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Solve the given problems by integration.If $a>0$ and $b>0,$ show that $\int_{1}^{a} \frac{d u}{u}+\int_{1}^{b} \frac{d u}{u}=\int_{1}^{a b} \frac{d u}{u}$.

Calculus 1 / AB

Chapter 28

Methods of Integration

Section 2

The Basic Logarithmic Form

Integrals

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

02:07

Make the $u$ -substitution…

03:44

Evaluate each definite int…

02:00

07:34

Use substitution to expres…

05:53

Rewriting Integrals

00:40

Evaluate the integral.

00:36

Evaluate the integral.…

02:55

(a) Make an appropriate $u…

09:32

Evaluate the integrals.

03:31

Show that $\int_{0}^{1} …

from zero to infinity of E to the negative axe over the square root of one minus E to native axe the X. And we're told here to let you be equal to one minus e actually negative X. Well, that implies that d'you is just going to be equal to, um e to the negative X decks. Okay, And now when x is equal, zero won't use equals zero. And when you as he coulda one, um, when you approach is one x approaches infinity, right? Since the limit as X approaches infinity of eats the native axes equals zero. So, um, our integral here just becomes we have the integral going now from this year Oto one of one over the square root of you. Thief. You okay? You go ahead and we take so limit as be approaches zero, which now makes us impact Integral to limit as be approaches zero of the Negro going from B toe, one off squirt of e one over spread of you. That's you. To the negative one half power. Do you okay? Is equal to live it as b approaches zero of we have to to you, to the one half. So you you to the one half evaluating from be toe one, which gives us the limit as beat approaches zero of two minus two times is going to be which is just equal to minus zero, which is equal to two so physically from convergence to to

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