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
Evaluate the integral by interpreting it in terms of areas.
$ \displaystyle \int^3_{-4} \biggl| \frac{1}{2}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
01:34
Frank Lin
00:33
Amrita Bhasin
Calculus 1 / AB
Chapter 5
Integrals
Section 2
The Definite Integral
Integration
Campbell University
Oregon State University
Baylor University
University of Michigan - Ann Arbor
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
$$\text { Evaluate the…
02:46
Evaluate the integral by i…
00:56
01:17
00:25
02:10
04:28
Compute the definite integ…
01:20
given the integral of the absolute value of one half X, we could go ahead and graft that and pull again the point when X is three. That would give us why is three halves over here and then plug in negative four. I would get us negative to absolute value is positive, too. And we want to find the area underneath. Each of these curves bounded between the curve and the X axis. And so to do that well, we have a triangle with a base of four. Height of two. So four times two is eight. Divided by two is four. So therefore, this triangle here has an area of four. Uh huh. And that's triangle here as an area of three times three halves, which was the nine halves. But then that divided by two. So let's go ahead and write that out. Nine hands. Dividing by two is the same thing is multiplying by 1/4. I mean, sorry, one half, and so that would end up being 9/4. So the area on the right side here is 9/4. Okay, Now we want to add these two together so we could just come up with common denominator. So we could change this for to 16 over four, and then we're finally ready to add the So let's do this. Cell circle needs to that we're gonna add in green. The area on the left side is four. Another way to say that is 16/4. The area on the right side is 9/4. Therefore, we end up with a total area of 25 fourth. Yeah, so that is our final evaluation for this integral using geometric area.
View More Answers From This Book
Find Another Textbook
03:11
Let P be polynomial of degree and let Q be polynomial of degree m (…