Find the volume of revolution when the region enclosed by the functions f(x)= e^x , g(x)=e^-x ,and the two lines is revolved about the x-axis

Solution
V=intergrate e^2x-e^-2x dx
From -1to1
1/2e^+e^-2x when you sub and cal. =0 why please help me

you don't say which two lines, but I infer that they are x = -1 and x=1.

The region is symmetric about the y-axis, so we can just double the area from 0 to 1.

using discs,
v = 2∫[0,1] π(R^2-r^2) dx
where R=e^x and r=e^-x
v = 2∫[0,1] π(e^2x - e^-2x) dx
This is just what you had.

Now, you know that
∫ e^u du = e^u
if we let
u = 2x
du = 2dx
but we only have dx, which is du/2. So,

∫e^(2x) dx = 1/2 ∫ e^u du

That's where the 1/2 comes from, and why the - changes to +, because you are multiplying by -1/2 in the 2nd term.

Not sure what the =0 problem is. e^0 = 1, though, if that helps.

To find the volume of revolution when the region enclosed by the functions f(x)= e^x, g(x)=e^-x, and the two lines is revolved about the x-axis, you need to use the method of cylindrical shells.

First, determine the interval of integration. From your question, it seems that the interval is from -1 to 1.

Next, calculate the volume using the formula:

V = ∫[a,b] 2πxf(x)dx

However, in this case, we are supposed to revolve the region about the x-axis, so we need to modify the formula slightly:

V = ∫[a,b] 2πx|f(x)-g(x)|dx

Therefore, the volume of revolution is given by:

V = ∫[-1,1] 2πx(e^x-e^-x)dx

Now, integrating this expression:
V = 2π ∫[-1,1] (x * e^x - x * e^-x)dx

To evaluate this integral, we can use the integration by parts method. Integrating by parts involves splitting the integral into two parts and applying the product rule of differentiation.

Let's consider the first part of the integral:

∫ (x * e^x)dx

To integrate this, we use integration by parts by choosing u = x and dv = e^x dx.
Taking the derivatives and evaluating, we have du = dx and v = e^x. Applying the integration by parts formula:

∫ (x * e^x)dx = x * e^x - ∫ e^xdx

Note that the resulting integral is easier to solve. Integrating e^x gives us:

∫ (x * e^x)dx = x * e^x - e^x + C1

Now let's consider the second part of the integral, which is:

-∫ (x * e^-x)dx

We use integration by parts again by choosing u = x and dv = e^-x dx.
Taking the derivatives and evaluating, we have du = dx and v = -e^-x. Applying the integration by parts formula:

-∫ (x * e^-x)dx = -x * e^-x + ∫ e^-xdx

Again, this integral is easier to solve. Integrating e^-x gives us:

-∫ (x * e^-x)dx = -x * e^-x - e^-x + C2

Now, substitute these values back into the original expression:

V = 2π [(x * e^x - e^x) - (-x * e^-x - e^-x)] evaluated from -1 to 1

Now, plug in the upper limit (1) into the equation:

V = 2π [(1 * e^1 - e^1) - (-1 * e^-1 - e^-1)]

We can simplify this further:

V = 2π [e - e^1 + e^-1 - e^-1] = 2π (0) = 0

So, the resulting volume of revolution is 0.