Consider the region bounded by the graphs of
y = ln(x), ...

Question

Consider the region bounded by the graphs of

y = ln(x), y = 0, and x = e.

a. Find the area of the region.

Accepted Answer

Welcome back everyone. I'm the area of the shaded region. In this problem, we want to find the area of the region bounded by the x axis. The curve y equals ln of 3 x and the vertical line x equals e. So let's remember that we can find the area as the integral from x1 up to x2 of the absolute value of f of x. D X In this problem, Notice that our region is above the x axis, right? So we are going to drop the absolute value and simply consider our function y equals ln of 3 x, which is f of x. What we want to do is understand that the upper limit of integration is x equals e based on the image. This is x2, and we're going to find x1, which is the intersection point between f x and the x axis. What we want to do is set fx equal to 0 because this is. The place where the y coordinate is equal to 0 and it essentially means that ln of 3 x is going to be equal to 0. Taking the anti log of each side, we can show that 3 x is e to the power of 0. Which is basically 1 and therefore x is equal to 1 divided by 3, and that's 1/3. So x1 is going to be 1/3. Therefore we can now show that the area is equal to the integral from 1/3 up toe off. LM of 3 XD X. Well, once we have our integral set up, we simply want to integrate. Now, what we can do is simply use the formula. Which Tells us that the integral of ln of a x d x is equal to x multiplied by ln of a x minus x plus a constant of integration c. This is the result from integration by parts. And using this formula. We can show that A is equal to 3. This is the coefficient in front of x. So we're going to have x, L, and of. 3 x minus x, right? Specifically, we're not introducing a constant of integration because we have a definite integral. We want to evaluate the result between 1/3 and e. Let's go ahead and do that. And in addition, we can factor out X, right? So we're going to have X. In parent ln of 3 x minus 1. Evaluated from 1/3 up to E. Which gives us e multiplied by ln of 3 e. 1 minus 1/3. In parentheses een of 3 multiplied by 1/3 minus 1. Well then, so from here we are going to simplify. In order to simplify. Let's use our first expression as follows, that would be e multiplied by in parentheses we have ln 3e, right? And using the properties of logarithms, we can rewrite it as ln of 3 plus ln ofe because we have a product. We are subtracting 1, and remember that L of E is 1, right? We get 1 minus 1, which is 0, we can cross out those two terms. And for the next part we are subtracting 1/3 in parentheses ln of 3/3, which is ln of 1. And Alan of 1 is equal to 0, right? -1, so we have 1/3 multiplied by -1. This is equal to e multiplied by ln of 3. 1/3, which is + 1/3. And this is our final answer. We can add square units since this is area. And we're going to label our final answer, which is ELN of 3. Plus 1/3 square units. Thank you for watching.

Consider the region bounded by the graphs ofy = ln(x), y = 0, and x = e.a. Find the area of the region.

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Consider the region bounded by the graphs of
y = ln(x), y = 0, and x = e.
a. Find the area of the region.