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Calculus 1 : Midpoint Riemann Sums

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Example Questions

Example Question #1031 : Calculus

Solve the integral

$\int_{2}^{5}{e{^{x}}}dx$

using the midpoint Riemann sum approximation with n=10 subintervals.

Possible Answers:

38.7644

468.3221

94.6324

140.4966

8.5848

Correct answer:

140.4966

Explanation:

Midpoint Riemann sum approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx (\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ]$

where is the number of subintervals and $f(m_{i})$ is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{n})=(\frac{5-2}{10})=0.3$ .

The approximate value at each midpoint is below.

Screen shot 2015 06 11 at 6.21.45 pm

The sum of all the approximate midpoints values is 468.3221 , therefore

$(\frac{b-a}{n})\left [ f({m_{1}})+f({m_{2}})+...+f({m_{n}}) \right ] =0.3(468.3221)\approx140.4966$

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Example Question #11 : Midpoint Riemann Sums

Let f(x)=x^3

What is the Midpoint Riemann Sum on the interval [0, 8] divided into four sub-intervals?

Possible Answers:

576

600

1600

992

Correct answer:

992

Explanation:

The interval [0,8] divided into four sub-intervals gives rectangles with vertices of the bases at

x=0,2,4,6,8

For the Midpoint Riemann sum, we need to find the rectangle heights which values come from the midpoint of the sub-intervals, or f(1), f(3), f(5), and f(7).

f(1)=1^3=1

f(3)=3^3=27

f(5)=5^3=125

f(7)=7^3=343

Because each interval has width 2, the approximated Midpoint Riemann Sum is

A=bh=hb=2(1+27+125+343)=992

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Example Question #1041 : Calculus

Using midpoint Riemann sum, approximate the area under the curve of $y = \frac{1}{2}x^2 + 6$ from [0,5] using 4 rectangular partitions.

Possible Answers:

70.9

71.025

62.7

41.3

55.4

Correct answer:

71.025

Explanation:

Four rectangles from x = 0 to x = 5 would give us a $\Delta x = \frac{5 - 0}{4} = \frac{5}{4} = 1.25$ .

Thus, there is a rectangle from x = 0 to x = 1.25 , a rectangle from to x = 2.50 , a rectangle from to x = 3.75 , and a rectangle from to x = 5 .

To find the area, evaluate the function at each midpoint of all the rectangles to get the height, then multiply it by the width of the rectangle and sum it all together.

Hence, evaluate f(x) at $f(\frac{5}{8}), f(\frac{15}{8}), f(\frac{25}{8}),$ and $f(\frac{35}{8})$ :

$A = 1.25(f(\frac{5}{8}) + f(\frac{15}{8}) + f(\frac{25}{8}) + f(\frac{35}{8}))$

$A = 1.25\left[ \left(\left(\frac{5}{8}\right)^2 + 6\right) + \left(\left(\frac{15}{8}\right)^2 + 6\right) + \left(\left(\frac{25}{8}\right)^2 + 6\right) + \left(\left(\frac{35}{8}\right)^2 + 6\right) \right]$

A = 71.025

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Example Question #2 : Numerical Integration

Solve the integral

$\int_{2}^{7}{\frac{1}{5x+3}}dx$

using the trapezoidal approximation with n=5 subintervals.

Possible Answers:

0.9548

0.7601

0.4301

0.8603

0.5233

Correct answer:

0.4301

Explanation:

Trapezoidal approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx T_{n}=(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]$

where is the number of subintervals and f(m) is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{7-2}{2*5})=0.5$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 8.19.15 pm

The sum of all the approximation terms is 0.86027754 , therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.5)*(0.86027754)=0.4301$

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Example Question #4 : Numerical Approximation

Solve the integral

$\int_{1}^{2}{x\sin (x)}dx$

using the trapezoidal approximation with n=5 subintervals.

Possible Answers:

28.7867

2.8787

2.5052

0.8415

Correct answer:

2.8787

Explanation:

Trapezoidal approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx T_{n}=(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]$

where is the number of subintervals and f(m) is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{2-1}{2*5})=0.1$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 8.32.39 pm

The sum of all the approximation terms is 28.7867 , therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.1)*(28.786699)=2.8787$

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Example Question #5 : Numerical Integration

Solve the integral

$\int_{0}^{4}{\sqrt{5+x^{^{3}}}}dx$

using the trapezoidal approximation with n=5 subintervals.

Possible Answers:

83.8900

33.5582

5.0000

14.3744

2.3547

Correct answer:

33.5582

Explanation:

Trapezoidal approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx =T_{n}=(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]$

where is the number of subintervals and f(m) is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{4-0}{2*5})=0.4$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 8.55.34 pm

The sum of all the approximation terms is 83.89553 , therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.4)*(83.89553)=33.5582$

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Example Question #2 : Numerical Integration

Solve the integral

$\int_{1}^{2}{\sqrt{ln(x)}}dx$

using the trapezoidal approximation with n=5 subintervals.

Possible Answers:

0.6174

11.7257

1.1726

0.8326

Correct answer:

1.1726

Explanation:

Trapezoidal approximations are solved using the formula

$\int_{a}^{b}f(x))dx\approx T_{n}=(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]$

where is the number of subintervals and f(m) is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{2-1}{2*5})=0.1$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 8.55.45 pm

The sum of all the approximation terms is 11.7257431 , therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.1)*(11.7257431)=1.1726$

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Example Question #2 : Numerical Approximation

Solve the integral

$\int_{1}^{5}{\frac{1}{x-7}}dx$

using Simpson's rule with 2n=10 subintervals.

Possible Answers:

-1.6667

-8.2400

-0.500

-1.0987

-0.1667

Correct answer:

-1.0987

Explanation:

Simpson's rule is solved using the formula

$\int_{a}^{b}f(x))dx =S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]$

where is the number of subintervals and f(m) is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{5-1}{10})=0.4$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 9.35.50 pm

The sum of all the approximation terms is -8.23995 therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.4)*(-8.23995)=-1.0987$

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Example Question #13 : How To Find Midpoint Riemann Sums

Solve the integral

$\int_{8}^{15}{ln(x)}dx$

using Simpson's rule with 2n=10 subintervals.

Possible Answers:

16.9852

8.6532

10.2289

10.6410

2.0794

Correct answer:

16.9852

Explanation:

Simpson's rule is solved using the formula

$\int_{a}^{b}f(x))dx=S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]$

where is the number of subintervals and f(m) is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{15-8}{10})=0.7$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 9.35.58 pm

The sum of all the approximation terms is 72.79378 therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.7)*(72.79378)=16.9852$

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Example Question #13 : Functions

Solve the integral

$\int_{-1}^{1}{\frac{1}{2+cos(x)}}dx$

using Simpson's rule with 2n=10 subintervals.

Possible Answers:

10.9830

10.5872

0.3948

0.7056

1.2340

Correct answer:

0.7056

Explanation:

Simpson's rule is solved using the formula

$\int_{a}^{b}f(x))dx =S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]$

where is the number of subintervals and f(m) is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{1+1}{10})=0.2$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 9.36.10 pm

The sum of all the approximation terms is 10.5840 therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.2)*(10.5840)=0.7056$

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Calculus 1 : Midpoint Riemann Sums

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1031 : Calculus

Example Question #11 : Midpoint Riemann Sums

Example Question #1041 : Calculus

Example Question #2 : Numerical Integration

Example Question #4 : Numerical Approximation

Example Question #5 : Numerical Integration

Example Question #2 : Numerical Integration

Example Question #2 : Numerical Approximation

Example Question #13 : How To Find Midpoint Riemann Sums

Example Question #13 : Functions

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