7. Let A(t) be the area of the region in the first quadra...

Question

7. Let A(t) be the area of the region in the first quadrant enclosed by the coordinate axes, the curve y=e^(-x), and the vertical line x=t, t>0. Let V(t) be the volume of the solid generated by revolving the region about the x-axis. Find the following limits.
a. lim(x→∞)A(t)

a. lim(x→∞)A(t)

Accepted Answer

Hello. In this video we are going to let a subt be the area of the region in the first quadrant bounded by the x and y axis. We have a vertical line defined as x is equal to t, where t is strictly a positive number greater than 0. And the curve y is equal to 3 e to the power of -2 x. We want to compute the limit ast approaches infinity of a of t. So let's go ahead and draw an image of the information that is given to us. So we have this region that is going to be bounded by the X and Y axis. One of the boundaries is defined as the curved y is equal to 3 e to the power of -2 x. That is going to be an exponential function that decreases from left to right and intercepts the point at 0.3. So this is the function Y is equal to 3 e to the power of -2 x. Now here, X is going to be a vertical line for some value of T greater than 0. So let's say that we let T equal to 2, for example, then we will have the line X is equal to 2. But here t is described as any positive number that is greater than 0, so we want to find the area when t is approaching positive infinity. Here we can go ahead and allow the lower boundary of the area to be defined as zero since it is bounded at the origin between the x and y axis, but the upper boundary is dependent on what value t we have. So the upper boundary of the of the integral is going to bet. So that means that we can set up our area function as the following. AFT is going to equal to the integral from 0 to T of the function 3e to the power of -2 XDX. Now that we've set up our area integral, we want to go ahead and approach this as an indefinite integral. Now again, t is approaching infinity, so we will have the limit ast approaches infinity. Of the integral from 0 tot of 3 e to the power of -2 x dx. Now, if we were to integrate, this will simplify to be the limit as T approaches infinity of -3 divided by 2 e to the power of -2 x, and this will be evaluated from 0 to T. By evaluating this area at each of the bounds, we will have the limit ast approaches infinity. Of -3 divided by 2 e to the power of -2 t. Minus. -3 divided by 2 e to the power of 0. Now this right hand side quantity. Is going to simplify to be positive 3 halves. And if we were to apply the limit to the left hand side, this left hand side will simplify to be zero. So that means that the final area ast approaches infinity is going to be 3 halves, and this is going to be the solution to the problem, with the overall solution being option B. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

7. Let A(t) be the area of the region in the first quadrant enclosed by the coordinate axes, the curve y=e^(-x), and the vertical line x=t, t>0. Let V(t) be the volume of the solid generated by revolving the region about the x-axis. Find the following limits.a. lim(x→∞)A(t)

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7. Let A(t) be the area of the region in the first quadrant enclosed by the coordinate axes, the curve y=e^(-x), and the vertical line x=t, t>0. Let V(t) be the volume of the solid generated by revolving the region about the x-axis. Find the following limits.
a. lim(x→∞)A(t)