Question

Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson's Rule to approximate the given integral with the specified value of\(n\). (Round your answers to six decimal places.)

\(\int_0^2 {\frac{{{e^x}}}{{1 + {x^2}}}} dx,\;\;\;n = 10\)

Short answer

1. Using Trapezoidal rule \(\int_0^2 {\frac{{{e^x}}}{{1 + {x^2}}}} dx,\;\;\;n = 10\) is \(2.660833\).
2. Using Midpoint rule \(\int_0^2 {\frac{{{e^x}}}{{1 + {x^2}}}} dx,\;\;\;n = 10\) is \(2.664377\).
3. Using Simpson’s rule \(\int_0^2 {\frac{{{e^x}}}{{1 + {x^2}}}} dx,\;\;\;n = 10\) is \(2.663244\).

Step-by-step solution

Definition:

Trapezoidal Rule is a rule that evaluates the area under the curves by dividing the total area into smaller trapezoids rather than using rectangles.

Simpson's Rule approximates the integral of a function between two limits\(a\,\& \,b\).It's based on knowing the area under a parabola, or a plane curve.

The midpoint rule approximates the definite integral using rectangular regions.

Computing interval width:(a)

For Trapezoidal rule:

Given integral is \(\int_0^2 {\frac{{{e^x}}}{{1 + {x^2}}}} dx,\;\;\;n = 10\).

Width of interval :

\(\begin{aligned}{c}\Delta x = \frac{{b - a}}{n}\\ = \frac{{2 - 0}}{{10}}\\ = 0.2\end{aligned}\)

Find table of values:

For the trapezoid rule, calculate values at the interval ends or boundaries.

\(x\)	\(f(x)\)
0	1
0.2	1.174426
0.4	1.286056
0.6	1.339793
0.8	1.357037
1	1.359141
1.2	1.360704
1.4	1.370000
1.6	1.391301
1.8	1.426804

Compute using Trapezoidal rule:

Trapezoid rule has \(n + 1 = 11\) terms.

\(\begin{aligned}{c}{T_{10}} = \int_0^2 {\frac{{{e^x}}}{{1 + {x^2}}}} dx\\ \approx \frac{{\Delta x}}{2}\left( {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 2f\left( {{x_3}} \right) + 2f\left( {{x_4}} \right) + 2f\left( {{x_5}} \right) + 2f\left( {{x_6}} \right)} \right.\left. { + 2f\left( {{x_7}} \right) + 2f\left( {{x_8}} \right) + 2f\left( {{x_9}} \right) + f\left( {{x_{10}}} \right)} \right)\\ \approx \frac{{0.2}}{2}(1 + 2(1.174426) + 2(1.286056) + 2(1.339793) + 2(1.357037) + 2(1.359141)\\ + 2(1.360704) + 2(1.37) + 2(1.391301) + 2(1.426804) + 1.477811)\\ \approx 2.660833\end{aligned}\)

Compute width of interval:(b)

For Midpoint rule:

Given integral is \(\int_0^2 {\frac{{{e^x}}}{{1 + {x^2}}}} dx,\;\;\;n = 10\).

Width of interval :

\(\begin{aligned}{c}\Delta x = \frac{{b - a}}{n}\\ = \frac{{2 - 0}}{{10}}\\ = 0.2\end{aligned}\)

Find table of values:

For the midpoint rule, calculate values for the function at the center of every interval. So we start at \(x = 0.1\) and increment by \(\Delta x\).

\(x\)	\(f(x)\)
\(0.1\)	\(1.094229\)
\(0.3\)	\(1.238403\)
\(0.5\)	\(1.318977\)
\(0.7\)	\(1.351512\)
\(0.9\)	\(1.358897\)
\(1.1\)	\(1.359351\)
\(1.3\)	\(1.364051\)
\(1.5\)	\(1.378981\)
\(1.7\)	\(1.407184\)
\(1.9\)	\(1.450302\)

Compute using Midpoint rule:

\(\begin{aligned}{c}{M_{10}} = \int_0^2 {\frac{{{e^x}}}{{1 + {x^2}}}} dx\\ \approx \Delta x\left( {f\left( {{x_0}} \right) + f\left( {{x_1}} \right) + f\left( {{x_2}} \right) + f\left( {{x_3}} \right) + f\left( {{x_4}} \right) + f\left( {{x_5}} \right) + f\left( {{x_6}} \right) + f\left( {{x_7}} \right)} \right)\\ \approx 0.2(1.094229 + 1.238403 + 1.318977 + 1.351512 + 1.358897 + 1.359351\\ + 1.364051 + 1.378981 + 1.407184 + 1.450302)\\ \approx 2.664377\end{aligned}\)

Find width of interval:(c)

Given integral is \(\int_0^2 {\frac{{{e^x}}}{{1 + {x^2}}}} dx,\;\;\;n = 10\).

Width of interval :

\(\begin{aligned}{c}\Delta x = \frac{{b - a}}{n}\\ = \frac{{2 - 0}}{{10}}\\ = 0.2\end{aligned}\)

Find table of values:

For Simpson’s rule calculate the value at interval ends or boundaries.

\(x\)	\(f(x)\)
0	1
\(0.2\)	\(1.174426\)
\(0.4\)	\(1.286056\)
\(0.6\)	\(1.339793\)
\(0.8\)	\(1.357037\)
1	\(1.359141\)
\(1.2\)	\(1.360704\)
\(1.4\)	\(1.370000\)
\(1.6\)	\(1.391301\)
\(1.8\)	\(1.426804\)
2	\(1.177811\)

Compute using Simpson’s rule:

Simpson's rule has \(n + 1 = 11\) terms.

\(\begin{aligned}{c}{S_{10}} = \int_0^2 {\frac{{{e^x}}}{{1 + {x^2}}}} dx\\ \approx \frac{{\Delta x}}{3}\left( {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + 2f\left( {{x_4}} \right)} \right.\left. { + 4f\left( {{x_5}} \right) + 2f\left( {{x_6}} \right) + 4f\left( {{x_7}} \right) + 2f\left( {{x_8}} \right) + 4f\left( {{x_9}} \right) + f\left( {{x_{10}}} \right)} \right)\\ \approx \frac{{0.2}}{3}(1 + 4(1.174426) + 2(1.286056) + 4(1.339793) + 2(1.357037)\\ + 4(1.359141) + 2(1.360704) + 4(1.37) + 2(1.391301) + 4(1.426804) + 1.477811)\\ \approx 2.663244\end{aligned}\)