Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Calculus

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 #62 : Differential Functions

Using the method of midpoint Reimann sums, approximate the integral $\int_{\pi}^{4\pi} \frac{sin(x)}{x^2}dx$ using three midpoints

Possible Answers:

$\frac{836}{225\pi}$

$\frac{836}{225}\pi$

$\frac{836}{75\pi}$

$\frac{836}{75\pi^2}$

$\frac{836}{225\pi^2}$

Correct answer:

$\frac{836}{225\pi}$

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 .

So the interval is $[\pi,4\pi]$ , the subintervals have length $\frac{4\pi-\pi}{3}=\pi$ , and since we are using the midpoints of each interval, the x-values are $[0.5\pi,1.5\pi,2.5\pi]$

$\int_{\pi}^{4\pi} \frac{sin(x)}{x^2}dx \approx \pi[\frac{sin(0.5\pi)}{(0.5\pi)^2}+\frac{sin(1.5\pi)}{(1.5\pi)^2}+\frac{sin(2.5\pi)}{(2.5\pi)^2}]$

$\int_{\pi}^{4\pi} \frac{sin(x)}{x^2}dx \approx \frac{836}{225\pi}$

Report an Error

Example Question #62 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_5^{11} \frac{ln^2(x)}{x^3}$ using three midpoints.

Possible Answers:

0.0572

0.0286

Correct answer:

0.0572

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 .

Our integral is $\int_5^{11} \frac{ln^2(x)}{x^3}$

So the interval is [5,11] , the subintervals have length $\frac{11-5}{3}=2$ , and since we are using the midpoints of each interval, the x-values are [6,8,10]

$\int_5^{11} \frac{ln^2(x)}{x^3} \approx 2[\frac{ln^2(6)}{6^3}+\frac{ln^2(8)}{8^3}+\frac{ln^2(10)}{10^3}]$

$\int_5^{11} \frac{ln^2(x)}{x^3} \approx 0.0572$

Report an Error

Example Question #63 : Differential Functions

Using the method of midpoint Reimann sums, approximate the integral $\int_{10\pi}^{16\pi}\frac{cos(x^2)}{x}dx$ using three midpoints.

Possible Answers:

0.469

-0.052

-0.235

-0.469

-0.104

Correct answer:

-0.104

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 .

The function given is $\int_{10}^{16}\frac{cos(x^2)}{x}dx$

So the interval is $[10\pi,16\pi]$ , the subintervals have length $\frac{16\pi-10\pi}{3}=2\pi$ , and since we are using the midpoints of each interval, the x-values are $[11\pi,13\pi,15\pi]$

$\int_{10}^{16}\frac{cos(x^2)}{x}dx \approx 2\pi[\frac{cos((11\pi)^2)}{11\pi}+\frac{cos((13\pi)^2)}{13\pi}+\frac{cos((15\pi)^2)}{15\pi}]$

Report an Error

Example Question #64 : Differential Functions

Using the method of midpoint Reimann sums, approximate the integral $\int_0^6 (x^2-x)dx$ using three midpoints.

Possible Answers:

Correct answer:

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 .

Our function is $\int_0^6 (x^2-x)dx$

So the interval is [0,6] , the subintervals have length $\frac{6-0}{3}=2$ , and since we are using the midpoints of each interval, the x-values are [1,3,5]

$\int_0^6 (x^2-x)dx\approx 2[(1^2-1)+(3^2-3)+(5^2-5)]$

$\int_0^6 (x^2-x)dx \approx 52$

Report an Error

Example Question #62 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_0^{12} \frac{x+4}{x+8}dx$ using three midpoints.

Possible Answers:

$\frac{2636}{945}$

$\frac{2636}{315}$

$\frac{659}{945}$

$\frac{10544}{315}$

$\frac{10544}{945}$

Correct answer:

$\frac{2636}{315}$

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 .

Our integral is $\int_0^{12} \frac{x+4}{x+8}dx$

So the interval is [0,12] , the subintervals have length $\frac{12-0}{3}=4$ , and since we are using the midpoints of each interval, the x-values are [2,6,10]

$\int_0^{12} \frac{x+4}{x+8}dx \approx 4[\frac{2+4}{2+8}+\frac{6+4}{6+8}+\frac{10+4}{10+8}]$

$\int_0^{12} \frac{x+4}{x+8}dx \approx \frac{2636}{315}$

Report an Error

Example Question #66 : Differential Functions

Using the method of midpoint Riemann sums, approximate the integral $\int_1^4 x^{x-1}dx$ using three midpoints.

Possible Answers:

12.0

28.1

75.0

37.3

14.0

Correct answer:

28.1

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 .

The integral we are given is $\int_1^4 x^{x-1}dx$

So the interval is [1,4] , the subintervals have length $\frac{4-1}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [1.5,2.5,3.5]

$\int_1^4 x^{x-1}dx \approx 1(1.5^{1.5-1}+2.5^{2.5-1}+3.5^{3.5-1})$

$\int_1^4 x^{x-1}dx \approx (1.5^{0.5}+2.5^{1.5}+3.5^{2.5})$

$\int_1^4 x^{x-1}dx \approx 28.1$

Report an Error

Example Question #67 : Differential Functions

The method of Riemann sums is a way of approximating integrals, often for functions which do not have a defined integral.

Picture an integral $\int_a^b f(x)dx$

A number of rectangular areas are defined, this number being defined by the amount of points or subintervals selected. Generally, these rectangles have uniform widths (that of the subinterval), given by the formula $\frac{b-a}{n}$ where n is the number of points/subintervals.

Each rectangle differs in height, however, with individual heights defined by the function value at a given point, f(x_i) .

The area of a particular rectangle is then $\frac{b-a}{n}f(x)dx$

Adding each of these rectangles together then gives the integral approximation.

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

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:

$\int_0^4 x^xdx$ ?

Possible Answers:

$n=7;I_i<I_{i+1}$

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

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

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

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

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.

Since the function x^x becomes progressively steeper, a better approximation for a fixed number of intervals would have the intervals grow thinner as increases.

Report an Error

Example Question #68 : Differential Functions

Using the method of midpoint Riemann sums, approximate the integral $\int_{10\pi}^{16\pi}e^{\frac{cos(x)}{x}}dx$ using three midpoints.

Possible Answers:

3.67

9.19

2.93

5.85

18.39

Correct answer:

18.39

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 .

The integral we're given is $\int_{10\pi}^{16\pi}e^{\frac{cos(x)}{x}}dx$

So the interval is $[10\pi,16\pi]$ , the subintervals have length $\frac{16\pi-10\pi}{3}=2\pi$ , and since we are using the midpoints of each interval, the x-values are $[11\pi,13\pi,15\pi]$

$\int_{10\pi}^{16\pi}e^{\frac{cos(x)}{x}}dx \approx 2\pi[e^{\frac{cos(11\pi)}{11\pi}}+e^{\frac{cos(13\pi)}{13\pi}}}+e^{\frac{cos(15\pi)}{15\pi}}}]$

$\int_{10\pi}^{16\pi}e^{\frac{cos(x)}{x}}dx \approx 18.39$

Report an Error

Example Question #69 : Differential Functions

Using the method of midpoint Riemann sums, approximate the area of the region between the functions $f(x)=x^{2x}$ and g(x)=x^x over the interval x=[1,2.5] using three midpoints.

Possible Answers:

19.28

9.27

18.55

11.23

37.10

Correct answer:

18.55

Explanation:

To find the area between two functions, the method traditionally used is to find the difference of the integral values between the larger and smaller function over the specified interval. However, neither of these functions can be integrated, so a method of approximation is useful.

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 told that we're integrating from 1 to 2.5, so the interval is [1,2.5] , the subintervals have length $\frac{2.5-1}{3}=0.5$ , and since we are using the midpoints of each interval, the x-values are [1.25,1.75,2.25] .

As $f(x)=x^{2x}$ is greater than g(x)=x^x over the given interval, our approximation looks as follows:

${\int_{1}^{2.5}(x^{2x}-x^x)dx \approx 0.5[(1.25^{2(1.25)}-1.25^{1.25})+(1.75^{2(1.75)}-1.75^{1.75})+(2.25^{2(2.25)}-2.25^{2.25})]}$

${\int_{1}^{2.5}(x^{2x}-x^x)dx \approx 18.55$

Report an Error

Example Question #1091 : Calculus

A Riemann Sum approximation of an integral $\int_a^b (f(x))dx$ follows the form

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

Where n is number of points/subintervals used to approximate the integral.

Knowing this, imagine a modified style of Riemann Sum, such 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:

$\int_{10}^{100} \frac{e^x}{x^2}dx$ ?

Possible Answers:

$n=20;I_i>I_{i+1}$

$n=10;I_i>I_{i+1}$

$n=20;I_i<I_{i+1}$

$n=10;I_i<I_{i+1}$

$n=11;I_i<I_{i+1}$

Correct answer:

$n=20;I_i>I_{i+1}$

Explanation:

The more points/intervals that are selected, the closer the Reimann sum approximation becomes to the actual integral value, so n=20 will give a closer approximation than n=10:

Reimannsumcomparison

Consider the function $\frac{e^x}{x^2}$ over the interval x=[10,100] . The steepness of this graph at a point can be found by taking the function's derivative.

The quotient rule of derivatives states:

$d\frac{u}{v}=\frac{vdu-udv}{v^2}$

So the derivative is

$\frac{x^2e^x-2xe^x}{x^4}$

$\frac{e^x(x-2)}{x^3}$

It might be intuitive to see that the steepness is positive and that it gets progressively greater over the specified interval, although to be precise, we may take the derivative once more to find the rate of change of this steepness and see if it's postive or negative:

$\frac{x^4(2xe^x+x^2e^x-2e^x-2xe^x)-4x^3(x^2e^x-2xe^x)}{x^8}$

$\frac{x^6e^x+6x^4e^x-4x^5e^x}{x^8}$

$\frac{e^x(x^2-4x+6)}{x^4}$

Since this derivative is positive over the specified interval, , and since the slope was shown to be positive as well, the slope is ever-increasing and growing steeper.

The intervals should in turn grow increasingly thinner.

Report an Error

View Calculus Tutors

Samuel
Certified Tutor

York University, Bachelor, Physics. York University, Master's/Graduate, Mathematics and Statistics.

View Calculus Tutors

Dalton
Certified Tutor

University of Massachusetts Amherst, Doctor of Philosophy, Mathematics.

View Calculus Tutors

Christopher
Certified Tutor

Rose-Hulman Institute of Technology, Bachelor of Science, Physics. University of Washington, Master of Science, Physics.

All Calculus 1 Resources

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

Popular Subjects

Calculus Tutors in Philadelphia, GMAT Tutors in New York City, Math Tutors in Houston, GMAT Tutors in Seattle, SAT Tutors in Houston, LSAT Tutors in Miami, Math Tutors in Chicago, Algebra Tutors in Los Angeles, GRE Tutors in Phoenix, English Tutors in Los Angeles

Popular Courses & Classes

ISEE Courses & Classes in San Francisco-Bay Area, LSAT Courses & Classes in Washington DC, MCAT Courses & Classes in Phoenix, GMAT Courses & Classes in San Diego, ISEE Courses & Classes in Miami, LSAT Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in Atlanta, GMAT Courses & Classes in Philadelphia, SAT Courses & Classes in Denver, Spanish Courses & Classes in Dallas Fort Worth

Popular Test Prep

MCAT Test Prep in Denver, ACT Test Prep in Dallas Fort Worth, GMAT Test Prep in Phoenix, MCAT Test Prep in San Francisco-Bay Area, SAT Test Prep in Atlanta, ISEE Test Prep in Houston, SAT Test Prep in Boston, MCAT Test Prep in San Diego, LSAT Test Prep in Dallas Fort Worth, GMAT Test Prep in San Diego

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 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #62 : Differential Functions

Example Question #62 : Midpoint Riemann Sums

Example Question #63 : Differential Functions

Example Question #64 : Differential Functions

Example Question #62 : Midpoint Riemann Sums

Example Question #66 : Differential Functions

Example Question #67 : Differential Functions

Example Question #68 : Differential Functions

Example Question #69 : Differential Functions

Example Question #1091 : Calculus

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: