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Solve the given problems by integration.An architect designs a wall panel that can be described as the firstquadrant area bounded by $y=\frac{50}{x^{2}+20}$ and $x=3.00 .$ If the area of the panel is $6.61 \mathrm{m}^{2}$, find the $x$ -coordinate (in $\mathrm{m}$ ) of the centroid of the panel.

Calculus 1 / AB

Chapter 28

Methods of Integration

Section 2

The Basic Logarithmic Form

Integrals

Campbell University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

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Section 76. Problem number 44. So here I am asked to find the area between a curve which you see given here, one over the square root of X square monies to X plus two area between that curve, the X axis when x goes from 0 to 3. So let's take a look at it first. So what you see here is the graph of the the initial curve. What you see here, I've got X goes from zero 23 And so I'm being asked to find this shaded area, which means, basically, I need to find the integral from 0 to 3 for this function, and it turns out to be about 2.3 to 5. So I'm gonna get an exact value, and it's approximation should be very close to ah, that number. So, in this case, what I'm doing, I'm using, um, integral look up tables. And so let's just take a look. I've been using the table that you see here that referenced, um, and so I go and I look one up and this is illustrates. You got a sort of be on top of your game here. There's a typo in this particular problem. So you see this Actually, what this one is is this is the integral of the one that I'm looking for One over a x weird plus b x plus c d x. They forgot to put the square root sign. And this works when a is greater than zero. So this is actually what I'm looking forward. So all the key reason that you're gonna have to really know how to integrate you can't rely always on integral table toe work it for you. So let's go back and let's just write down What? That solution Ah, would be, um so according to our interval look up table, I know that the indefinite integral one over square root X squared plus B x plus See the X is going to be equal to one over the square root of a natural log to a X plus B plus two square root of a and in the square root of a X squared plus BX plus c And then all of this, plus a constant. And this is when a is greater than zero. So in my case, I need to do this particular problem where a coefficient of the quadratic term is one be coefficient of the linear term is negative. Two. And see the constant term is positive, too, so I would need to substitute those values in. So just using this formula, you'll have one over the square root of one natural log two times one times acts plus negative, too, plus two times one, um, and then the square root of X squared, minus two X plus two. And then all of this is evaluated from 0 to 3. So let's clean this up just a little bit. This is the natural log of two X minus two plus two square root of X squared on its two X plus two. Evaluate from 0 to 3. So let's do this evaluation. Calculus is over. We're into arithmetic now, so substituted three into that expression so you'll have the natural log. And when you substitute a three, you have six minus two plus two. You substituted three. You're gonna X square. That's 99 minus six is 33 plus two is five that when you substitute a zero, you'll have natural log zero minus two plus two, and then the square root of to. So what we have now we've got the natural log of four plus two, Route five minus the natural log. And I've got to root two minus two. If I put all of these together, this is the natural log four plus two, Route five divided by 22 minus two. In fact, or two out of the numerator and denominator, You're gonna get to two plus Route five over to, um Tu minus one. These 2/2. That's going to be one. So this final answer is going to be the natural log two, plus Route five over route to minus one. Substitute that into your calculator. That is about 2.32 50 And then how does that compare? You see, 2.3 to 50 So we were correct in making that work. So again, the key here Waas. I needed to graph it to actually see what's happening between zero and three to tell me that the integration will indeed find the area under the curve. Then I write down this integral and I go to my integral table. And in this case, there was an error in my integral tables and hopefully years will not have that. Um, correct that Arab used the formula substitute, do the arithmetic and get down to the final answer here.

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