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Sketch the region enclosed by the curves and find its area.$$x=1 / y, x=0, y=1, y=e$$

$$A=1$$

Calculus 1 / AB

Calculus 2 / BC

Chapter 6

APPLICATIONS OF THE DEFINITE INTEGRAL IN GEOMETRY, SCIENCE, AND ENGINEERING

Section 1

Area Between Two Curves

Integrals

Integration

Applications of Integration

Area Between Curves

Volume

Arc Length and Surface Area

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don't let ah, the different notation kind of scare you. Um, X equals one over. Why? Still looks like y equals one of her, actually. And think about cross multiplying and dividing still looks the same. S o X equals zero. Is this vertical line. So why access and then y equals one. Is that why it was one? And I y calls it he is up here. I'm saying we were looking for this area, and it's just in terms of why. So you're horizontal lines of your bounds. Um, your upper function is one over Why? And then your lower function is is the ex eagle zero. But you don't actually have to write that out cause of trying you zero does not change the value of that eso as you're looking at this. Um, when you do the anti derivative of one over why, it's just natural log of why from one to e um, And then as you plug it in, you get natural of e minus natural log of one. And I like to think everybody in calculus would know natural law gov is equal to one natural log of one is equal to zero. The one minus zero is one his aunt's. We want

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