Accumulation problems: motion, flow, a parameter and an extremum

The velocity $\displaystyle v(t)$ is the rate of change of the position $\displaystyle s(t)$, so the integral of the velocity is the change in position – the displacement. The distance a body travelled is the area between the graph of the velocity and the axis, without sign. In the example below a body moves with velocity $\displaystyle {v(t)}={t^2-4t+3}$ metres a second, $\displaystyle 0\le t\le4$:

• The displacement: $\displaystyle {\int_0^4\left(t^2-4t+3\right)dt}={\left[\frac{t^3}{3}-2t^2+3t\right]_0^4}={\frac{64}{3}-32+12}={\frac{4}{3}}$ metres.
• The velocity changes sign at $\displaystyle {t}={1}$ and at $\displaystyle {t}={3}$: the body moves forward, goes back, and moves forward again.
• $\displaystyle {\int_0^1v\,dt}={\frac{4}{3}}$, $\displaystyle {\int_1^3v\,dt}={-\frac{4}{3}}$, $\displaystyle {\int_3^4v\,dt}={\frac{4}{3}}$.
• The distance: $\displaystyle {\frac{4}{3}+\frac{4}{3}+\frac{4}{3}}={4}$ metres.

In the example below water flows into a tank for 10 minutes, at the rate $\displaystyle {f(t)}={5-0.5t}$ litres a minute:

• The amount of water that flowed in during the first $\displaystyle x$ minutes: $\displaystyle {W(x)}={\int_0^x\left(5-0.5t\right)dt}={5x-\frac{x^2}{4}}$.
• In the 10 minutes, $\displaystyle {W(10)}={50-25}={25}$ litres flowed in.
• $\displaystyle {W'(x)}={5-0.5x}$: the derivative of the amount is the rate – the fundamental theorem.
• The rate falls, so more flows in during the first 5 minutes than in the last five: $\displaystyle {W(5)}={18.75}$, and $\displaystyle {25-18.75}={6.25}$.

When the area is given and a limit is not known, the integral gives an equation. In the example below the area under the graph of $\displaystyle {y}={x^2}$ between $\displaystyle {x}={0}$ and $\displaystyle {x}={k}$ ($\displaystyle k>0$) is 9:

• $\displaystyle {\int_0^kx^2\,dx}={\frac{k^3}{3}}$.
• $\displaystyle {\frac{k^3}{3}}={9}$, so $\displaystyle {k^3}={27}$ and $\displaystyle {k}={3}$.

An area one of whose limits changes is a function, and it can have an extremum. The fundamental theorem gives the derivative without calculating the integral. The example below finds the $\displaystyle k$ for which the area under the graph of $\displaystyle {f(x)}={4x-x^2}$ over $\displaystyle [k,k+1]$, $\displaystyle 0\le k\le3$, is greatest:

• $\displaystyle {A(k)}={\int_k^{k+1}f(x)\,dx}={F(k+1)-F(k)}$, where $\displaystyle F$ is an antiderivative of $\displaystyle f$.
• $\displaystyle {A'(k)}={f(k+1)-f(k)}={4(k+1)-(k+1)^2-4k+k^2}={3-2k}$.
• $\displaystyle {A'(k)}={0}$ at $\displaystyle {k}={1.5}$; before it $\displaystyle A'>0$, and after it $\displaystyle A'<0$: a maximum.
• The piece $\displaystyle [1.5,2.5]$ is symmetric about $\displaystyle {x}={2}$, the vertex of the parabola – the piece where the graph is highest.

A body moves along a line with velocity $\displaystyle {v(t)}={6-2t}$ metres a second, $\displaystyle 0\le t\le5$.

What is the body’s displacement after 5 seconds?

What distance did the body travel in the 5 seconds?

A shopping centre opens at 9:00.
The board below shows the rate at which people come in, in people a minute, against the time in minutes since it opened.

How many people came into the shopping centre by 12:00?

The area enclosed by the graph of $\displaystyle {y}={x^2}$ and the line $\displaystyle {y}={kx}$, $\displaystyle k>0$, is 36.

The graphs meet at $\displaystyle {x}={0}$ and at $\displaystyle {x}={k}$. Write the area as an expression in $\displaystyle k$.

Find $\displaystyle k$.

The function $\displaystyle {f(x)}={6x-x^2}$ is given.
$\displaystyle A(k)$ is the area under the graph over $\displaystyle [k,k+2]$, $\displaystyle 0\le k\le4$.

Write $\displaystyle A(k)$ without an integral.

Find the $\displaystyle k$ for which $\displaystyle A(k)$ is greatest: solve $\displaystyle {A'(k)}={0}$.