Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Midpoint Riemann Sums

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #101 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{0.25}^{1.75} (3^{x^2}-2^{x^3})dx$ using three midpoints.

Possible Answers:

1.94

1.35

2.03

0.97

2.69

Correct answer:

1.35

Explanation:

A Reimann 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 approximating $\int_{0.25}^{1.75} (3^{x^2}-2^{x^3})dx$

So the interval is [0.25,1.75] , the subintervals have length $\frac{1.75-0.25}{3}=0.5$ , and since we are using the midpoints of each interval, the x-values are [0.5,1,1.5]

$\int_{0.25}^{1.75} (3^{x^2}-2^{x^3})dx \approx 0.5[(3^{0.5^2}-2^{0.5^3})+(3^{1^2}-2^{1^3})+(3^{1.5^2}-2^{1.5^3})]$

$\int_{0.25}^{1.75} (3^{x^2}-2^{x^3})dx \approx 1.35$

Report an Error

Example Question #102 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{0.95}^{2.15} sin(2^{tan(x)})dx$ using three midpoints.

Possible Answers:

1.02

1.61

-0.93

0.18

-0.37

Correct answer:

-0.37

Explanation:

A Reimann 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 approximating $\int_{0.95}^{2.15} sin(2^{tan(x)})dx$

So the interval is [0.95,2.15] , the subintervals have length $\frac{2.15-0.95}{3}=0.4$ , and since we are using the midpoints of each interval, the x-values are [1.15,1.55,1.95]

$\int_{0.95}^{2.15} sin(2^{tan(x)})dx \approx 0.4[sin(2^{tan(1.15)})+sin(2^{tan(1.55)})+sin(2^{tan(1.95)})]$

$\int_{0.95}^{2.15} sin(2^{tan(x)})dx \approx -0.37$

Report an Error

Example Question #103 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate $\int_{0.14}^{0.78} tan(e^{-x})dx$ using two midpoints.

Possible Answers:

2.093

0.670

0.484

4.723

1.511

Correct answer:

0.484

Explanation:

A Reimann 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 approximating $\int_{0.14}^{0.78} tan(e^{-x})dx$

So the interval is [0.14,0.78] , the subintervals have length $\frac{0.78-0.14}{2}=0.32$ , and since we are using the midpoints of each interval, the x-values are [0.30,0.62]

$\int_{0.14}^{0.78} tan(e^{-x})dx \approx 0.32[tan(e^{-0.30})+tan(e^{-0.62})]$

$\int_{0.14}^{0.78} tan(e^{-x})dx \approx 0.484$

Report an Error

Example Question #101 : Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate the integral $\int_1^2 ln(1.3^{x^{2.4}})$ using four midpoints.

Possible Answers:

3.096

1.548

0.735

0.229

0.916

Correct answer:

0.735

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 approximating $\int_1^2 ln(1.3^{x^{2.4}})$

So the interval is [1,2] , the subintervals have length $\frac{2-1}{4}=0.25$ , and since we are using the midpoints of each interval, the x-values are [1.125,1.375,1.625,1.875]

$\int_1^2 ln(1.3^{x^{2.4}})\approx 0.25[ln(1.3^{1.125^{2.4}})+ln(1.3^{1.375^{2.4}})+ln(1.3^{1.625^{2.4}})+ln(1.3^{1.875^{2.4}})]$

$\int_1^2 ln(1.3^{x^{2.4}})\approx 0.735$

Report an Error

Example Question #104 : How To Find Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate $\int_{0.03}^{0.96} 4tan(cos(\frac{1}{x}))dx$ using three midpoints.

Possible Answers:

3.043

0.760

0.881

0.404

1.212

Correct answer:

0.760

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 approximating $\int_{0.03}^{0.96} 4tan(cos(\frac{1}{x}))dx$

So the interval is [0.03,0.96] , the subintervals have length $\frac{0.96-0.03}{3}=0.31$ , and since we are using the midpoints of each interval, the x-values are [0.185,0.495,0.805]

$\int_{0.03}^{0.96} 4tan(cos(\frac{1}{x}))dx \approx 0.31[4tan(cos(\frac{1}{0.185}))+4tan(cos(\frac{1}{0.495}))+4tan(cos(\frac{1}{0.805}))]$

$\int_{0.03}^{0.96} 4tan(cos(\frac{1}{x}))dx \approx 0.760$

Report an Error

Example Question #101 : How To Find Midpoint Riemann Sums

Let

f(x)=x

What is the Midpoint Riemann Sum on the interval [0,10] divided into five sub-intervals?

Possible Answers:

Correct answer:

Explanation:

The interval [0,10] divided into four sub-intervals gives rectangles with vertices of the bases at

x=0,2,4,6,8,10

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) , f(7) , and f(9) .

f(1)=1

f(3)=3

f(5)=5

f(7)=7

f(9)=9

Because each sub-interval has a width of , the Midpoint Riemann sum is

A=bh=2(1+3+5+7+9)=50

Report an Error

Example Question #101 : How To Find Midpoint Riemann Sums

Let

f(x)=x^3-4x^2+4x

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

Possible Answers:

372

124

200

448

Correct answer:

448

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)=1^3-4*1^2+4*1=1

f(3)=3^3-4*3^2+4*3=3

f(5)=5^3-4*5^2+4*5=45

f(7)=7^3-4*7^2+4*7=175

Because each sub-interval has a width of 2, the Midpoint Riemann sum is

A=bh=2(1+3+45+175)=448 .

Report an Error

Example Question #108 : 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:

$\int_0^{0.5} \frac{1}{2}^{x^2-2x+1}dx$ ?

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

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 $f(x)=\frac{1}{2}^{x^2-2x+1}$ , find the first derivative:

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

$f'(x)=(2x-2)\frac{1}{2}^{x^2-2x+1}ln(\frac{1}{2})$

The term in the parenthesis is what may lend to variability in the sign. Over the interval x=[0,0.5] the parenthetical is negative, so the derivative is positive, since $ln(\frac{1}{2})=-0.693$ , and the product of two negatives is positive. Now consider the second derivative:

d(uv)=udv+vdu

$f''(x)=\frac{1}{2}^{x^2-2x+2}ln(\frac{1}{2})+(2x-2)^2\frac{1}{2}^{x^2-2x+1}ln^2(\frac{1}{2})$

The second term will always be positive, and the first term will always be negative. On the interval x=[0,0.5] , it can be found, however, that the second derivative will be positive. If calculating this is to nettlesome, plot this function on the interval:

Save 3

Since the first and second derivatives have opposite signs, the function is getting flatter. The intervals should in turn get wider.

Report an Error

Example Question #102 : 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:

$\int_0^{\frac{\pi}{2}} cos(\frac{x^2}{\pi})dx$ ?

Possible Answers:

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

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

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

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

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

Considering $f(x)=cos(\frac{x^2}{\pi})$ take the first derivative:

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

$f'(x)=-\frac{2x}{\pi}sin(\frac{x^2}{\pi})$

Within the interval $x=[0,\frac{\pi}{2}]$ this function is negative. Now consider the second derivative:

d(uv)=udv+vdu

$f''(x)=-\frac{2}{\pi}sin(\frac{x^2}{\pi})-\frac{4x^2}{\pi^2}cos(\frac{x^2}{\pi})$

Over the interval $x=[0,\frac{\pi}{2}]$ , the second derivative is negative as well.

Since the signs match, the slope gets steeper, and the intervals should get thinner.

Report an Error

Example Question #102 : 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:

$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} e^{cos(x)}dx$ ?

Possible Answers:

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

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

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

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

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

Considering $f(x)= e^{cos(x)}$ , find the first derivative:

d(e^u)=e^udu

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

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

Within the interval $x=[\frac{\pi}{3},\frac{\pi}{2}]$ , the function only takes negative values. Now, take the second derivative:

d(uv)=udv+vdu

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

On the interval $x=[\frac{\pi}{3},\frac{\pi}{2}]$ , this function takes on positive values.

Since the signs of the derivatives are opposing, the function gets flatter over the course of this interval, and the subintervals should thus get wider.

Report an Error

View Calculus Tutors

Abhisek
Certified Tutor

University of Florida, Bachelor of Engineering, Computer Science.

View Calculus Tutors

Francis
Certified Tutor

University of South Florida-Main Campus, Bachelor of Science, Computer Science.

View Calculus Tutors

Magdy
Certified Tutor

Cairo University, Egypt, Bachelor of Science, Electrical Engineering. New Mexico State University-Main Campus, Doctor of Phil...

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Algebra Tutors in Washington DC, Spanish Tutors in Dallas Fort Worth, Algebra Tutors in Chicago, SSAT Tutors in Seattle, GMAT Tutors in Houston, GMAT Tutors in Miami, MCAT Tutors in Los Angeles, Math Tutors in Houston, Calculus Tutors in Phoenix, GRE Tutors in Phoenix

Popular Courses & Classes

ISEE Courses & Classes in Denver, LSAT Courses & Classes in Los Angeles, LSAT Courses & Classes in Seattle, LSAT Courses & Classes in San Diego, MCAT Courses & Classes in Chicago, LSAT Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in Denver, ACT Courses & Classes in Miami, SAT Courses & Classes in Los Angeles, SSAT Courses & Classes in Dallas Fort Worth

Popular Test Prep

LSAT Test Prep in San Diego, ACT Test Prep in Miami, ACT Test Prep in Seattle, ISEE Test Prep in Seattle, LSAT Test Prep in Philadelphia, SAT Test Prep in San Diego, LSAT Test Prep in Los Angeles, GRE Test Prep in Denver, GRE Test Prep in Philadelphia, GMAT Test Prep in Denver

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 1 : Midpoint Riemann Sums

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #101 : How To Find Midpoint Riemann Sums

Example Question #102 : How To Find Midpoint Riemann Sums

Example Question #103 : How To Find Midpoint Riemann Sums

Example Question #101 : Midpoint Riemann Sums

Example Question #104 : How To Find Midpoint Riemann Sums

Example Question #101 : How To Find Midpoint Riemann Sums

Example Question #101 : How To Find Midpoint Riemann Sums

Example Question #108 : How To Find Midpoint Riemann Sums

Example Question #102 : How To Find Midpoint Riemann Sums

Example Question #102 : Midpoint Riemann Sums

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: