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=e^−2x, and x=ln 4

Accepted Answer

Welcome back, everyone. Find the area of the region bounded by Y equals eats the power of X, Y equals eats the power of negative X, and X equals LN 5. What we want to do for this problem is simply draw a very rough sketch, right? So let's begin by doing that. Let's introduce a coordinate plane. And let's draw exponential functions E rates the power of X and E raises the power of negative X. Understanding that if X is equal to 0, then each Y is equal to 1. So they share a common point at x equals 0 Y equals 1. Now, the main difference is that an exponential function, Y equals E raises the power of X is increasing, right, with an increase in X, while The exponential function E raises the power of negative X is decreasing. So what we can do is simply draw a rough sketch of that. And now we are going to also draw a vertical line X equals of 5, right? That would be our next function to consider. And now we can define our region. It is bounded by two curves. Up until we reach the vertical line X equals LN of 5, right? We already know that our intersection point is at X equals 0 and then our region is bounded up to X of LN 5. What we can do is simply identify our area as the integral from the intersection point of 0 up to. X equals a line of 5. That's our upper limit of integration. And now let's recall that what we have to do is simply take the upper curve, which is E raises the power of X, and subtract the lower curve, which is E raises the power of negative X to obtain our integra. And that's the whole setup. We have our integral from 0 up to LN 5, of E to the power of X minus E to the power of negative XDX. We're integrating with respect to X. And now let's go ahead and integrate the integral of eats the power of axes, eats the power of X, and we're subtracting the integral of eats the power of negative X, right, which is negative E raises the power of negative X. And now let's evaluate the result from 0 up to 11 of 5. We can also simplify that E er raises the power of X plus E erase the power of negative X. Evaluated from 0 up to 11 of 5. Let's evaluate the result at the upper limit, that's E to the power of LN of 5 plus E to the power of negative LN of 5, and then we subtract. E raised the power of 0 plus E raise the power of negative 0. Basically, again, it's the power of 0, right? Let's go ahead and simplify. Ease the power of Elan 5 is simply 5. E raises the power of negative. LN 5 can be written as E raises the power of LN of 1/5. So we're basically adding 1/5, and then we're subtracting. Eats the power of 0 and again it's the power of 0, right? So we're subtracting 1 + 1 which is 2. That's, I'm sorry, that's 5 minus 2, which is 3. Plus 15th. We can find the common denominator, which is 53 multiplied by 5, plus 1 is 16. So we end up with 16 divided by 5 square units. That'll be the area of the region bounded by those curves. Thank you for watching.

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

The region bounded by y=e^x, y=e^−2x, and x=ln 4

Key Concepts

Understanding the Region Bounded by Curves

Definite Integration for Area Calculation

Properties of Exponential Functions

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