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Evaluate the double integral $\int_{R} \int f(x, y) d A$ over the indicated region $R$, for the given function $f$. $f(x, y)=\frac{1}{1+y^{2}}, R$ is the region bounded by $x=0, y=x$ and $y=2$.

$$\frac{1}{2} \ln 5$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 6

Double Integrals

Partial Derivatives

Johns Hopkins University

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Harvey Mudd College

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Lectures

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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Evaluate the double integr…

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for this problem we are asked to evaluate the double integral of f X Y equals 1/1 plus Y squared over the region bounded by X equals zero, Y equals x and y equals two. So we'll want to take a look at how to set up our integral by first plotting out the region. Now we can see that this is a pretty simple region both horizontally and vertically. So let's say that we are treating this as a horizontally simple region. In which case then then excuse me, this will be the integral from 0 to 2 of the integral. From at the lower bound. That would be Y equals X. Upper bound is Y equals two of 1/1 plus Y squared dy dx. Actually I'm going to back up here because we actually have a much easier integral if we treat this as a vertically simple region. So in that case we'd be integrating still now we're going to be integrating outside why? From 0-2. And inside we would be integrating X from zero to Why? Then we have 1/1 plus Y squared dx Dy so integrating first over X. That's a pretty simple integral soul. Just write down the result directly. We get y over one plus y squared. Dy Now we can integrate this using u substitution. Writing U equals one plus y squared. In which case we'll find that the results will be 1/2 times the natural algorithm of one plus y squared Evaluated from 0 to 2. In which case this will then become 1/2 times lawn. Actually, I'll write it like this 1/2 times one of one plus four. So lawn of five minus lawn of one plus zero, so minus lawn of one, but lawn of one is just zero, So this becomes 1/2 times lawn of five oh.

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