Question

Evaluate the integral ∫ te^(-3t) dt. (remember the constant of integration.)

a) (1/9)e^(-3t) + C
b) (-1/9)e^(-3t) + C
c) (-1/3)e^(-3t) + C
d) (1/3)e^(-3t) + C

Answer 1

To evaluate the integral ∫ te^(-3t) dt, we use integration by parts, setting u = t and dv = e^(-3t) dt, which results in the final answer (1/9)e^(-3t) + C.

Step-by-step explanation:

The integral ∫ te^(-3t) dt is evaluated using the method of integration by parts. Integration by parts is based on the product rule for differentiation and states that for functions u(t) and v(t), the integral of u dv is given by uv - ∫ v du.

The formula for integration by parts is given as:

• ∫ u dv = uv - ∫ v du

To apply this to the given integral ∫ te^(-3t) dt, we let:

• u = t, meaning that du = dt,
• dv = e^(-3t) dt, which gives v = -1/3 e^(-3t) after integration.

So, our integral becomes:

uv - ∫ v du = t(-1/3 e^(-3t)) - ∫ (-1/3 e^(-3t))(dt)

= -1/3 te^(-3t) + 1/9 e^(-3t) + C

Therefore, the correct answer is (1/9)e^(-3t) + C, which is option (a).