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Calculus 1 : Midpoint Riemann Sums

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Example Questions

Example Question #111 : Differential Functions

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as I_i ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$

Which of the following parameters would give the closest integral approximation of the function:

$\int_{ln(\frac{\pi}{2})}^{ln(\pi)} sin(e^{x})dx$ ?

Possible Answers:

$n=8,I_i>I_{i+1}$

$n=8,I_i<I_{i+1}$

$n=16,I_i<I_{i+1}$

$n=16,I_i>I_{i+1}$

Correct answer:

$n=16,I_i>I_{i+1}$

Explanation:

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=16 will give a closer approximation than n=8:

Reimannsumcomparison

Now, assuming the amount of points/intervals are fixed, but the actual widths can differ, it would be prudent to have narrower intervals at those regions where the graph is steepest. Relatively flatter regions of the function do not lead to the approximation overshooting or undershooting as much as the steeper areas, as can be seen in the above figure.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

Considering the function $f(x)=sin(e^{x})$ , take the first derivative:

d(sin(u))=cos(u)du

d(e^u)=e^udu

$f'(x)=e^xcos(e^{x})$

Over the interval $x=[ln(\frac{\pi}{2}),ln(\pi)]$ this function is negative due to the cosine term. Now take the second derivative:

d(uv)=udv+vdu

$f''(x)=e^xcos(e^{x})-e^{2x}sin(x)$

This function is also negative over the interval $x=[ln(\frac{\pi}{2}),ln(\pi)]$ .

Since both derivatives share a sign, the function grows steeper over the specified interval. The subintervals should therefore get progressively narrower.

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Example Question #111 : How To Find Midpoint Riemann Sums

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as I_i ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$

Which of the following parameters would give the closest integral approximation of the function:

$\int_{0}^{1} e^{x^2-4x+4}$ ?

Possible Answers:

$n=8;I_i>I_{i+1}$

$n=16;I_i<I_{i+1}$

$n=8;I_i<I_{i+1}$

$n=16;I_i>I_{i+1}$

Correct answer:

$n=16;I_i<I_{i+1}$

Explanation:

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=16 will give a closer approximation than n=8 :

Reimannsumcomparison

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from . Since indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

Considering $f(x)=e^{x^2-4x+4}$ find the first derivative:

d(e^u)=e^udu

$f'(x)=(2x-4)e^{x^2-4x+4}$

On the interval x=[0,1] , the slope is negative due to the parenthetical term. Now, Consider the second derivative:

d(uv)=udv+vdu

$f''(x)=2e^{x^2-4x+4}+(2x-4)^2e^{x^2-4x+4}$

Since the parenthetical term is squared, on the interval x=[0,1] , the second derivative is positive.

As the first and second derivatives have opposite signs, the function itself is flattening on the specified interval. The intervals should get progressively wider.

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Example Question #112 : How To Find Midpoint Riemann Sums

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as I_i ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$

Which of the following parameters would give the closest integral approximation of the function:

$\int_2^4 2^{x^2-3x-4}dx$ ?

Possible Answers:

$n=8;I_i<I_{i+1}$

$n=16;I_i<I_{i+1}$

$n=16;I_i>I_{i+1}$

$n=8;I_i>I_{i+1}$

Correct answer:

$n=16;I_i>I_{i+1}$

Explanation:

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=16 will give a closer approximation than n=8 :

Reimannsumcomparison

Consider then the function $f(x)=2^{x^2-3x-4}$ and find its first derivative:

d(a^u)a=uln(a)du

$f'(x)=(2x-3)2^{x^2-3x-4}ln(2)$

The only way this function can be negative is if the term in the parentheses is negative, which occurs for x<1.5 . On the interval x=[2,4] the slope is therefore positive. Now, consider the second derivative:

d(uv)=udv+vdu

$f''(x)=2^{x^2-3x-3}ln(2)+(2x-3)^22^{x^2-3x-4}ln^2(2)$

Since the parenthetical term is squared, it will always be positive, and the second derivative will always be positive.

Both derivatives share the same sign, so the function is getting steeper.

The intervals should get progressively thinner.

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Example Question #114 : Midpoint Riemann Sums

Use the midpoint rule and n=4 to estimate the following integral: $\int_{0}^{4} e^x \:dx$ . Round to four decimal places.

Possible Answers:

54.5982

52.3867

53.5982

52.4283

51.4283

Correct answer:

51.4283

Explanation:

Write the midpoint rule.

$\int_{a}^{b} f(x)\:dx \approx \Delta x [f(x_1)+f(x_2)+...+f(x_n)]$

where x_1, x_2 ... x_n are the midpoints of the intervals.

Determine $\Delta x$ by the following formula. This is the width of the subintervals.

$\Delta x = \frac{b-a}{n} = \frac{4-0}{4} = 1$

[0,1],[1,2],[2,3],[3,4]

Evaluate the function at each interval's midpoints and substitute them into the midpoint rule.

f(0.5), f(1.5),f(2.5), f(3.5)

$\int_{a}^{b} f(x)\:dx \approx(1) [f(0.5)+ f(1.5)+f(2.5)+ f(3.5)]$

$\int_{a}^{b} f(x)\:dx \approx(1) [e^{0.5}+e^{1.5}+e^{2.5}+e^{3.5}]$

$\int_{a}^{b} f(x)\:dx \approx [1.64872+4.481689+12.18249+33.11545]\approx 51.4283$

Therefore:

$\int_{0}^{4} e^x \:dx \approx 51.4283$

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Example Question #115 : Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate the integral $\int_{10}^{30} e^{\frac{1}{e^{x}}}$ using four midpoints.

Possible Answers:

33.41

4.05

13.29

20.00

11.67

Correct answer:

20.00

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{10}^{30} e^{\frac{1}{e^{x}}}$

So the interval is [10,30] , the subintervals have length $\frac{30-10}{4}=5$ , and since we are using the midpoints of each interval, the x-values are [12.5,17.5,22.5,27.5]

$\int_{10}^{30} e^{\frac{1}{e^{x}}}\approx (5)[e^{\frac{1}{e^{12.5}}}+e^{\frac{1}{e^{17.5}}}+e^{\frac{1}{e^{22.5}}}+e^{\frac{1}{e^{27.5}}}]$

$\int_{10}^{30} e^{\frac{1}{e^{x}}}\approx 20.00$

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Example Question #116 : Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate $\int_{-5}^{5} e^{\frac{1}{x+1}}dx$ using five midpoints.

Possible Answers:

4.33

12.84

8.67

6.42

Undefined.

Correct answer:

12.84

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{-5}^{5} e^{\frac{1}{x+1}}dx$

So the interval is [-5,5] , the subintervals have length $\frac{5-(-5)}{5}=2$ , and since we are using the midpoints of each interval, the x-values are [-4,-2,0,2,4]

$\int_{-5}^{5} e^{\frac{1}{x+1}}dx \approx (2)[e^{\frac{1}{-4+1}}+e^{\frac{1}{-2+1}}+e^{\frac{1}{0+1}}+e^{\frac{1}{2+1}}+e^{\frac{1}{4+1}}]$

$\int_{-5}^{5} e^{\frac{1}{x+1}}dx \approx 12.84$

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Example Question #117 : Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate $\int_{0.02}^{0.08} 13^{\sqrt[3]{x}}dx$ using three midpoints.

Possible Answers:

6.55

0.019

0.131

7.67

0.153

Correct answer:

0.153

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{0.02}^{0.08} 13^{\sqrt[3]{x}}dx$

So the interval is [0.02,0.08] , the subintervals have length $\frac{0.08-0.02}{3}=0.02$ , and since we are using the midpoints of each interval, the x-values are [0.03,0.05,0.07]

$\int_{0.02}^{0.08} 13^{\sqrt[3]{x}}dx \approx (0.02)[13^{\sqrt[3]{0.03}}+13^{\sqrt[3]{0.05}}+13^{\sqrt[3]{0.07}}]$

$\int_{0.02}^{0.08} 13^{\sqrt[3]{x}}dx \approx 0.153$

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Example Question #118 : Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate $\int_{-0.01}^{0.15} 123^{\sqrt{x}}dx$ using four midpoints.

Possible Answers:

0.032

0.015

1.972

0.578

0.881

Correct answer:

0.578

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{-0.01}^{0.15} 123^{\sqrt{x}}dx$

So the interval is [-0.01,0.15] , the subintervals have length $\frac{0.15-(-0.01)}{4}=0.04$ , and since we are using the midpoints of each interval, the x-values are [0.01,0.05,0.09,0.13]

$\int_{-0.01}^{0.15} 123^{\sqrt{x}}dx \approx (0.04)[123^{\sqrt{0.01}}+123^{\sqrt{0.05}}+123^{\sqrt{0.09}}+123^{\sqrt{0.13}}]$

$\int_{-0.01}^{0.15} 123^{\sqrt{x}}dx \approx 0.578$

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Example Question #119 : Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate $\int_{0.9}^{1.5} 3^{x^2-x+4}dx$ using three midpoints.

Possible Answers:

54.18

67.26

97.12

83.90

143.39

Correct answer:

67.26

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{0.9}^{1.5} 3^{x^2-x+4}dx$

So the interval is [0.9,1.5] , the subintervals have length $\frac{1.5-0.9}{3}=0.2$ , and since we are using the midpoints of each interval, the x-values are [1,1.2,1.4]

$\int_{0.9}^{1.5} 3^{x^2-x+4}dx \approx (0.2)[3^{1^2-1+4}+3^{1.2^2-1.2+4}+3^{1.4^2-1.4+4}]$

$\int_{0.9}^{1.5} 3^{x^2-x+4}dx \approx 67.26$

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Example Question #120 : Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate $\int_{0.3}^{0.57} \frac{x^2+x}{x^3-x}dx$ using two midpoints.

Possible Answers:

-0.184

-0.559

-0.397

-0.211

-0.485

Correct answer:

-0.485

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{0.3}^{0.57} \frac{x^2+x}{x^3-x}dx$

So the interval is [0.3,0.57] , the subintervals have length $\frac{0.57-0.3}{2}=0.135$ , and since we are using the midpoints of each interval, the x-values are [0.3675, 0.5025]

$\int_{0.3}^{0.57} \frac{x^2+x}{x^3-x}dx \approx (0.135)[\frac{0.3675^2+0.3675}{0.3675^3-0.3675}+\frac{0.5025^2+0.5025}{0.5025^3-0.5025}]$ $\int_{0.3}^{0.57} \frac{x^2+x}{x^3-x}dx \approx -0.485$

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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 #111 : Differential Functions

Example Question #111 : How To Find Midpoint Riemann Sums

Example Question #112 : How To Find Midpoint Riemann Sums

Example Question #114 : Midpoint Riemann Sums

Example Question #115 : Midpoint Riemann Sums

Example Question #116 : Midpoint Riemann Sums

Example Question #117 : Midpoint Riemann Sums

Example Question #118 : Midpoint Riemann Sums

Example Question #119 : Midpoint Riemann Sums

Example Question #120 : Midpoint Riemann Sums

All Calculus 1 Resources

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