how do
I solve for this V=πintegral from1 to 2 (10-e^x)^2dx
integral (100 - 20 e^x + e^2x ) dx ??
100 x - 20 e^x + .5e^2x ??? at x = 2 - at x = 1
but remember that π out front that I forgot.
To solve for the given integral V=π∫(10-e^x)^2 dx from 1 to 2, follow these steps:
Step 1: Simplify the integrand.
Start by expanding the expression (10-e^x)^2 using algebraic rules. Simplify and rewrite the expression as (100 - 20e^x + (e^x)^2).
Step 2: Evaluate the integral.
Now, integrate the simplified expression (100 - 20e^x + (e^x)^2) using the power rule of integration. For each term, apply the power rule to obtain an antiderivative.
The integral of 100 dx is 100x.
To integrate -20e^x, use the substitution u = e^x:
-20∫e^x dx = -20u.
After integrating, substitute back u = e^x:
-20e^x.
To integrate (e^x)^2, use the power rule:
∫(e^x)^2 dx = ∫e^(2x) dx.
Now, use the substitution v = 2x:
∫e^(2x) dx = ∫(1/2)e^v dv.
After integrating, substitute back v = 2x:
(1/2)e^(2x).
So the integral becomes:
V = π[100x - 20e^x + (1/2)e^(2x)] evaluated from 1 to 2.
Step 3: Apply the limits of integration.
Substitute the upper limit (2) and lower limit (1) into the expression obtained in step 2:
V = π[(100 * 2) - 20e^2 + (1/2)e^(4)] - [100 - 20e + (1/2)e^2].
Step 4: Simplify and calculate.
Calculate the value of the expression to obtain the final result.
So, by following these steps, you can solve for V in the given integral.