Find the area of the region described in the following exercis...

Question

Find the area of the region described in the following exercises.

The region in the first quadrant bounded by y=5/2−1/x and y=x

Accepted Answer

Welcome back, everyone. Determine the area of the region bounded by the curves Y equals 3 minus 4 divided by X and Y equals X divided by 2. For this problem we want to find the area, and first of all, we're interested in the intersection points. According to the provided graph, we can see that those are X equals 2 and X equals 4, and we want to check if they are accurate. So, the first function Y equals 3 minus 4 divided by X at X equals 2. Gives us 3 minus 4 divided by 2, which is 1, right? So this is an accurate point. And now for the second point, we are essentially considering our function Y equals X divided by 2. We still want to check X equals 2, so 2 divided by 2 is 1, and we also have the same coordinate of Y, right? So we have verified the first intersection point, and now for the second intersection point, if X equals 4, Y equals 3 minus 4 divided by 4, which is 2. And considering our next curve, Y equals X divided by 2. We got 4 divided by 2, which is also 2. So we have the correct intersection points. And now let's recall that area equals an integral. In this case, from 2 to 4, those are our intersection points, and our integran contains a difference between two curves. We have to take the top one, which is 3 minus 4 divided by X, and subtract the bottom one. Which is the blue one in the form of X divided by 2, and we're integrating with respect to X. Over this interval, our red curve remains on top all the time, right? So we don't need to split our integral into two parts. So now let's go ahead and integrate. The integral of 3DX is 3X. Then we're subtracting 4, which is our constant in the integral of 1 divided by x the axis lm of the absolute value of X. Then we're going to subtract one half, which is our constant for the 3rd part, and the integral of X is X2 divided by 2. We want to calculate the result from 2 up to 4. And because X is positive, we can drop the absolute value for the natural log, and we can now apply the evaluation theorem. We're going to get 3 multiplied by 4 minus 4 LN 4 minus 1/4 multiplied by 42, and then we're subtracting the value of this function at X equals 2. So 3 multiplied by 2 minus 4 LN of 2, minus 1/4 multiplied by 2 squad. Let's simplify. This is 12 minus 4 LN of 4 minus 16 divided by 4, so we're subtracting 4 minus 6 plus 4 LN of 2, and then we're adding 1/4 multiplied by 4, so that's 1, right? Now, let's combine the like terms. 12 minus 4 is 88 minus 6 is 2, and 2 + 1 is equal to 3. And now constraining our LN parts, we have minus 4 LN of 4 plus 4 LN of 2. What we can do is simply factor out 4, and we get 3 + 4 in C. Alan of 2 minus Alan of 4. Using the properties of logarithms, we can show that this is. Or in parent LN of 2 divided by 4, right, or basically LN of 12, which is 2 raises the power of -1. And now let's simplify. We're going to remove res because now we have a single logarithm. In applying the properties of logarithms, we can bring down -1 because it's our exponent. So we get 3 minus 4 LM of 2 units. And specifically, because this is an area we're going to include square units, right? Well, then we have our final answer, thank you for watching.

Find the area of the region described in the following exercises.The region in the first quadrant bounded by y=5/2−1/x and y=x

Watch next

Find the area of the region described in the following exercises.

The region in the first quadrant bounded by y=5/2−1/x and y=x