find the volume of the solid whose base is bounded by y=e^(-x), y=3cos(x), and x=0 and whose cross sections cut by planes perpendicular to the x-axis are squares

the answer is 3.992 units cubed but can someone explain to me how to get this answer using integrals?

To find the volume of the solid, we need to integrate the area of each cross section as it varies along the x-axis.

First, let's consider a small interval between x and x + dx. The width of the cross section at this interval is simply dx. The length of the cross section can be determined by finding the distance between the curves y = e^(-x) and y = 3cos(x) at the given x-value.

To find the length, we subtract the value of y = e^(-x) from y = 3cos(x). Therefore, the length of the cross section is given by L(x) = 3cos(x) - e^(-x).

Now, we can calculate the area of each cross section A(x) by squaring the length: A(x) = [3cos(x) - e^(-x)]^2.

To determine the total volume, we need to integrate the area function A(x) over the interval [0, c], where c is the x-value where the curves intersect.

So, the volume (V) is given by the integral:

V = ∫[from 0 to c] [3cos(x) - e^(-x)]^2 dx

Evaluating this integral will give us the volume of the solid.

Using numerical methods, this integral evaluates to approximately V = 3.992 units cubed.