Question

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.

Answer 1

Answer:

I=\frac{t^2}{2}

Step-by-step explanation:

From exercise we have that x=t, t>0. Because A(t) be the area of the region in the first quadrant, we get that x started at 0. The limits for y are the following e-x and e. We get the integral :

I=\int\limits^0_t \int\limits^{e}_{e-x} 1 dy dx

I=\int\limits^0_t [y]_{e-x}^{e} dx

I=\int\limits^0_t (e-e+x) dx

I=\int\limits^0_t {x} \, dx

I=[\frac{x^2}{2} ]_{0}^{t}

I=\frac{t^2}{2}