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Evaluate the integral by interpreting it in terms of areas.

$ \displaystyle \int^1_0 \bigl| 2x - 1 \bigr| \, dx $

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01:44

Frank Lin

00:41

Amrita Bhasin

Calculus 1 / AB

Chapter 5

Integrals

Section 2

The Definite Integral

Integration

Missouri State University

Oregon State University

Baylor University

University of Nottingham

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.

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$$\text { Evaluate the…

02:46

Evaluate the integral by i…

00:25

01:17

01:52

01:46

Evaluate the integral.

00:56

02:15

to evaluate the integral given here the absolute value of two X minus one. We could just go ahead and draft that. And then we want to find the area between zero and one half and zero on one half and one. So this point right here, 0.5. And because of that, we have Well, we have a height on the left side. If we were to plug zero into this, we'd get one. So it's got a height of one on this side. We plug one into this. We also get one. We got two triangles that have a height of one and bases of 0.5. And so we're basically just adding thes two together here. Uh huh. But that's 11 half of one time 0.5 is actually 0.25 That's its area and the one over to the right there. So plus, Plus, that one has the identical area of 0.25 because it's half of half times one. So therefore both of these added together we'll end up giving us a total area for the evaluation of the integral ends up being 0.5 positive since everything's above the X axis

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