Find the area of the “triangular” region bounded on the left b...

Question

Find the area of the “triangular” region bounded on the left by x + y = 2, on the right by y = x², and above by y = 2.

Accepted Answer

Welcome back everyone. Find the area of the shaded region. A 27 divided by 2 minus 4 square root of 6. B 27 divided by 2 + 4 square root of 6. C 27 divided by 2 minus 2 square root of 6. And D 27 minus 4 square root of 6. For this problem, our region is bounded by y equals x 2 minus 2. X + y equals 4 and the horizontal line y equals 4. So what we're going to do is identify the area using the integral between two y values, y1 and y2. Let's remember that we take the right curve f of y and subtract the left curve g of y, and we're integrating with respect to y. So in this case, we can show that F of Y is the right curve, which is the line, right? We will call it F of Y. And for the left curve, that would be the parabola, G of Y. Let's identify the limits of integration, so we will have integral from y1. This is going to be the lower y value and specifically this is the horizontal line Y equals 4. And now the upper limit of integration is going to be the intersection point between the two curves. This is going to be y equals 7. We basically have it on the graph. We're going to subtract f of y minus g of y. So what we want to do is express each function as x y, right? Starting with the line. We have x + y equals 4, right, so we can show that x is equal to 4 minus y. We can basically subtract y from both sides. So we get 4 minus y, and we're going to subtract g off y. We know that y is equal to x 2 minus 2, so that Y + 2 is going to be x 2, then we can take the square root of both sides to show that x is equal to plus or minus square root of y + 2. And because we are within The negative x values. Notice that our region is between -3 and 0 in terms of the x axis, we're going to take the negative solutions. So we're subtracting the negative square root of Y + 2. And we're integrating with respect to why. Well then, so now that we have our integral set up. We can Start integrating. Let's simplify before we integrate, so we have integral from 4 to 7 of or minus y plus square root of. Y + 2. DY. Now we can integrate and specifically the integral of 4DY is going to be 4 Y. We're subtracting the integral of y d y, that'd be y square divided by 2 using the power rule, and then we're integrating the square root term, right, so we can rewrite it as y plus 2 raise to the power of 1/2. So, here, we can basically use to use substitution. And in this case, if we set u equals y + 2, we will see that DU is going to be Dy, right? So essentially we can just apply the power rule here, understanding that according to the reverse chain rule, we will still have a slope of 1 in front of y, right? And this will be essentially the same as the integral of y to the power of 1/2, but instead of y, we just have a linear function y + 2. So how do we integrate? Well, essentially we're going to apply the power rule which gives us y + 2, raise to the power of 1/2 plus 1, and that's 3 halves divided by 3 halves. And according to the reverse chain rule, we're dividing by the slope of the linear term. So in that case, that's division by one, meaning nothing really changes, right? Well done, and now what we're going to do is calculate the result from 4 to 7. We can also simplify instead of division by 3 halves, we can write multiplication by 2/3. And now we are going to perform the calculations, so let's go ahead and do that. We get 4 multiplied by 7 minus 72 divided by 2. + 2/3 Multiplied by. 7 + 2, that's basically a 9 to the power of 3 halves, and we're going to subtract 4 multiplied by 4 minus 42 divided by 2 plus 2/3 multiplied by 4 + 2 gives us 6 to the power of 3 halves. Now, let's go ahead and simplify. This is going to be. 28 minus 49 divided by 2 + 18 Minus 16 minus 8 + 4 square root of 6. And now we can combine the terms, right? So that would be 43 Divided by 2. -8 minus 4 square root of 6. And this is equal to 27 divided by 2 minus 4 square root of 6. And that's our final answer, 27 divided by 2 minus. Or square root of 6, which essentially corresponds to the answer of choice a. Thank you for watching.

Find the area of the “triangular” region bounded on the left by x + y = 2, on the right by y = x², and above by y = 2.

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