Accumulation at a constant rate and of a step function

Integral calculus, grade 11 (5 units) Lesson 3 of 20Learning

Accumulation at a constant rate and of a step function

A constant rate

When a quantity changes at a constant rate, what accumulates is the rate times the time. In the example below a car drives along a straight road at a constant speed of 80 km/h:

The speed is the rate of change of the distance: how many kilometres are added every hour.
After 3 hours the car has covered $\displaystyle {80\cdot3}={240}$ km.
On the graph of the speed, $\displaystyle 80\cdot3$ is the area of the rectangle under the graph, from $\displaystyle {t}={0}$ to $\displaystyle {t}={3}$ .
After $\displaystyle t$ hours it has covered $\displaystyle {s(t)}={80t}$ km: the accumulated distance is a linear function of the time.

On the board below the $\displaystyle x$ -axis is the time in hours and the $\displaystyle y$ -axis is the speed in km/h:

\displaystyle v(x)=80

A step function

When the rate is constant in each stretch of time but changes from stretch to stretch, its graph is a step function, and the accumulation is the sum of the rectangles. In the example below a car drives for two hours at 60 km/h and then for an hour at 90 km/h:

In the first two hours: $\displaystyle {60\cdot2}={120}$ km.
In the third hour: $\displaystyle {90\cdot1}={90}$ km.
After 3 hours: $\displaystyle {120+90}={210}$ km – the area under the graph of the speed.

\displaystyle v(x)

The distance accumulated after $\displaystyle t$ hours rises with slope 60 until $\displaystyle {t}={2}$ , and then with slope 90 – steeper. The slope of the graph of the distance is the speed:

\displaystyle s(x)

A negative rate

A negative rate makes the accumulated quantity smaller, so an area below the $\displaystyle x$ -axis counts as negative. In the example below a cyclist rides for 5 hours at 25 km/h eastwards, and then for 3 hours at 15 km/h back westwards:

Take east as positive, so in the last three hours the velocity is $\displaystyle -15$ km/h.
In the first five hours he moved $\displaystyle {25\cdot5}={125}$ km away.
In the last three hours his distance from the starting point changed by $\displaystyle {(-15)\cdot3}={-45}$ km.

\displaystyle v(x)

The displacement – the position relative to the starting point – is the sum of the areas with their signs: $\displaystyle {125-45}={80}$ km.
The distance he travelled is the sum of the areas without the signs: $\displaystyle {125+45}={170}$ km.
On the graph of the displacement: a rise until $\displaystyle {t}={5}$ (a maximum of 125), and then a fall. The maximum is at the moment the velocity changed sign.

\displaystyle s(x)

Accumulation in other settings

The same idea works for any quantity whose rate of change is given:

Water in a pool: a rate of filling (m³ a minute) is positive, a rate of emptying negative; the area is the change in the amount of water.
Money in an account: income a week is positive, spending negative; the area is the change in the balance.
In every case: the accumulated quantity is the area between the graph of the rate and the $\displaystyle x$ -axis, with its sign.

Exercise 1

A tank held 10 litres of water.
In the first five minutes water flowed into it at 4 litres a minute.
In the next three minutes the tap was closed.
In the last six minutes water drained from it at 2 litres a minute.

\displaystyle r(x)

How much water was in the tank after 5 minutes?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

How much water was in the tank after 14 minutes?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

When was there the most water in the tank? Explain.

Exercise 2

A body moves along a line.
The board below shows its velocity, in metres a second, against the time in seconds.

\displaystyle v(x)

What is the body’s displacement after 10 seconds?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

What distance did the body travel in the 10 seconds?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

At what moment does the body return to its starting point?

Exercise 3

The rate is constant: $\displaystyle {f(x)}={2}$ .
$\displaystyle F(x)$ is the accumulation of $\displaystyle f$ from $\displaystyle {x}={1}$ to $\displaystyle x$ : the area under the graph of $\displaystyle f$ between 1 and $\displaystyle x$ .
Write $\displaystyle F(x)$ on the board.

\displaystyle f(x)=2

Formulas: integral calculus, grade 11 (5 units)

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Basic integrals

The indefinite integral – all the antiderivatives
$\displaystyle \int f(x)\,dx=F(x)+C$ $\displaystyle F^{\prime}(x)=f(x)$
A power
$\displaystyle \int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C$ $\displaystyle n\neq-1$
A constant
$\displaystyle \int k\,dx=kx+C$
$\displaystyle \int\frac{1}{x^{2}}\,dx=-\frac{1}{x}+C$
One over a root
$\displaystyle \int\frac{1}{\sqrt{x}}\,dx=2\sqrt{x}+C$
A root
$\displaystyle \int\sqrt{x}\,dx=\frac{2}{3}x\sqrt{x}+C$

The rules of integration

A constant factor
$\displaystyle \int k\cdot f(x)\,dx$ $\displaystyle =$ $\displaystyle k\int f(x)\,dx$
A sum and a difference
$\displaystyle \int\left(f(x)\pm g(x)\right)dx$ $\displaystyle =$ $\displaystyle \int f(x)\,dx\pm\int g(x)\,dx$
A linear inner function
$\displaystyle \int f(mx+b)\,dx=\frac{1}{m}F(mx+b)+C$ $\displaystyle m\neq0$
A composite function times the inner derivative
$\displaystyle \int f(u(x))\cdot u^{\prime}(x)\,dx=F(u(x))+C$
A power of a function times its derivative
$\displaystyle \int\left(u(x)\right)^{n}\cdot u^{\prime}(x)\,dx=\frac{\left(u(x)\right)^{n+1}}{n+1}+C$ $\displaystyle n\neq-1$

Trigonometric functions (in radians)

$\displaystyle \int\cos x\,dx=\sin x+C$
$\displaystyle \int\sin x\,dx=-\cos x+C$
$\displaystyle \int\frac{1}{\cos^{2}x}\,dx=\tan x+C$
From $\displaystyle {\cos2x}={1-2\sin^2x}$
$\displaystyle \sin^{2}x$ $\displaystyle =$ $\displaystyle \frac{1-\cos2x}{2}$
From $\displaystyle {\cos2x}={2\cos^2x-1}$
$\displaystyle \cos^{2}x$ $\displaystyle =$ $\displaystyle \frac{1+\cos2x}{2}$
From $\displaystyle {\sin2x}={2\sin x\cos x}$
$\displaystyle \sin x\cos x$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sin2x$
From $\displaystyle {\sin^2x+\cos^2x}={1}$
$\displaystyle \tan^{2}x$ $\displaystyle =$ $\displaystyle \frac{1}{\cos^{2}x}-1$

The accumulation function and the definite integral

The accumulation function of $\displaystyle f$ from $\displaystyle a$
$\displaystyle F_{a}(x)=\int_{a}^{x}f(t)\,dt$
The fundamental theorem
$\displaystyle F_{a}^{\prime}(x)=f(x)$
The definite integral, $\displaystyle F$ an antiderivative of $\displaystyle f$
$\displaystyle \int_{a}^{b}f(x)\,dx=F(b)-F(a)$
The same limit
$\displaystyle \int_{a}^{a}f(x)\,dx=0$
Swapping the limits
$\displaystyle \int_{b}^{a}f(x)\,dx$ $\displaystyle =$ $\displaystyle -\int_{a}^{b}f(x)\,dx$
Joining intervals
$\displaystyle \int_{a}^{b}f(x)\,dx+\int_{b}^{c}f(x)\,dx$ $\displaystyle =$ $\displaystyle \int_{a}^{c}f(x)\,dx$

Areas and accumulation

An area above the $\displaystyle x$ -axis
$\displaystyle S=\int_{a}^{b}f(x)\,dx$ $\displaystyle f(x)\geq0$
An area below the $\displaystyle x$ -axis
$\displaystyle S=-\int_{a}^{b}f(x)\,dx$ $\displaystyle f(x)\leq0$
The area between two graphs
$\displaystyle S=\int_{a}^{b}\left(f(x)-g(x)\right)dx$ $\displaystyle f(x)\geq g(x)$
The displacement from the velocity
$\displaystyle s(b)-s(a)=\int_{a}^{b}v(t)\,dt$