Separate the integrals]}} \\[5pt]
&= \frac{1}{4}\left[ u – \texttt{ln}|u| \right] + C
&\color{gray}{\text{[Simplify the first integrand and integrate]}} \\[5pt]
&= \frac{1}{4} \left[ 2x+1 – \texttt{ln}|2x+1| \right] + C
&\color{gray}{\text{[Substitute back to x]}} \\[5pt]
&= \frac{x}{2} + \frac{1}{4} -\frac{1}{4} \texttt{ln}|2x+1|+ C
&\color{gray}{\text{[Distribute the fraction]}} \\[5pt]
&= \color{green}{\frac{x}{2}  -\frac{1}{4} \texttt{ln}|2x+1|+ C}
&\color{gray}{\text{[Include the fraction as part of the C constant]}} \\[5pt]
\end{align*}
\)

Let \(u=x^2\), and \(du=2xdx\). Substituting these values into the integral leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \frac{x^3}{x^2-4} dx
&= \int \frac{x^2 \cdot x}{x^2-4} dx
&\color{gray}{\text{[Rewrite the integrand]}} \\[5pt]
&= \int \left( \frac{\color{red}{u}}{\color{red}{u-4}} \right) \left( \color{red}{\frac{du}{2}} \right)
&\color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{2} \int \frac{u}{u-4} du
&\color{gray}{\text{[Rewrite the integrand]}} \\[5pt]
&= \frac{1}{2} \int \frac{u-4+4}{u-4} du
&\color{gray}{\text{[Add 0 by doing 4 – 4]}} \\[5pt]
&= \frac{1}{2} \left( \int \frac{u-4}{u-4} du\: +4\int \frac{1}{u-4} du \right)
&\color{gray}{\text{[Separate the integrals strategically]}} \\[5pt]
&= \frac{1}{2} \left( \int du\: +4\int \frac{1}{u-4} du \right)
&\color{gray}{\text{[Simplify the first integrand]}} \\[5pt]
&= \frac{1}{2} ( u\: +4\texttt{ln}|u-4|) + C
&\color{gray}{\text{[Integrate]}} \\[5pt]
&= \frac{1}{2} ( x^2\: +4\texttt{ln}|x^2-4|) + C
&\color{gray}{\text{[Substitute back]}} \\[5pt]
&= \color{green}{\frac{x^2}{2}\: +2\texttt{ln}|x^2-4| + C}
&\color{gray}{\text{[Distribute the fraction]}} \\[5pt]
\end{align*}
\)

Let \(u=2x^3+1\), and \(du=6x^2dx\). Substituting these values into the integral leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int x^2\texttt{e}^{2x^3+1} dx
&= \int \texttt{e}^{2x^3+1}(x^2dx)
&\color{gray}{\text{[Rewrite the integrand]}} \\[5pt]
&= \int \texttt{e}^\color{red}{u} \left( \color{red}{\frac{du}{6}} \right)
&\color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{6}\int \texttt{e}^udu
&\color{gray}{\text{[Rewrite the integrand]}} \\[5pt]
&= \frac{1}{6} \texttt{e}^u\:+ C
&\color{gray}{\text{[Integrate]}} \\[5pt]
&= \color{green}{\frac{1}{6} \texttt{e}^{2x^3+1}\:+ C}
&\color{gray}{\text{[Substitute back]}} \\[5pt]
\end{align*}
\)

Let \(u=\texttt{e}^{2x}\), and \(du=2\texttt{e}^{2x}dx\). Substituting these values into the integral leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \frac{\texttt{e}^{2x}}{1+e^{2x}} dx
&= \int \frac{\color{red}{\frac{du}{2}}}{\color{red}{1+u}}
&\color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{2} \int \frac{1}{1+u} du
&\color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{2}\texttt{ln}|1+u|\: + C &\color{gray}{\text{[Integrate]}} \\[5pt]
&= \color{green}{\frac{1}{2}\texttt{ln}|1+e^{2x}|\: + C}
&\color{gray}{\text{[Substitute back]}} \\[5pt]
\end{align*}
\)

Let \(u=\texttt{ln}(x)\), and \(du=\frac{dx}{x}\). Substituting these values into the integral leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int^\texttt{e}_1 \frac{\texttt{ln}(x)}{x} dx
&= \int \frac{\color{red}{\frac{du}{2}}}{\color{red}{1+u}}
&\color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{2} \int \frac{1}{1+u} du
&\color{gray}{\text{[Rewrite the integrand]}} \\[5pt]
&= \frac{1}{2}\texttt{ln}|1+u|\: + C
&\color{gray}{\text{[Integrate]}} \\[5pt]
&= \color{green}{\frac{1}{2}\texttt{ln}|1+\texttt{e}^{2x}|\: + C}
&\color{gray}{\text{[Substitute back]}} \\[5pt]
\end{align*}
\)

Let \(u=\texttt{cos}(x)+1\), and \(du=-\texttt{sin}(x)dx\). Substituting these values into the integral leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \frac{\texttt{sin}(x)}{1+\texttt{cos}(x)} dx
&= \int \frac{\color{red}{-1}}{\color{red}{u}} du
&\color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= -\texttt{ln}|u|\: + C
&\color{gray}{\text{[Integrate]}} \\[5pt]
&= \color{green}{-\texttt{ln}|\texttt{cos}(x)+1|\: + C}
&\color{gray}{\text{[Substitute back]}} \\[5pt]
\end{align*}
\)

Let \(u=\texttt{e}^x\), and \(du=\texttt{e}^xdx\). Substituting these values into the integral leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \frac{\texttt{e}^x}{1+\texttt{e}^{2x}} dx
&= \int \frac{\color{red}{du}}{\color{red}{1+u^2}} du
&\color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \texttt{arctan}(u)\: + C
&\color{gray}{\text{[Integrate]}} \\[5pt]
&= \color{green}{\texttt{arctan}(\texttt{e}^x)\:+ C}
&\color{gray}{\text{[Substitute back]}} \\[5pt]
\end{align*}
\)

Transform the current integrand using algebraic manipulations:

\(\normalsize
\begin{align*}\displaystyle
\int \frac{1}{\texttt{e}^x+\texttt{e}^{-x}} dx
&= \int \frac{1}{1+\frac{1}{\texttt{e}^x}} dx
&\color{gray}{\text{[Use power rules]}} \\[5pt]
&= \int \frac{1}{\frac{\texttt{e}^{2x}+1}{\texttt{e}^x}} dx
&\color{gray}{\text{[Perform the sum]}} \\[5pt]
&= \int \frac{\texttt{e}^x}{\texttt{e}^{2x}+1} dx
&\color{gray}{\text{[Reverse the fraction]}} \\[5pt]
\end{align*}
\)

 

Let \(u=\texttt{e}^x\), and \(du=\texttt{e}^xdx\). Substituting these values into the integral leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \frac{\texttt{e}^x}{\texttt{e}^{2x}+1} dx
&= \int \frac{\color{red}{du}}{\color{red}{u^2+1}}
&\color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \texttt{arctan}(u)\:+ C
&\color{gray}{\text{[Integrate]}} \\[5pt]
&= \color{green}{\texttt{arctan}(\texttt{e}^x)\: + C}
&\color{gray}{\text{[Substitute back]}} \\[5pt]
\end{align*}
\)

Let \(u=\sqrt{x}\), \(du=\frac{1}{2\sqrt{x}}dx\), and \(x+1=u^2+1\). Substituting these values into the integral leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \frac{1}{\sqrt{x}(x+1)} dx
&= \int \frac{\color{red}{2 \cdot du}}{\color{red}{u^2+1}}
&\color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= 2\int \frac{1}{u^2+1} du
&\color{gray}{\text{[Rearrange the integrand]}} \\[5pt]
&= 2\texttt{arctan}(u)\: + C
&\color{gray}{\text{[Integrate]}} \\[5pt]
&= \color{green}{2\texttt{arctan}(\sqrt{x})\: + C}
&\color{gray}{\text{[Substitute back]}} \\[5pt]
\end{align*}
\)

Let \(u=x^2\), and \(du=2xdx\). Substituting these values into the integral leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \frac{x}{\sqrt{1-x^4}} dx
&= \int \frac{\color{red}{\frac{du}{2}}}{\sqrt{1-\color{red}{u^2}}}
&\color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{2} \int \frac{1}{\sqrt{1-u^2}} du
&\color{gray}{\text{[Rearrange the integrand]}} \\[5pt]
&= \frac{1}{2} \texttt{arcsin}(u)\: + C
&\color{gray}{\text{[Integrate]}} \\[5pt]
&= \color{green}{\frac{1}{2}\texttt{arcsin}(x^2)\: + C}
&\color{gray}{\text{[Substitute back]}} \\[5pt]
\end{align*}
\)

Transform the integrand using algebraic manipulations:

\(\normalsize
\begin{align*}\displaystyle
\int \frac{x^2}{x^2+1} dx
&= \int \frac{x^2+1-1}{x^2+1} dx
&\color{gray}{\text{[Add 0 by doing 1-1]}} \\[5pt]
&= \int \frac{x^2+1}{x^2+1} dx\:- \int \frac{1}{x^2+1} dx
&\color{gray}{\text{[Separate the integrals]}} \\[5pt]
&= \int dx\:- \int \frac{1}{x^2+1} dx
&\color{gray}{\text{[Simplify the first integrand]}} \\[5pt]
&= \color{green}{x\:- \texttt{arctan}(x)\:+ C}
&\color{gray}{\text{[Integrate]}} \\[5pt]
\end{align*}
\)