Simple applications

A change of variable, not of method. Up to now $x$ has always been the variable of integration. In applications it is often more natural to integrate with respect to whatever variable the problem is phrased in — commonly time, denoted $t$, in problems of motion. The method is unchanged; only the letter changes.

The core idea: recovering a function from its rate of change. This section is about using integration, not developing new integration technique. Given the derivative of an unknown function — described in words via a rate, a growth/decay rate, a marginal quantity, or a phrase like "varies", "increases", "decreases" — the task is to integrate that derivative to recover the original function, and then use a given initial condition to pin down the one particular antiderivative the problem actually wants (exactly the mechanism of Theorem 11.1, §11.2).

Illustration 1 — recovering $f(x)$ from $f'(x)$ and a point condition. If $f'(x)=3x^2-4x+5$ and $f(1)=3$, then integrating both sides with respect to $x$:

Substituting the known point, $f(1)=3 \Rightarrow (1)^3-2(1)^2+5(1)+c=3 \Rightarrow c=-1$, so $f(x)=x^3-2x^2+5x-1$.

Illustration 2 — velocity to distance (a train's journey). A train leaves Madurai Junction at 3pm ($t=0$) with velocity $v(t)=20t+50$ km/h. Since velocity is the rate of change of position, $\dfrac{ds}{dt}=20t+50$, so

The distance covered is zero when $t=0$, so $c=0$, giving $s=10t^2+50t$. At 5pm, $t=2$ hours, so $s=10(2)^2+50(2)=140$ km.

Illustration 3 — a two-variable rate (weight as a function of height). If the rate of change of a person's weight $w$ (kg) with respect to height $h$ (cm) is $\dfrac{dw}{dh}=4.364\times10^{-5}h^2$, integrating gives $w=4.364\times10^{-5}\!\left(\dfrac{h^3}{3}\right)+c$. Since weight is zero when height is zero, $c=0$; substituting $h=150$ gives $w\approx49$ kg.

Illustration 4 — a fractional-power rate (tree growth) and solving the resulting equation both ways. A tree's height increases at $\dfrac{dh}{dt}=\dfrac{18}{\sqrt t}=18t^{-1/2}$ cm/year, with $h=5$ at $t=0$. Integrating, $h=18\!\left(2t^{1/2}\right)+c=36\sqrt t+c$; the initial condition gives $c=5$, so $h=36\sqrt t+5$. This one relation now answers two different kinds of question: substitute a known $t$ to get $h$ (at $t=4$: $h=36\sqrt4+5=77$ cm), or substitute a known $h$ and solve for $t$ (at $h=149$: $\sqrt t=\frac{149-5}{36}=4\Rightarrow t=16$ years).

Illustration 5 — a full accel → velocity → position chain (the braking-bike problem of §11.1, solved). Take the direction of motion as positive; a slowing bike has acceleration in the opposite sense, i.e. retardation, so $a=\dfrac{dv}{dt}=-8\text{ m/s}^2$. Integrating once,

at the instant braking starts, $t=0,\ v=24$, giving $c_1=24$, so $v=-8t+24$. Since $v=\dfrac{ds}{dt}$, integrate once more:

$s = \int(-8t+24)\,dt = -4t^2+24t+c_2;$ …