Integration: the reverse of differentiation

Three situations where a rate of change decides how much has collected: a tank filling, a cup of tea cooling and a pulsing pump. The pictures come first; the working in each solution comes from the same numbers.

$50\text{ litres at }t = 0$$110\text{ litres at }t = 2$$210\text{ litres at }t = 4$$\text{front view}$

Maths In Real Life 1

A tank filling from a tap

A tap does not always run at a steady speed. If you know how fast the volume is changing at every moment, integrating that rate gives the volume in the tank at any time.

Water flows from a tap into a plain open tank with straight vertical sides. The volume of water in the tank is $V$ litres at time $t$ minutes after the tap is opened, and $\dfrac{dV}{dt} = 10 t + 20$, measured in litres per minute. The tank holds $50$ litres of water when $t = 0$.

• (a)Find $V$ in terms of $t$, and the volume of water in the tank when $t = 4$.
• (b)Find the volume of water added to the tank between $t = 2$ and $t = 4$.

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Colour keythe rule applied to a termthe antiderivative or final result

(a)

$\displaystyle V\;= \int \left(10 t + 20\right)\,dt$

A rate of change is a derivative, so reverse it: $V$ is the integral of $\frac{dV}{dt} = 10 t + 20$.

$\displaystyle \int 10 t\,dt\;= \textcolor{#010105}{\frac{10t^{1 + 1}}{1 + 1}}$

Add $1$ to the power, then divide by the new power. The constant multiplier stays in front.

$\displaystyle = \textcolor{#010106}{5 t^{2}}$

Simplify.

$\displaystyle \int 20\,dt\;= \textcolor{#010106}{20 t}$

The integral of a constant $k$ is $kt$.

$\displaystyle V\;= \textcolor{#010106}{5 t^{2} + 20 t} + c$

Add the integrals of the terms and write one constant of integration $c$ at the end.

$\displaystyle V\;= 5 t^{2} + 20 t + c$

The integral with its constant $c$.

$\displaystyle 50 = 5 \times \left(0\right)^{2} + 20 \times \left(0\right) + c$

Use the starting value: $V = 50$ when $t = 0$.

$\displaystyle c\;= \textcolor{#010106}{50}$

Solve for $c$.

$\displaystyle V\;= \textcolor{#010106}{5 t^{2} + 20 t + 50}$

Put the constant back.

$\displaystyle V\;= 5 \times \left(4\right)^{2} + 20 \times \left(4\right) + 50$

Substitute $t = 4$.

$\displaystyle = \textcolor{#010106}{210}$

Evaluate.

(b)

$\displaystyle \int_{2}^{4} \left(10 t + 20\right)\,dt = \left[5 t^{2} + 20 t\right]_{2}^{4}$

Integrate with the usual rules but leave out the constant $c$: it cancels in a definite integral.

$\displaystyle = \left(5 \times \left(4\right)^{2} + 20 \times \left(4\right)\right) - \left(5 \times \left(2\right)^{2} + 20 \times \left(2\right)\right)$

Substitute the upper limit $4$, then the lower limit $2$, and subtract (upper minus lower).

$\displaystyle = \textcolor{#010106}{100}$

Evaluate.

Answer

(a) $V = 5 t^{2} + 20 t + 50$; $V = 210$ litres

(b) $100\text{ litres}$

$\text{80 °C at pouring}$$\text{front view}$

Maths In Real Life 2

A cup of tea cooling

A hot drink loses heat fastest at the start and more and more slowly as it nears the temperature of the room. The rate of cooling is known, and integrating it recovers the temperature itself, with the starting temperature fixing the constant.

Not in G2 (G3 Additional Maths only): this uses the exponential $e^{ax + b}$. A cup of tea is poured at time $t = 0$, when its temperature is $80$ °C. At $t$ minutes after pouring, the temperature $T$ °C is changing at a rate of $\dfrac{dT}{dt} = - \frac{5 e^{- \frac{t}{20}}}{2}$, measured in °C per minute.

• (a)Find $T$ in terms of $t$, and hence the temperature of the tea $20$ minutes after it is poured. Give the exact value and then the value to 1 decimal place.

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Colour keythe antiderivative or final result

(a)

$\displaystyle T\;= \int - \frac{5 e^{- \frac{t}{20}}}{2}\,dt$

A rate of change is a derivative, so reverse it: $T$ is the integral of $\frac{dT}{dt} = - \frac{5 e^{- \frac{t}{20}}}{2}$.

$\displaystyle = \textcolor{#010106}{50 e^{- \frac{t}{20}}} + c$

The integral of $e^{ax+b}$ is $\frac{1}{a}e^{ax+b}$: divide by the coefficient of $x$, which is $- \frac{1}{20}$. The constant multiplier stays in front.

$\displaystyle T\;= 50 e^{- \frac{t}{20}} + c$

The integral with its constant $c$.

$\displaystyle 80 = 50 \times e^{- \frac{\left(0\right)}{20}} + c$

Use the starting value: $T = 80$ when $t = 0$.

$\displaystyle c\;= \textcolor{#010106}{30}$

Solve for $c$.

$\displaystyle T\;= \textcolor{#010106}{30 + 50 e^{- \frac{t}{20}}}$

Put the constant back.

$\displaystyle T\;= 30 + 50 \times e^{- \frac{\left(20\right)}{20}}$

Substitute $t = 20$.

$\displaystyle = \textcolor{#010106}{50 e^{-1} + 30}$

Evaluate.

$\displaystyle \approx 48.4$

Round at the very end, to 1 decimal place.

Answer

(a) $T = 30 + 50 e^{- \frac{t}{20}},\ \ t \ge 0$; $T = 50 e^{-1} + 30 \approx 48.4$ °C (1 d.p.): the tea has cooled but is still hotter than the temperature it is heading towards

$\text{after 3 min}$$\text{after 6 min}$$\text{front view}$

Maths In Real Life 3

A pulsing garden pump

A small pump delivers water in pulses: the flow rises to a peak and falls back to zero. The flow rate is known, so integrating it gives the volume delivered, and a definite integral gives the volume delivered over a whole pulse.

Not in G2 (G3 Additional Maths only): this uses $\sin$, $\cos$ or $\sec^{2}$ of a linear bracket. A hand pump fills an empty water butt. For $0 \le t \le 6$, where $t$ is the time in minutes after the pump starts, the pump delivers water at a rate of $\dfrac{dV}{dt} = 6 \sin{\left(\frac{\pi t}{6} \right)}$, measured in litres per minute, where the angle in the sine is in radians and $V$ litres is the volume in the butt.

• (a)Find $V$ in terms of $t$, and the volume in the butt after $3$ minutes. Give the exact value and then the value to 1 decimal place.
• (b)Evaluate $\int_{0}^{6} 6 \sin{\left(\frac{\pi x}{6} \right)}\,dx$ to find the volume the pump delivers in the whole pulse of $6$ minutes. Give the exact value and then the value to 1 decimal place.

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Colour keythe antiderivative or final result

(a)

$\displaystyle V\;= \int 6 \sin{\left(\frac{\pi t}{6} \right)}\,dt$

A rate of change is a derivative, so reverse it: $V$ is the integral of $\frac{dV}{dt} = 6 \sin{\left(\frac{\pi t}{6} \right)}$.

$\displaystyle = \textcolor{#010106}{- \frac{36 \cos{\left(\frac{\pi t}{6} \right)}}{\pi}} + c$

The integral of $\sin(ax+b)$ is $-\frac{1}{a}\cos(ax+b)$: the sign changes and you divide by $a$, here the coefficient of $x$ is $\frac{\pi}{6}$. The constant multiplier stays in front.

$\displaystyle V\;= - \frac{36 \cos{\left(\frac{\pi t}{6} \right)}}{\pi} + c$

The integral with its constant $c$.

$\displaystyle 0 = - \frac{36 \times \cos{\left(\frac{\pi \times \left(0\right)}{6} \right)}}{\pi} + c$

Use the starting value: $V = 0$ when $t = 0$.

$\displaystyle c\;= \textcolor{#010106}{\frac{36}{\pi}}$

Solve for $c$.

$\displaystyle V\;= \textcolor{#010106}{- \frac{36 \cos{\left(\frac{\pi t}{6} \right)}}{\pi} + \frac{36}{\pi}}$

Put the constant back.

$\displaystyle V\;= - \frac{36 \times \cos{\left(\frac{\pi \times \left(3\right)}{6} \right)}}{\pi} + \frac{36}{\pi}$

Substitute $t = 3$.

$\displaystyle = \textcolor{#010106}{\frac{36}{\pi}}$

Evaluate.

$\displaystyle \approx 11.5$

Round at the very end, to 1 decimal place.

(b)

$\displaystyle \int_{0}^{6} 6 \sin{\left(\frac{\pi t}{6} \right)}\,dt = \left[- \frac{36 \cos{\left(\frac{\pi t}{6} \right)}}{\pi}\right]_{0}^{6}$

Integrate with the usual rules but leave out the constant $c$: it cancels in a definite integral.

$\displaystyle = \left(- \frac{36 \times \cos{\left(\frac{\pi \times \left(6\right)}{6} \right)}}{\pi}\right) - \left(- \frac{36 \times \cos{\left(\frac{\pi \times \left(0\right)}{6} \right)}}{\pi}\right)$

Substitute the upper limit $6$, then the lower limit $0$, and subtract (upper minus lower).

$\displaystyle = \textcolor{#010106}{\frac{72}{\pi}}$

Evaluate.

$\displaystyle \approx 22.9$

Round at the very end, to 1 decimal place.

Answer

(a) $V = - \frac{36 \cos{\left(\frac{\pi t}{6} \right)}}{\pi} + \frac{36}{\pi},\ \ 0 \le t \le 6$; $V = \frac{36}{\pi} \approx 11.5$ litres (1 d.p.): the pump is still running, so the butt fills further

(b) $\frac{72}{\pi} \approx 22.9$ litres (1 d.p.): the whole pulse delivers twice the volume of its first half

Differentiation takes a curve and gives its gradient function. Integration goes the other way: given a gradient function, it finds the curve (or any quantity) it came from. Because several curves have the same gradient function, the answer always carries an unknown constant. Writing $\int \ldots\,dx$ means “find a function whose derivative is what is inside”.

A sum is integrated one term at a time. A plain number is a constant term: the integral of a constant $k$ is $kx$, because $\dfrac{d}{dx}(kx) = k$.

Fractions and negative numbers behave exactly as they do in differentiation: divide by the new power, and keep the sign and the multiplier.

Many curves share one gradient function: they are the same shape, shifted up or down by different values of $c$. The table shows this, and shows how one given point picks out the single curve you want.

The rule $\int x^{n}\,dx = \dfrac{x^{n+1}}{n+1} + c$ works for any rational power $n$, positive, negative or fractional, except one special value that is met later (a G3 topic). The skill is rewriting the function as a sum of powers of $x$ first.

Some functions are not yet a sum of powers. A fraction with a single term on the bottom is split over that term, and brackets are multiplied out first. Neither a quotient nor a product of two factors can be integrated as it stands.

A bracket $(ax + b)$ raised to a power looks like $x^{n}$, but it is the chain rule that has been used on the way down: differentiating $(ax + b)^{n+1}$ brings out the extra factor $a$. To undo it, divide by $a$ as well.