14–25. {Use of Tech} Areas of regions Determine the area of th...

Question

14–25. {Use of Tech} Areas of regions Determine the area of the given region.

The region bounded by y = ln x,y = 1, and x = 1

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to calculate the area of the region bounded by Y is equal to the natural log of X, Y is equal to -1, and X is equal to 0.5. So, the first thing we want to do is to sketch a graph with our three curves on it. And here we're gonna be focusing on the 4th quadrant. So, we'll draw our X and Y axis so that we can see the 4th quadrant. And the first curve that we want to but is Y is equal to the natural log of X. And so that is a fundamental curve, but we can just draw that directly, and we call that the natural log of X crosses the X axis at X equal to 1, and the natural log of X approaches minus infinity as X approaches 0. The next curve that we want to draw is Y is equal to -1. And that is just gonna be a horizontal line. At Y is equal to -1. We'll sketch that. Curve on our graph. And the last curve that we want to plot is X is equal to 0.5. And here, That is going to be a vertical line at X equal to 0.5. And so the region that we're interested in finding the area of is the region that is bounded by those three curves. And if we look at our sketch here, we see that that region is bounded above by the natural log of X, bonded below by Y is equal to -1, bonded on the right by X is equal to 0.5, and on the left is bounded by the intersection point of the natural log of X, and Y is equal to -1. So, what we want to do is to determine the X value for that intersection point. So to do that, we'll set the natural log of X equal to -1. And now we just need to solve 4 X. So inverting the natural log function, we will get that X is equal to E raised to the minus 1 power, which we can rewrite as being equal to 1 divided by E. So now we have all the information that we need to set up our integral for the area. So, our area, which will denote as A, is going to be equal to the integral. From 1 divided by E to 1 divided by 2. Since our region that we're interested in. Goes from X equal to 1 divided by E to X equal to 1/2. Then for the integran, we'll have the quantity, and here we want to have the curve that bounds a region from above, minus the curve that bounds a region from below. So the curve that bounds a region from above is going to be the natural log of X minus the curve that bounds the region from below, which is going to be -1. In quantity. Multiplied by DX. So, before we do any integration, we'll just simplify this expression. So this is going to be equal to the integral from 1 divided by E to 1 divided by 2 of the quantity of the natural log of X plus 1 in quantity DX. And so now we can just integrate this term by term. So that means our area is going to be equal to, and here for the first term, we're going to be integrating the natural log of X with respect to X, and recall that when you integrate the natural log of X with respect to X, you get X multiplied by the natural log of X. Minus X. And then we'll have plus, and for the next term, we're going to integrate 1 with respect to X, and we recall that when you integrate 1 with respect to X, you get X. Then that whole quantity has to be evaluated from 1 divided by E to 1 divided by 2. Now, before we substitute in our limits of integration, we're going to simplify our expression. Since when we combine the minus X term with the plus x term, that will combine to zero. So, simplify the expression here, this is going to be equal to X multiplied by the natural log of X, and this needs to be evaluated from 1 divided by E to 1 divided by 2. Now substituting in the limits of integration, this is going to be equal to. 1/2 multiplied by the natural log of 1/2. Minus 1 divided by E multiplied by the natural log of quantity of 1 divided by E in quantity. So we can use our properties of logarithms to simplify this expression. So, in doing so, this is going to be equal to minus. 1/2 multiplied by the natural log of 2. Plus 1 divided by E. And we can actually combine these two terms with a common denominator. In doing so, this is going to be equal to. 2 minus E multiplied by the natural log of 2, and then that quantity is going to be divided by the quantity of 2 multiplied by E. So this here is the answer that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

14–25. {Use of Tech} Areas of regions Determine the area of the given region.The region bounded by y = ln x,y = 1, and x = 1

Watch next

14–25. {Use of Tech} Areas of regions Determine the area of the given region.

The region bounded by y = ln x,y = 1, and x = 1