Exercises 83–86 are about the infinite region in the first qua...

Question

Exercises 83–86 are about the infinite region in the first quadrant between the curve y = e^(-x) and the x-axis.

83. Find the area of the region.

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find the area of the region in the first quadrant bounded by the curve y is equal to 3 multiplied by e to the power of 3 x and the x axis. OK, so it appears for this particular prong we're ultimately asked to find what the area is of the region that's located in the first quadrant that is bounded specifically by this curvey and the x axis. So now that we know what we're ultimately trying to solve for, we're trying to solve for this area value, our first step that we need to take is we need to note that the region is bounded by the curve. Y is equal to 3 multiplied by e to the power of -3 x, and we also need to note that it's bounded by the x axis, which will mean that y will be equal to 0. And for the y axis, that will mean that X is equal to 0. And since the curve y is equal to 3 multiplied by e to the power of 3 x is always positive for all x, that will mean that the area which will denote as A extends infinitely to the right from x is equal to 0 to positive infinity. So that means based on our prior knowledge we could write the formula for the area as specifically the area will be given by the improper integral A. The area A is equal to the integral. Evaluated from 0 to positive infinity of 3 multiplied by e to the power of -3 XDX. Fantastic, we are on a roll. And we will need to evaluate the improper integral by replacing the infinite limit with a variable which we'll call T. And by taking the limit as T approaches positive infinity, so you go ahead and write that A, the area is equal to the limit as T approaches positive infinity of the integral evaluated from 0 tot of 3 multiplied by e to the power of -3 xDX. Fantastic. So now, Our first step that we need to take at this point is we need to evaluate the definite interval. So now that we have our expression for the area, the areas that we're ultimately trying to solve for, our first step now, once again, is to evaluate the definite interval. So in order to do that, based on what we should be able to recognize, we can use a simple U substitution. For the indefinite integral, the integral of 3 multiplied by e to the power of -3 x dx. So using our U substitution method, we can say that you will be equal to -3 x. So that will mean that DU by DX will be equal to -3. Which will mean that 1/3 DU will be equal to VX. Awesome. So what we need to do now is we need to take the integral. So we're taking the integral of. 3 multiplied by e to the power of 3 x dx, which will be equal to the integral of e to the power of u multiplied by negative duu, which will be equal to the integral of e to the power of u d u which will be equal to e to the power of u plus c where c represents some constant variable. And when we substitute back in our value for U is equal to -3X, we can go ahead and write that the integral of 3 multiplied by e to the power of 3XDX will be equal to e to the power of -3X + C. So now at this point, we need to evaluate our definite integral. So we need to take the integral evaluated from 0 tot of 3 multiplied by e to the power of -3 xDX, fantastic, which will be equal to parentheses of e to the power of 3x. Evaluate from 0 to T. Which will be equal to the power of -3T. In parentheses minusses of e to the power of 3 multiplied by 0. Which when we perform this evaluation, we will find that it's simply equal to e to the power of 3 T. Plus 1 fantastic. So we are on a roll here. We are almost done, I promise. So now our second step now is we need to take the limit. So we will now solve for the area a by taking the limit once again ast approaches infinity. So the area a is equal to the limit ast approaches infinity. Of 1 minus e to the power of -3t. Which will mean that A is equal to 1 minus the limit as T approaches infinity of e to the power of -3 T. And we need to note once again that as T approaches infinity, that will mean that the exponent -3T will approach negative infinity. So therefore, as a result, e to the power of -3 T. Will be equal to 1 divided by. E, so it's gonna be 1 divided by E to the power of 3 T, and this will approach 0. So that means we could safely say that the limit as T approaches 0 of E to the power of -3T will be equal to 0. And we need to substitute this limit back into our expression, so that will mean that A, the area is equal to 1 minus 0, so that will mean that A is simply equal to 1, and our units will be square units because for the area we deal with square units, so A will be equal to 1 square unit. And that's it. That is our final answer. Hooray, we did it, so. Let's put a box around our final answer so we can identify it. So once again, based on what we've learned together, we could say that our final answer for the area of this specific region is once again one square unit, and that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Exercises 83–86 are about the infinite region in the first quadrant between the curve y = e^(-x) and the x-axis.83. Find the area of the region.

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Exercises 83–86 are about the infinite region in the first quadrant between the curve y = e^(-x) and the x-axis.
83. Find the area of the region.