Integration accumulation and discounting

A flow rate is not a stock. Revenue arriving at 5 currency units per hour for three hours contributes 15 currency units. When the rate varies, an integral adds the small contributions. The same framework reconstructs total cost from marginal cost and values a stream under an explicitly chosen discount model.

1 Riemann sums and the definite integral

Partition $[a,b]$ into $a=t_0<\cdots<t_n=b$, choose sample points $\xi_i\in[t_{i-1},t_i]$, and form

$$\sum_{i=1}^n f(\xi_i)(t_i-t_{i-1}).$$

A bounded function is Riemann integrable if all such sums approach the same number as the largest subinterval length tends to zero. This number is $\int_a^b f(t)\,dt$. A continuous function on a closed bounded interval is integrable: uniform continuity makes the oscillation on every sufficiently short subinterval small, so upper and lower sums differ by at most $(b-a)$ times that oscillation.

The integral is signed accumulation. Below-axis contributions are negative; total unsigned area is $\int|f|$, not generally $|\int f|$. The average value over an interval of positive length is $(b-a)^{-1}\int_a^b f$.

2 The fundamental theorem

Theorem. If $f$ is continuous on $[a,b]$ and $F(x)=\int_a^x f(t)\,dt$, then $F'(x)=f(x)$ for $a<x<b$. If $G'=f$ on an interval containing $[a,b]$, then $\int_a^b f=G(b)-G(a)$.

Proof. The difference quotient for $F$ is the average of $f$ over the oriented interval from $x$ to $x+h$. Its difference from $f(x)$ is bounded by the largest $|f(t)-f(x)|$ on that interval, which tends to zero by continuity. Thus $F'=f$. Then $(G-F)'=0$, so the mean value theorem makes $G-F$ constant; evaluate at $a$ and $b$. $\square$

If marginal cost is $C'(q)=4+0.04q$ and fixed cost is $C(0)=100$, then

$$C(q)=100+\int_0^q(4+0.04s)\,ds=100+4q+0.02q^2.$$

The integration constant cannot be inferred from a marginal rate alone.

3 Substitution and integration by parts

If $u=g(t)$ and the functions have the needed continuous derivatives, the chain rule yields

$$\int_a^b f(g(t))g'(t)\,dt=\int_{g(a)}^{g(b)} f(u)\,du.$$

The transformed limits matter, including their order. The product rule similarly gives

$$\int_a^b u(t)v'(t)\,dt=[uv]_a^b-\int_a^b u'(t)v(t)\,dt.$$

These formulas do not justify ignoring a singularity inside the interval. For example, $\int_{-1}^1 1/t\,dt$ is not an ordinary Riemann integral, nor a convergent improper integral, even though a symmetric principal value exists.

4 A continuous discounting model

Assume a deterministic continuously compounded discount rate $r>0$ per year. A payment at time $t$ is multiplied by $e^{-rt}$. A continuous cash-flow rate $c(t)$ on $[0,T]$ then has model present value

$$V=\int_0^T c(t)e^{-rt}\,dt.$$

For constant $c$, integration gives $V=c(1-e^{-rT})/r$. Units check: currency/year times years is currency. As $r\to0$, the limit is $cT$, agreeing with undiscounted accumulation. For a nonnegative growing-rate model $c(t)=c_0e^{gt}$ with $c_0\ge0$, the formula is $c_0[1-e^{-(r-g)T}]/(r-g)$ when $r\ne g$, and $c_0T$ when $r=g$.

For $c_0>0$, the infinite-horizon integral converges exactly when $r>g$. In that case it equals $c_0/(r-g)$. If $c_0=0$, the flow and every finite- or infinite-horizon integral are zero for all rate values; the positive-flow convergence restriction does not apply. The conclusion follows by evaluating the finite integral and taking $T\to\infty$. If $r=g$ and $c_0>0$, it grows linearly; if $r<g$, it grows exponentially. These are conditional model calculations, not claims that a real cash flow or discount rate is known.

5 Graded practice

E1 Foundation. Find $C(q)$ when $C'(q)=3q^2+2$ and $C(0)=7$. Compute the cost accumulated from $q=1$ to $q=3$.

E2 Technique. Evaluate $\int_0^1 2t e^{t^2}\,dt$ and $\int_0^1 t e^t\,dt$.

E3 Application. A model flow is $c(t)=12e^{0.02t}$ currency units per year and the discount rate is 0.05 per year. Give its finite-$T$ and infinite-horizon model values.

E4 Interpretation. A net flow is $f(t)=2t-2$ on $[0,2]$. Find net accumulation and total absolute flow.

6 Full solutions

S1. $C(q)=7+q^3+2q$. The accumulated cost is $C(3)-C(1)=(7+27+6)-(7+1+2)=30$.

S2. With $u=t^2$, the first integral is $\int_0^1e^u\,du=e-1$. For the second, integration by parts with $u=t$, $v'=e^t$ gives $[te^t]_0^1-\int_0^1e^t\,dt=e-(e-1)=1$.

S3. $r-g=0.03$, so $V_T=400(1-e^{-0.03T})$ currency units. The infinite-horizon value is 400 because $r>g$. Mixing percentage numbers 5 and 2 with decimal rates would change the model incorrectly.

S4. The signed integral is $[t^2-2t]_0^2=0$. The absolute integral splits at $t=1$: $\int_0^1(2-2t)dt+\int_1^2(2t-2)dt=1+1=2$. Zero net flow need not mean no activity.

Summary

Integration accumulates a rate and reconstructs an antiderivative only up to an initial value. The fundamental theorem links these viewpoints. Discounted-flow formulas require explicit rate conventions, consistent time units and, for an infinite horizon, a convergence check.