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 bounded by y=2 / 1 + x^2 and y=1

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says to find the area of the region bounded by Y is equal to 3, divided by the quantity of 1 + X2 in quantity, and Y is equal to 2. And we'll give 4 possible choices as our answers. For choice A, we have 3 pi divided by 2 minus the square of 2 square units. For choice B, we have 3 pi divided by 2, plus the square of 2 square units. For C, we have 2 multiplied by the quantity of 3 multiplied by the inverse tangent of the squared 2 divided by 2 minus the square of 2 in quantity square units. And for choice D, we have 2 multiplied by the quantity of 3 multiplied by the inverse tangent of the square of 2 divided by 2, plus the square of 2 in quantity square units. So we're asked to find the area of the region bounded by our two curves, and so what we need to do is to find the intersection points of our two curves. That means that we will set Y equal to 2 in the first function, and that means that we will have 2 is equal to 3 divided by the quantity of 1 plus X squared in quantity. So, we need to solve this for X. So the first step is to multiply both sides of the equation by the quantity of 1 plus X squad in quantity. In doing so, we will have 2 + 2 X 2 is equal to 3. Now we'll subtract 2 from both sides of the equation, so we'll have 2 x 2 is equal to 1. Dividing both sides by 2, we will get X2 is equal to 1 divided by 2, and then taking the square root of both sides, we'll get that X is equal to plus or minus 1 divided by the square root of 2, which we can rewrite as being equal to, plus or minus the square root of 2 divided by 2. So these two boundary points are going to give us our limits of integration. So, the area of the region, which we'll denote as A, is going to be equal to the integral of our first function. So it's going to be the integral from minus the square root of 2 divided by 2 to the square root of 2 divided by 2 of 3 divided by the quantity of 1 plus X squared and quantity DX. And then we'll subtract the integral for the second function. So we'll have minus the integral from minus the square root of 2 divided by 2, to the square root of 2 divided by 2 of 2D X. Then we just need to evaluate each one of these integrals. So doing so, we'll see that our area is going to be equal to. For the first integral, we can factor out a 3 in front of the integral since its constants will have 3, and then this gets multiplied by the integral of 1, divided by the quantity of 1 plus X squared. And if we recall that that is just going to be equal to the inverse tangent. Of X, and this needs to be evaluated from minus the square root of 2 divided by 2 to the square root of 2 divided by 2. Then we'll have minus, and here we're integrating 2 with respect to X, which is going to give us 2 X, and that needs to be evaluated from minus the square root of 2 divided by 2 to the square root of 2 divided by 2. So substituting in our limits of integration, we see that this is going to be equal 2. 3, multiplied by the quantity of the inverse tangent of the square root of 2, divided by 2. Minus the inverse tangent. Of minus the square root of 2, divided by 2. In quantity, then we'll have minus 2 multiplied by the quantity, and here substituting in our limits of integration for a second interval. We'll have the square root of 2 divided by 2 minus the quantity of minus square root of 2, divided by 2. In quantity So now we just need to simplify this expression. So this is going to be equal to 3, multiplied by the quantity of the inverse tangent of the square root of 2, divided by 2. Minus, and here we can use our Trick properties, since the inverse tangent of a negative argument is just equal to minus the inverse tangent of that argument, so we'll have minus the inverse tangent. Of the square root of 2, divided by 2. In quantity Then we'll have minus 2, and combining our next two terms, we'll have just the square root of 2. And so simplifying this even further, we see that this is going to be equal to. 3, multiplied by the quantity of 2 multiplied by the inverse tangent of the square root of 2 divided by 2 in quantity, minus 2 multiplied by the square root of 2. And so we can factor out a 2 out of each of those terms, and so our final answer is going to be equal to. 2, multiplied by the quantity of 3, multiplied by the inverse tangent. Of the square root of 2, divided by 2. Minus the square root of 2, in quantity. So that is the answer that we were looking for for this problem. And if we compare our answer to our answer choices, we see that the correct answer choice corresponds to answer choice C. So, I hope that this explanation has been useful, and I'll see you in the next video.

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

The region bounded by y=2 / 1 + x^2 and y=1

Key Concepts

Definite Integrals

Finding Points of Intersection

Area Between Curves

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