- #1
philipc
- 57
- 0
just wonder if someone could help me solve this integral,
thanks
(e^-2x) / (1+e^-x)
thanks
(e^-2x) / (1+e^-x)
philipc said:just wonder if someone could help me solve this integral,
thanks
(e^-2x) / (1+e^-x)
The integral (e^-2x) / (1+e^-x) represents the area under the curve of the function (e^-2x) / (1+e^-x) from x=0 to x=∞.
2.The general formula for solving integrals involving exponential functions is: ∫e^ax dx = (1/a)e^ax + C, where a is a constant and C is the constant of integration.
3.To solve the integral (e^-2x) / (1+e^-x), we can use the substitution method. Let u = 1+e^-x, then du=-e^-x dx. Substituting into the integral, we get ∫(e^-2x) / (1+e^-x) dx = ∫(1/u) (-du) = -ln(u) + C = -ln(1+e^-x) + C.
4.The constant of integration represents the unknown value that is added to the result of the integral. This is because when we take the derivative of a constant, it equals 0. In the case of indefinite integrals, we must always include the constant of integration in the solution.
5.Since the function (e^-2x) / (1+e^-x) approaches 0 as x approaches ∞, the integral is convergent. This means that the area under the curve is finite and can be calculated using the given formula.