Practice Problems: Simpson's Rule (1/3)

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.

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

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

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Practice Problems

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

\(\displaystyle \int (2x+1)^2dx\)
\(\displaystyle \int x\sqrt{2x+3}dx\)
\(\displaystyle \int \frac{x}{2x+1}dx\)
\(\displaystyle \int \frac{x^3}{x^2-4}dx \)
\(\displaystyle \int \texttt{e}^{x^2+1} dx \)

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

\(\displaystyle \int {\frac{1}{\sqrt{x}(x+1)}} dx \)
\(\displaystyle \int {\frac{x}{\sqrt{1-x^4}}} dx \)
\(\displaystyle \int {\frac{x^2}{x^2+1}} dx \)

Suggestive Solution Guide

Problem 1: Solution

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

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

Problem 2: Solution

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

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

Problem 3: Solution

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

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

Problem 4: Solution

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

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

Problem 5: Solution

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

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

Problem 6: Solution

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

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

Problem 7: Solution

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

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

Problem 8: Solution

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

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

Problem 9: Solution

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

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

Problem 10: Solution

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:

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

Problem 11: Solution

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:

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

Problem 12: Solution

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

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

Problem 13: Solution

Transform the integrand using algebraic manipulations:

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

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