Independent solution

How to solve this Continuous Cash Flows question

Setup

Setup

Let dC(t) be the payment made during a short interval at time t. A payment made at t accumulates for 4 minus t years.

dC(t)=100e0.5t dtdC(t)=100e^{0.5t}\,dt
a(t,4)=e0.08(4−t)a(t,4)=e^{0.08(4-t)}

Model

Model

Integrate the accumulated value of every payment over the deposit window; the exponential terms combine before integration.

V4=∫13100e0.5te0.08(4−t) dtV_4=\int_1^3 100e^{0.5t}e^{0.08(4-t)}\,dt
V4=100e0.32∫13e0.42t dtV_4=100e^{0.32}\int_1^3e^{0.42t}\,dt

Compute

Compute

Evaluating the antiderivative at the two endpoints gives an account value just below 657.

V4=100e0.320.42(e1.26−e0.42)V_4=\frac{100e^{0.32}}{0.42}\left(e^{1.26}-e^{0.42}\right)
V4=656.9096V_4=656.9096

Answer

Answer

Rounded to the nearest listed amount, the accumulated value is 657, so the keyed answer is E.

V4≈657(E)\boxed{V_4\approx657\quad\text{(E)}}