Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Question

Answered step-by-step

Express the integral as a limit of Riemann sums. Do not evaluate the limit.

$ \displaystyle \int^5_2 \biggl( x^2 + \frac{1}{x} \biggr) \, dx $

Video Answer

Solved by verified expert

This problem has been solved!

Try Numerade free for 7 days

Like

Report

Official textbook answer

Video by Stephen Hobbs

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

02:02

Frank Lin

00:46

Amrita Bhasin

Calculus 1 / AB

Chapter 5

Integrals

Section 2

The Definite Integral

Integration

Missouri State University

Campbell University

Harvey Mudd College

Baylor University

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

0:00

Express the integral as a…

03:15

Express the integral as a …

01:37

09:09

Evaluate the integral by c…

06:55

06:49

getting this integral as a limit of sums. We could go ahead and start off by putting the limit as our variable M approaches infinity. And then we're going to some from eyes one up to end. Now, the first thing that we calculate, we look at two and five and then we go ahead and do be minus a over and basically get like the width of each of our rectangles. B is five minus. A is to over end and this turns into three over end and so we could go ahead and put that out front of the sun there. And then the next thing we do is we start from the lower limit and then we add that amount with its i f component. And so from there we could go ahead and write that next part. As to which is just like we said, our lower limit, the starting amount Waas. This helps us find the width of the the, uh, rectangle here three i over. And now all of that is basically this and so that's why it needs to be squared. And then we have plus one over that same amount and so once you find what X is replaced with in one area the same thing in the next. Let's make sure we leave enough room there, so that'll be three I over and and then, lastly, the DX is replaced with a Delta X there. So basically, that's the idea except Delta X. We've already kind of written as this, and so that would be our final limit in this case here.

View More Answers From This Book

Find Another Textbook