135. Find the area of the “triangular” region in the first qua...

Question

135. Find the area of the “triangular” region in the first quadrant that is bounded above by the curve y = e^(2x), below by the curve y = e^x, and on the right by the line x = ln(3).

Accepted Answer

Welcome back everyone. Find the area of the region in the first quadrant bounded above by y equals e to the power of 2 x plus e to the power of x. Below bye y equals e to the power of x + 1. And on the right by x equals eon of 3. A 4 minus eon of 3, B 4 plus eon of 3, C 9 halves minus 11 of 3, and D 9 halves plus eon of 3. So for this problem, let's remember that whenever we want to find the area bounded by two curves, we basically integrate the difference between the two functions, right? So we take the one that is Above and subtract the one that is below. So what we have to do is simply identify the limits of integration. We know that. The upper limit of integration is X equals 1 of 3 because it's on the right. So we want to find the intersection point on the left. And for that purpose, we're going to Set e to the power of 2 x plus e to the power of x. Equal to eat the power of x plus 1. Notice that we can immediately cancel out e to the power of x, which leaves us with e to the power of 2 x equals 1, and it basically means that x is equal to 0. What we can now do is define our area as the integral from 0, that's the lower limit of integration, up to ln of 3. Notice that indeed this is the first quadrant x greater than 0, and each of our functions. I also greater than 0, and then we take the difference. We take the function that is above, so that's e to the power of 2 x plus e to the power of x, and we subtract e to the power of x plus 1. Now let's go ahead and simplify, and then we can integrate. That's integral from 0 to a land of 3 of e to the power of 2 x minus 1 dx. And that we can integrate the integral of e to the power of 2 x is e to the power of 2 x divided by 2 according to the reverse chain rule, and the integral of minus 1 dx is basically a negative x, right? Now let's evaluate the result from 0 up to LN of 3 using the fundamental theorem of calculus, part 2. So we basically get one half we can factor out that constant e to the power of 2 ln of 3 minus. Alan of 3. Minus 0 we get 1/2 each to the power of 0 minus 0. Well done, and now let's apply the properties of logarithms to ln of 3 can be written as ln of 32, which is ln of 9. Right, and we can then get 1/2 multiplied by each to the power of eon of 9, which is simply 9. We're then subtracting eon of 3, subtracting 1/2 eats the power of 0, which is 1/2, and then we are basically adding 0, which is 0, right? From here, let's go ahead and simplify, so we get 9 divided by 2, which is 9 halves. 1/2 minus 1 of 3. And this is equal to 9 minus 1. That's 8 divided by 2 gives us 4 minus 1 of 3, right? So the correct answer is 4 minus 1 of 3 square units, which corresponds to the answer of choice a. Thank you for watching.

135. Find the area of the “triangular” region in the first quadrant that is bounded above by the curve y = e^(2x), below by the curve y = e^x, and on the right by the line x = ln(3).

Watch next