Problem

Use Property 8 to estimate the value of the integ…

00:28

Problem 59 Easy Difficulty

Use Property 8 to estimate the value of the integral.

$ \displaystyle \int^1_0 x^3 \,dx $

Answer

$0 \leq \int_{0}^{1} x^{3} d x \leq 1$

Discussion

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Video Transcript

Well, let's go ahead and answer this question. It says the integral off X cubed between 0 to 1. It has to be between zero and one inclusive. Prove this using property eight. Okay, It's actually quite intuitive group. Basically, what you have to say is, if you have a new integration from A to B of ffx, it is always going to be between these two values M times B minus A and large and times B minus A When F of X is between large m and small m. What do they mean by that? Well, let's pretend that ffx looks something like this okay between A and B. Let's say that the lower bound of this is M and M doesn't necessarily be the point at f of A as just as long as F of X is, um, larger than small. M n e m would satisfy. Let's say that the upper bound iss large m So right now you see that f of X is between small M and large m. So what is M times B minus a equal to well, this portion is the area under this curve off em. So it's this red rectangle. Similarly, if I take a look at this value, it's the area off the blue rectangle. So what can I say about the area off the integral? The area under this curve? Well, it's very clear that it's going to be between or larger than the blue rectangle, but smaller than the red rectangle. So that's what Property eight is talking about. Okay, so all you need to do to show this is that the rectangle, the red rectangle, is one. The blue rectangle is zero. Why did I know that X cubed looks like this crime 01 when X is equal to zero, the left end. It's right here. So X cubed is greater than or equal to zero. Is it less than or equal to one? Yeah. If I plug in X is equal to one. The height is one X cubed is less than one. So that distance from 0 to 1 is one minus zero, which is equal to one. So to be a little bit careful here, I can say that this is zero times one minus zero, less than or equal to the area between zero toe one of X cubed the X and one times one minus zero, which basically represents the area under the red rectangle. The area under the blue rectangle which is sandwiching the area off. Ex cute. And this is how you prove this statement using Property eight.

Yuki H.

University of California, Berkeley

Topics

Integrals

Integration

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 2

The Definite Integral

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