Evaluate the integral. (Remember the constant of integration.)
integral of te^(-4t)dt
To evaluate the integral of te^(-4t)dt, we can use integration by parts. The formula for integration by parts is ∫u dv = uv - ∫v du.
Let's choose u = t and dv = e^(-4t)dt.
Then, du = dt and v = -1/4 * e^(-4t).
Using the formula for integration by parts, we have:
∫te^(-4t)dt = (uv - ∫v du)
= t * (-1/4 * e^(-4t)) - ∫(-1/4 * e^(-4t)) dt
= -1/4 * te^(-4t) + 1/4 * ∫e^(-4t) dt.
Now, let's evaluate the remaining integral ∫e^(-4t) dt.
This integral is the integral of e^(-4t) with respect to t. We can evaluate it by using the formula for integrating exponential functions, which is ∫e^(kx) dx = (1/k) * e^(kx) + C.
In this case, k = -4, so we have:
∫e^(-4t) dt = (1/-4) * e^(-4t) + C.
Substituting this back into our original equation, we have:
-1/4 * te^(-4t) + 1/4 * ∫e^(-4t) dt
= -1/4 * te^(-4t) + 1/4 * [(1/-4) * e^(-4t) + C]
= -1/4 * te^(-4t) - 1/16 * e^(-4t) + C.
Therefore, the integral of te^(-4t)dt is given by:
∫te^(-4t)dt = -1/4 * te^(-4t) - 1/16 * e^(-4t) + C, where C is the constant of integration.