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=e^x, y=2e^−x+1, and x=0

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the area of the region bounded by Y is equal to 3, multiplied by E X, Y is equal to 5 multiplied by E minus X plus 2, and X equal to 0. So we began by creating a sketch of our three curves. So here we've drawn our X and Y axis, and here we're just focusing on the first quadrant. The first curve that we want to plot is Y is equal to 3 multiplied by E to the X. So this has a Y intercept of Y equal to 3, and so we're going to begin at that Y intercept point and just get an exponentially increasing curve. For the second curve, we have Y is equal to 5 multiplied by E to the minus X plus 2. This has a Y intercept of 7. And so here we'll begin at that Y intercept point, and we'll sketch an exponentially decreasing curve that is going to be approaching the horizontal line of Y is equal 2. And for our final curve, we have X equal to 0, and that just happens to be our Y axis. And so, what we want to do is determine the area of the region bounded by those three curves. So based upon our sketch, we see that that area is bounded above by Y is equal to 5 multiplied by E minus X plus 2. It's bounded below by Y is equal to 3, multiplied by EX. It's bounded on the left by X equal to 0, and it's bounded to the right by the intersection point between the curves Y is equal to 3, multiplied by E to the X and Y is equal to 5 multiplied by E minus X plus 2. So, the first thing we want to do is find the X value for that intersection point. So that means that that we need to set those two Y values for those two curves equal to each other. So, we're gonna set. 3, multiplied by ETD X equal to. 5 multiplied by E to the minus X plus 2. And so, we want to solve this equation for X. So the first step is to multiply both of the equation by E to the X. In doing so, we'll have 3 multiplied by E to the 2 X is equal to 5, plus 2 multiplied by E X. And we want to collect everything on the left-hand side of the equation, so we'll subtract 5, as well as subtract 2 multiplied by E to the X from both sides of the equation. In doing so, we'll have 3 multiplied by E to the 2 X. -2 multiplied by ETD X. Minus 5 is equal to 0. And we can actually factor this expression. So this factors into the quantity of 3 multiplied by E X minus 5 in quantity, multiplied by the quantity of E X plus 1 in quantity, and that's going to be equal to 0. So now we want to set each factor equals to 0 and solve for X. So for the first factor, we'll have 3 multiplied by E to the X, minus 5 is equal to 0. So, the first step is to add 5 to both sides of the equation, and then dividing both sides of the equation by 3. We'll get ET the X is equal to 5 divided by 3, and taking the natural log of both sides of the equation, we'll get that X is equal to the natural log of the quantity of 5 divided by 3 in quantity. Now for the other factor, we'll have that e to the X plus 1 is equal to 0, subtracting one from both sides we'll get that e to the X is equal to -1. However, E to the X can never be negative. So that means that this equation has no real solution. So we only have one X value, which is going to be X is equal to the natural log of the quantity of 5 divided by 3 in quantity, which agrees with our sketch since we only have one point of intersection between the two curves. So now we just need to set up the interval to find our area. So our area, which will denote as A, is going to be equal to the integral from 0 to the natural log of quantity of 5 divided by 3 in quantity of the quantity, and here we'll have the curve that bounds our region area from above, which is 5 multiplied by E to the minus X. Plus 2, then we'll have minus the curve that bounds our region from below, which is going to be 3, multiplied by E to the X in quantity DX. Now we just need to perform the integration, and we can actually integrate term by term here. That means that our area is going to be equal to. The quantity of minus 5 multiplied by E to the minus X, that recall that when you integrate E to the minus X, you get minus E to the minus X, then we'll have plus. And here we'll be integrating 2 with respect to X, which is just give us 2 X. Then we'll have minus 3, multiplied by the integral of E to the X, which is just to the X. In quantity, and we need to evaluate this expression from 0 to the natural log of quantity of 5 divided by 3 in quantity. So substituting in our lips of integration, we see that our area is going to be equal to minus 5 multiplied by E to the minus natural log of quantity of 5 divided by 3 in quantity. Plus 2 multiplied by the natural log of the quantity of 5 divided by 3 in quantity. -3, multiplied by E raise to the natural log of quantity of 5 divided by 3 in quantity, then we'll have minus the quantity of -5 multiplied by E to the 0. Plus 2 multiplied by 0, minus 3 multiplied by E to the 0 in quantity. So now we just need to simplify this expression. So this is going to be equal to minus 5. Multiplied by E, raised to the quantity, and here we'll pull the minus sign inside the natural log argument, and when we do that, that means that we need to invert the argument inside the natural log function. So we'll have the natural log of quantity of 3 divided by 5 in quantity. Then we'll have plus 2 multiplied by the natural log. Of the quantity of 5 divided by 3 in quantity, then we'll have minus 3, and here we can simplify it, since we have E raise to the natural log, we'll just get the argument of the natural log, which is going to be the quantity of 5 divided by 3 in quantity, then we can simplify the remaining terms of Plus 5 minus 0. Plus 3. So again, we need to simplify this expression. So, that means our area is going to be equal to. We'll have -5, multiplied by, and here again, we have E raises to the natural log of an argument, so we just get back to that argument, which is going to be 3 divided by 5. And then we'll have plus 2 multiplied by the natural log of the quantity of 5 divided by 3 in quantity. And then adding up the remaining terms, we're just going to get plus 3. And finally, you'll find this expression. This is just going to be equal to 2 multiplied by the natural log of quantity of 5 divided by 3, this will have square units, since this is an area. So this here is the answer that we are looking for for this problem. 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=e^x, y=2e^−x+1, and x=0

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

The region bounded by y=e^x, y=2e^−x+1, and x=0