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Find the values of $ c $ such that the area of th…

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Problem 58 Hard Difficulty

(a) Find the number a such that the line $ x = a $ bisects the area under the curve $ y = \frac{1}{x^2} $, $ 1 \le x \le 4 $.

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 6

Applications of Integration

Section 1

Areas Between Curves

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Video Transcript

find a number. A such set. The Line X equals a by sex. The area under the given curve Why equals one over X squared. And then we have this domain that X is between one and four. So let's first get a feel for this function again using the domain that X is between one and four. We can on Lee feed our function. Why, with X values in that domain. So how about one, two, three and four and so why becomes one too? Is 1/4 three is one night and four is 1/16. So let's plot that blue function there on our X y table here. So one in one, the scales a little bit, um, not uniformed, but that's okay to end 1/4 and three and one knife is around there, say, and four and 1/16. Let's put that there. All right, then we have the access between one and four. So we've got a left bound and a right bound, Um, right before our eyes, X equals one would be the left bound. We'll call that l B and X equals four. That's the right bounds will call that our B and, well, just plot those lines right here. And it's, um it's kind of gonna cover our one, but that's okay. We know that X equals one, and then X equals four. All right. And then let's connect these dots as well. The blue guys. So we're building this, um, enclosed area between these curves, and then the last thing we need is to know. Well, what is the what is tthe e lower bound? What's on the bottom? And so, um, we can take the limit as X approaches infinity of why and see where that tends to. So that is Theo Limit as X approaches. Infinity of one over X squared, which is zero. So the lower bound is going to be Y equals zero. We'll do that in red. I was there. Okay, so we've got this green enclosed region that we want to find. Um, some line X equals a that bisects that green region. Um, So what we want to do here is fine. What is the actual area of this region? Then we're gonna find what half of the area is, and then find that line X equals a find that a value that yields half of Theis area of this region. So since we first want to find the area of this region, we're going to take an integral We're going to use our, um, functions and look in our bounds that we've discussed here. And so we would have that the area of the green region is equal to the integral from, um, for the bounds, we're gonna take our left bound minus our, uh, excuse me. We're gonna take our left down in our right bound. We're going to use those guys. So the, um smaller one, the left bound to the right, bound. And then we're going Thio, subtract the top function minus the bottom function. So blue minus red. So that's one over X squared minus zero DX. And so that gives us, of course, the integral from 1 to 4 of one over X squared DX. And then I like Teoh, Right? This, um with X in the numerator and then just use negative exponents. So that makes it more clear what I need to do when I integrate so negative x to the minus one. Evaluated from 1 to 4. And then that's negative. One over x evaluated from 1 to 4. So then plugging those in upper bound first or the the top down first minus one over four minus a minus one over one, which is just one. So this gives us negative 1/4 plus four over four, which is three over four. So that is tthe e entire area of the green region. So then half of that area is, of course, 1/2 times the full area, which is three over eight. So this is half of the area of that green region, and what we want to do now is fine. The value for a that will yield. Um, when we take the integral to try to find half of the area will yield indeed, half of the area. So we're going to suppose that we have some line X equals A, and we're going thio tried thio um find, uh, that a value. So let's take the, um, right half of the area that has resulted in our supposed line X equals A. We're gonna taken Inderal and do pretty much the same thing that we just did, except now our left bound. Instead of being one, we're going to start it at a and then go to our right bound being for So we will have that half the area which we know is 3/8 is equal to the inner girl from a 24 of our top function minus our bottom functions. So one over X squared, minus zeros. We don't even need to write that D X, which we know is negative one over X and we're evaluating that from a 24 So that gives us negative 1/4 minus a minus one over a and this yields. Then, of course, we can simplify that negative one over four plus one over a. So we have three over eight equals negative 1/4 plus one over a course for solving for a here. So plus 1/4 um, plus 2/8 right? Same thing is 1/4 was cancel. We have 5/8 equals one over a and then flip around both sides just like so. So we have that a equals eight over five, and that is the value of that line. So the line X equals eight over five by sex that green region, so that we have half of the area on each side

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Catherine Ross

Kayleah Tsai

Caleb Elmore

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Kayleah Tsai

Caleb Elmore

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Catherine Ross

Kayleah Tsai

Caleb Elmore

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals