Linearity Of Accumulation Functions

Why Linearity Matters

An accumulation function records the net amount contributed by a rate between a fixed starting point and a variable endpoint. In many models, the total rate is made up of several components. For example, a total rate might combine production and loss, or contributions from several sources.

Linearity lets us calculate the accumulation of that combined rate using the accumulations of its components. The essential rules are additivity and constant scaling. The basepoint must be the same for every accumulation, and the scaling factors must be constants.

Accumulation Functions and Linearity

Let $J$ be an interval, let $a\in J$ be a fixed basepoint, and suppose $f$ is integrable on every closed subinterval of $J$. The accumulation function generated by $f$, with basepoint $a$, is the function $A_f$ defined by

A_f(x)=\int_a^x f(t)\,dt,\qquad x\in J.

The integration variable $t$ is a dummy variable: the input to the accumulation function is $x$, the variable endpoint. If $x<a$, the integral is oriented in the reverse direction, so $\int_a^x f(t)\,dt=-\int_x^a f(t)\,dt$.

Definition: For a fixed basepoint $a$, the accumulation function of a rate $f$ is $A_f(x)=\int_a^x f(t)\,dt$. When comparing or combining accumulation functions, the same basepoint is used unless stated otherwise.

If $f$ and $g$ are integrable on the relevant intervals and $c$ is a constant, then the accumulation function of $f+g$ is the sum of their accumulation functions, and the accumulation function of $cf$ is $c$ times the accumulation function of $f$. The coefficients are constants, not functions of the endpoint.

Theorem: Suppose $f$ and $g$ are integrable on every closed subinterval of $J$, $a\in J$, and $c,d\in\mathbb{R}$ are constants. For $x\in J$, define all accumulation functions using the same basepoint $a$. Then $$ A_{f+g}(x)=A_f(x)+A_g(x) \qquad\text{and}\qquad A_{cf}(x)=cA_f(x). $$ Consequently, $$ A_{cf+dg}(x)=cA_f(x)+dA_g(x). $$

Why the Rule Works

The rule follows from the linearity of the definite integral. The key idea can be seen from Riemann sums: on each subinterval, the sampled value of a sum of rates is the sum of the sampled values, and a constant multiple of a rate gives the same constant multiple of each sampled value. Taking the limit preserves these equalities.

Start with an endpoint to the right.
Fix $x\ge a$ and take a partition of $[a,x]$ with sample points $t_i^*$ and subinterval widths $\Delta t_i$.

Write the Riemann sum.
The sum for $f+g$ is $\sum_i (f(t_i^*)+g(t_i^*))\Delta t_i$. Distributing each term gives $\sum_i f(t_i^*)\Delta t_i+\sum_i g(t_i^*)\Delta t_i$.

Take limits.
Because $f$ and $g$ are integrable, these sums converge to their definite integrals. Thus $\int_a^x(f(t)+g(t))\,dt=\int_a^x f(t)\,dt+\int_a^x g(t)\,dt$. Likewise, the Riemann sum for $cf$ is $c\sum_i f(t_i^*)\Delta t_i$, so its limit is $c\int_a^x f(t)\,dt$.

Include endpoints to the left.
If $x<a$, apply the result on $[x,a]$ and use $\int_a^x h(t)\,dt=-\int_x^a h(t)\,dt$. The same linearity identities follow for this orientation.

This proves the two identities for every $x\in J$. Applying them successively gives the formula for $cf+dg$. No assumption that the rates are positive is needed: definite integrals are signed, so negative contributions are included automatically.

Worked Examples

Worked Example: Adding Two Rate Functions

Let $f(t)=t+2$ and $g(t)=3-t$, and use basepoint $0$. Find the accumulation function for the combined rate, then check it at $x=2$.

First add the rates:

$$ f(t)+g(t)=(t+2)+(3-t)=5. $$

Therefore the combined accumulation is

A_{f+g}(x)=\int_0^x 5\,dt=5x.

Now calculate the individual accumulation functions:

A_f(x)=\int_0^x(t+2)\,dt =\left[\frac{t^2}{2}+2t\right]_0^x =\frac{x^2}{2}+2x, $$ $$ A_g(x)=\int_0^x(3-t)\,dt =\left[3t-\frac{t^2}{2}\right]_0^x =3x-\frac{x^2}{2}.

Adding them gives

A_f(x)+A_g(x) =\left(\frac{x^2}{2}+2x\right) +\left(3x-\frac{x^2}{2}\right) =5x.

At $x=2$, the values are $A_f(2)=2+4=6$, $A_g(2)=6-2=4$, and $A_{f+g}(2)=5(2)=10$. The check is $6+4=10$, in agreement with the direct integral.

Worked Example: Scaling and Combining Rates

Let $r_1(t)=2t$ and $r_2(t)=t^2$, with basepoint $1$. Find the accumulation of $4r_1-3r_2$ at $x=3$.

By constant scaling and additivity,

A_{4r_1-3r_2}(3)=4A_{r_1}(3)-3A_{r_2}(3).

Compute each accumulation from the common basepoint:

A_{r_1}(3)=\int_1^3 2t\,dt =\left[t^2\right]_1^3 =3^2-1^2 =9-1 =8, $$ $$ A_{r_2}(3)=\int_1^3 t^2\,dt =\left[\frac{t^3}{3}\right]_1^3 =\frac{27}{3}-\frac{1}{3} =\frac{26}{3}.

Substitute these values:

4A_{r_1}(3)-3A_{r_2}(3) =4(8)-3\left(\frac{26}{3}\right) =32-26 =6.

Check directly by forming the combined rate, $4r_1(t)-3r_2(t)=8t-3t^2$, and integrating:

\int_1^3(8t-3t^2)\,dt =\left[4t^2-t^3\right]_1^3 =(36-27)-(4-1) =9-3 =6.

The direct calculation agrees with the linearity calculation.

Worked Example: A Linear Combination of Three Rates

Let $f(t)=2$, $g(t)=t$, and $h(t)=1-t$, all with basepoint $0$. Find the accumulation of $2f-g+3h$ at $x=2$.

First compute each accumulation:

A_f(2)=\int_0^2 2\,dt=[2t]_0^2=4, $$ $$ A_g(2)=\int_0^2 t\,dt =\left[\frac{t^2}{2}\right]_0^2 =\frac{4}{2}-0 =2, $$ $$ A_h(2)=\int_0^2(1-t)\,dt =\left[t-\frac{t^2}{2}\right]_0^2 =(2-2)-0 =0.

Use the same coefficients to combine these values:

A_{2f-g+3h}(2)=2A_f(2)-A_g(2)+3A_h(2) =2(4)-2+3(0) =6.

For a direct check, the combined rate is

$$ 2f(t)-g(t)+3h(t) =2(2)-t+3(1-t) =7-4t. $$

Its accumulation is

\int_0^2(7-4t)\,dt =\left[7t-2t^2\right]_0^2 =(14-8)-0 =6.

Thus the result from the individual accumulations matches the direct integral.

Conditions to Keep in View

Linearity is simple to apply, but its conditions matter. The functions being combined must be integrable on the intervals used, and their accumulation functions must refer to the same basepoint. Each coefficient must be constant with respect to the integration variable and the endpoint.

The rule does not say that an arbitrary factor depending on $t$ can be pulled outside the integral. For instance, with $f(t)=1$, basepoint $0$, and $x=2$,

\int_0^2 t f(t)\,dt =\int_0^2 t\,dt =2, \qquad 2\int_0^2 f(t)\,dt =2\int_0^2 1\,dt =4.

The incorrect step would be treating the variable factor $t$ as a constant. In general, a factor that varies across the interval remains inside the integral.

Common Pitfall: Mixing Basepoints

The identity $A_{f+g}=A_f+A_g$ assumes that all three accumulation functions use the same basepoint. A concrete example shows why this matters. Let $f(t)=g(t)=1$. Define $A_f$ using basepoint $0$, but define $A_g$ using basepoint $1$. At $x=2$,

A_f(2)=\int_0^2 1\,dt=2, \qquad A_g(2)=\int_1^2 1\,dt=1.

Adding these gives $3$. But the accumulation of $f+g=2$ from basepoint $0$ is

\int_0^2 2\,dt=4.

So the claim that this mixed-basepoint sum equals the accumulation of $f+g$ is false. To use linearity directly, first define every accumulation with the same basepoint.

Why the Rule Is Useful

A complicated rate can often be described as a sum of simpler rates, each multiplied by a constant. Linearity then allows the total accumulation to be assembled from the component accumulations. This can simplify exact calculations, organize models with multiple sources, and reveal cancellation between positive and negative contributions.

It also gives a useful interpretation: accumulation respects the same constant-weighted combinations as the rates themselves. If a rate is $cf+dg$, its accumulated value at each endpoint is $c$ times the accumulation of $f$, plus $d$ times the accumulation of $g$, provided the basepoint is shared.

Check Your Understanding

Use the same-basepoint condition in each question, and show the calculations that support your answers.

If $A_f(x)=\int_2^x f(t)\,dt$ and $A_g(x)=\int_2^x g(t)\,dt$, write the accumulation function for $5f-2g$.
For $f(t)=1$ and $g(t)=t$, with basepoint $0$, calculate $A_{f+g}(3)$ both directly and by adding $A_f(3)$ and $A_g(3)$.
Why must a coefficient in the scaling rule be constant with respect to the integration variable?
Let $f(t)=3t$ and $g(t)=t^2$, with basepoint $1$. Express $A_{2f+g}(x)$ in terms of $A_f(x)$ and $A_g(x)$.
Give one reason why accumulation functions with different basepoints cannot be combined using the linearity rule without adjustment.

Linearity Of Accumulation Functions

What You'll Learn

Why Linearity Matters

Accumulation Functions and Linearity

Why the Rule Works

Worked Examples

Worked Example: Adding Two Rate Functions

Worked Example: Scaling and Combining Rates

Worked Example: A Linear Combination of Three Rates

Conditions to Keep in View

Common Pitfall: Mixing Basepoints

Why the Rule Is Useful

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Linearity Of Accumulation Functions

What You'll Learn

Why Linearity Matters

Accumulation Functions and Linearity

Why the Rule Works

Worked Examples

Worked Example: Adding Two Rate Functions

Worked Example: Scaling and Combining Rates

Worked Example: A Linear Combination of Three Rates

Conditions to Keep in View

Common Pitfall: Mixing Basepoints

Why the Rule Is Useful

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Previous

Next