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Find the number $ b $ such that the line $ y = b $ divides the region bounded by the curves $ y = x^2 $ and $ y = 4 $ into two regions with equal area.

$y=4^{2 / 3}$

Applications of Integration

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Numerade Educator

Missouri State University

Campbell University

Baylor University

So what we want to do here, since we want to find to regions with equal area, is we want to split our region that's enclosed between the red and green graph and half. And what's nice is our parabola is symmetric, so we can split it smack dab right through the center along the Y axis right there. And just consider one of these pieces to find half the area of this region. So let's consider the right hand side. So let's computer integral with respect to why so our right hand function will call it f of why. And so we're gonna convert our green y equals X squared to a function in terms of why instead of X. So we just saw for why they're so f of why is equal to the square root of why Andi, we're choosing the positive square root because we're in the first quadrant. So when we plug in positive why values will come out with positive X values as a result, and then our left hand function is gonna be our blue function that we're using to split our region in half. And that is tthe e y axis, which is X equals zero Our. In other words, we call that G of why equals zero Now for our bounds, we have our lower bound that's the X axis, and then our upper bound is going to be our red y equals four. So with the X axis, that's why equals zero. And then why equals four is our upper bound, So to create are integral for 1/2 of the area of this region that's equal to the integral from 0 to 4 of our left hand function, or are excuse me a right hand function minus our left hand function, So f of y minus ci of why? Which is the square root of Why? Why? To the 1/2 minus zero? We don't even need to write that. We just have d y. So evaluating this we have 2/3. Why, to the three halves evaluated from 0 to 4 and plugging in our bounds, we have 2/3 of four to the three halves minus Well, when we plugged in zero, we just get zero. So we don't need to write that 2/3 times wth e um square root of four cute, which is the square root of 64 which we know is eight. So this is 2/3 times eight, which is 16 over three. So that is the value of half of the area of our region. Great. So what we want to do now is find that be value where we can find a line. Why equals B that will split our region, Um, in half as well, such that both of the areas are equivalent. So we're going to do very similar process here. This is still f of why equals thes square root of why on the left hand side, we call this G of y that's equal to the negative square root of why and our bounds here for our bottom half of the area, while the lower bound is again the X axis. And so that's going to be that. Why equals zero. Our upper bound is gonna be our blue line. Why equals B? We don't know what B is yet. We're going to find that in this process. So to construct the integral for this, we'll have that half of the area which we now know is actually 16 over three. That's equal to the integral fromthe lower bound zero to the upper bound B of our f of why minus g of why? Which is the square root of why minus the negative square root of why So why did the 1/2 minus and minus Why to the 1/2? Do you? Why? So that's equal to the integral from zero to be of two. Why to the 1/2? Do you? Why, which is equal Thio? 4/3. Why? To the three halves evaluated from zero to be and then we plug in our bounds. We get 4/3 be to the three halves minus. When we plug in zero, we just get zero. So we don't need to worry about that. And that's equal. Thio again 16 over three And now we just sell for B. So let's multiply both sides by three over four goodbye, goodbye and goodbye and four goes into 16 4 times. So we have four equals B to the three halves and we want Thio isolate Be so using properties of exponents those go away So we have that be is equal to four to the two thirds. Now we can check our work Thio verify that that is actually the correct be constant for our line y equals beat. So let's look at and consider, um d top half of this enclosed region and using our be that we found four to the 2/3. Let's calculate the integral and verify that this area in here can be calculated Using B to B equals four to the 2/3 and in fact, results in six over 13. So our, um, our bounds for our actor region upper half the lower bound is gonna be y equals b, which is why equals we know now, four to the 2/3 and our our upper bound is going to be our red y equals four. So when we want to check, we want to verify that 16 over three, which is our 1/2 of our area, is equal to the integral from four to the 2/3 24 off our f of y minus g of way. So that is why to the 1/2 minus minus. Why to the 1/2 do you? Why, which is the integral from four to the 2/3 to the 4 to 4 of two? Why? To the 1/2 do you want, and that's equal to 4/3. Why? To the three halves evaluated from four to the 2/3 24 and so plugging in our upper and lower bounds. We have 4/3 times four to the three halves, minus 4/3 times four to the 2/3 23 house. And then we want to simplify this 4/3 times four to the three halves. Well, we already know that this guy is the square root of 64 which is eight. So eight times for his 32 over three. And then here the buy properties of exponents. These guys cancel each other out, and we have minus four times four over three. So minus 16 over three. And yes, indeed, that's equal to 16 over three. So we've got that. The line that divides our region in half is why equals four to the 2/3. That is the B value that satisfies this problem.

Missouri State University

Applications of Integration