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The applications with no shape in them: $W = \int F\,dx$, the pressure on a wall, and the height of the rectangle with the same area.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to write the work done by a varying force as an integral and say why force times total distance overstates it, slice a pumping problem horizontally so that each slab has a single lift distance and write that distance the right way round, set up a hydrostatic force integral with depth measured below the surface, compute an average value as a height rather than an area and state the mean value theorem for integrals with its continuity hypothesis, and recognise from the units alone when a total is an integral of a rate.
Areas, volumes by plates, volumes by tubes, lengths and surfaces were five different-looking calculations and one method: say what a slice contributes, then integrate.
This lesson finishes the unit with the applications that have no picture to slice — work done by a varying force, the pressure on a dam, the average value of a function, a total accumulated from a rate. The method does not change at all. What changes is that the slice is a piece of a physical quantity rather than a piece of a shape, and nobody hands you the picture. Every application of the integral is the same move: write what one thin slice contributes, then add the slices up. Get the slice right and the integral writes itself; reach for a remembered formula instead and the first unfamiliar region defeats you.
Work is force times distance when the force is constant, and $\int F\,dx$ when it is not. Its unit is the joule: a newton-metre.
Weight density $\rho g$ is weight per unit volume — about $9800$ newtons per cubic metre for water. In a pumping problem it converts a slab's volume into the force needed to lift it.
The average value of $f$ on $[a,b]$ is $\bar f = \dfrac{1}{b-a}\int_a^b f$. It is a height, not an area: the height of the rectangle on $[a,b]$ with the same area as the region under the curve.
Work against a varying force. Over a displacement so short that $F$ does not appreciably change, the work is $F(x)\,dx$. So $$W = \int_a^b F(x)\,dx.$$ For a spring obeying Hooke's law $F = kx$, stretching from $0$ to $d$ costs $\tfrac12 kd^2$ — half of $F(d)\cdot d$, because the force rose linearly from nothing.
Pumping a fluid. Here different slices travel different distances, which is what makes it an integral rather than a multiplication. A horizontal slab at height $y$, of cross-sectional area $A(y)$ and thickness $dy$, weighs $\rho g A(y)\,dy$ and must be lifted $(H - y)$ to reach a spout at height $H$: $$W = \int \rho g\,A(y)\,(H - y)\,dy.$$ Slice horizontally, because a slab is what has a single lift distance.
Hydrostatic force. Pressure at depth $h$ is $\rho g h$, the same in every direction. A horizontal strip of a vertical wall, at depth $h$ and of width $w(h)$, feels $\rho g h\,w(h)\,dh$: $$F = \int \rho g\,h\,w(h)\,dh.$$ Again the strip is horizontal, because pressure is constant along it.
Average value. $$\bar f = \frac{1}{b-a}\int_a^b f(x)\,dx.$$ The mean value theorem for integrals says that if $f$ is continuous, some $c$ in $(a,b)$ has $f(c) = \bar f$: a continuous function actually attains its average. Continuity is needed — a step function need not.
A total from a rate. If $r(t)$ is a rate of change, $\int_a^b r\,dt$ is the total change. The units are the giveaway: rate times time is quantity.
Another way: picture
The average value drawn on the graph: the curve, the region under it, and a horizontal line at height $\bar f$ cutting through the curve. The rectangle under that line has exactly the area of the region — what the curve rises above the line somewhere, it falls below it elsewhere, and the two amounts are equal. The mean value theorem for integrals is the statement that the line really does cross the curve.
Another way: steps
A tank of water weighing $W$ newtons, pumped to a spout $H$ metres above the bottom, does not take $WH$ joules to empty. It takes less, because the water at the top has hardly any distance to travel.
The correct statement is that the work equals $W$ times the distance the centre of mass rises. For a uniform column of water of depth $H$ pumped to the top, the centre of mass starts at $H/2$ and ends at $H$, so the work is $W \cdot H/2$ — half the naive answer, and exactly what $\int_0^H \rho g A (H - y)\,dy$ gives when $A$ is constant.
That check is worth keeping, because it is independent of the integral: work out where the centre of mass starts and ends, multiply by the total weight, and compare. If the two disagree, the lift distance has been written the wrong way round — $y$ instead of $H - y$ — which is by far the commonest error in these problems and produces a plausible number.
For a tank that is not a prism the centre of mass is no longer at half depth, and the integral is the only route. But the constant-area case gives you a calibrated instinct for what size of answer to expect.
Force times total distance, when the force varies. That is the whole point of the integral, and it overstates a spring's work by exactly a factor of two.
The lift distance written as $y$ instead of $H - y$. The slab at the bottom travels furthest. Check the extreme cases: at $y = H$ the distance should be zero.
Slicing a pumping problem vertically. A vertical column does not have a single lift distance, so its contribution cannot be written. Slice horizontally.
Confusing the average value with the average of the endpoints. $\tfrac12(f(a) + f(b))$ is the trapezoidal estimate, not the average value, and they agree only for a linear function.
Reporting an average as an area. Dividing by the width is not an optional tidying step; it is what makes the answer a height, with the units of $f$.
Measuring depth from the wrong place for hydrostatic pressure. Pressure depends on depth below the surface, not height above the bottom. Both appear in these problems and they are different variables.
Dropping the units. In this lesson the units are usually the only check available, and they catch a rate mistaken for a total instantly.
Learners often come away believing that $W = Fd$ was replaced by an integral because it was incorrect. It was not. It is exactly right whenever the force is constant, and the integral reduces to it in that case — $\int_a^b F\,dx = F(b-a)$ for constant $F$.
What the integral does is handle the case where $F$ changes as you go, by applying $W = Fd$ on pieces small enough that it is true on each. That is the same relationship every application in this unit has to its elementary formula: area of a rectangle becomes $\int(f-g)dx$, volume of a cylinder becomes $\int A\,dx$, length of a segment becomes $\int ds$. In each case the simple formula is used on a slice and then summed.
Seeing it that way is what makes a new application doable. Nobody will teach you the formula for the work done winding a heavy chain, or the force on a trapezoidal dam gate, or the average concentration of a drug over six hours. But if you can say what a short piece of chain, a thin strip of gate, or a minute of the dose contributes, you can write the integral — and that is the entire content of this unit, restated.
The question to ask, always, is: what does one slice contribute, and what makes it differ from its neighbour?
A cylinder of radius $2$ m and height $5$ m, full of water, pumped over the top. Slice horizontally: a slab at height $y$ has volume $4\pi\,dy$ and weighs $9800 \cdot 4\pi\,dy$ newtons.
Horizontal slabs, because each has one lift distance.
It must be lifted $5 - y$ metres: at the bottom that is $5$, at the top it is $0$, which is the check that the expression is the right way round.
Test the extremes before integrating.
$W = \displaystyle\int_0^5 9800 \cdot 4\pi\,(5 - y)\,dy = 39200\pi\left[5y - \tfrac{y^2}{2}\right]_0^5 = 490000\pi$ joules. The centre-of-mass check: the weight is $9800 \cdot 20\pi = 196000\pi$ newtons and it rises $2.5$ m, giving the same number.
Two routes, one answer.
The average of $f(x) = x^2$ on $[0, 3]$: $\displaystyle\int_0^3 x^2 dx = 9$, and the width is $3$, so $\bar f = 3$.
Integrate, then divide by the width.
By the mean value theorem for integrals there is a $c$ in $(0,3)$ with $c^2 = 3$, namely $c = \sqrt3 \approx 1.73$.
A continuous function attains its average.
Note that $\sqrt3$ is not the midpoint $1.5$: the function is attaining its average later than halfway, because it grows faster in the second half. The average of the endpoints, $\tfrac12(0 + 9) = 4.5$, is a different and larger number, and for any convex function it always overestimates.
The force varies, so the work is $\displaystyle\int_{0.1}^{0.3}40x\,dx$, not a product.
Write the slice: force at $x$, times $dx$.
$= 20\left[x^{2}\right]_{0.1}^{0.3} = 20(0.09 - 0.01) = 1.6$ joules.
Note it is not stretching from zero: both limits matter.
Check: the average force over that range is $40 \times 0.2 = 8$ N and the distance is $0.2$ m, giving $1.6$ J. The average-force shortcut works because the force is linear, and fails as soon as it is not.
A spring obeys $F(x) = 2x$ newtons at extension $x$ metres. How much work is done stretching it from $0$ to $3$ metres?
Answer:
Find the average value of $f(x) = 3x^{2}$ on $[\,0, 6\,]$.
Answer:
Match each quantity to what one slice of it contributes. (This one is about the area of a surface of revolution, among others.)
| $(f - g)\,dx$ | $\pi R^{2}\,dx$ | $\sqrt{1 + (y')^{2}}\,dx$ | $F(x)\,dx$ | |
|---|---|---|---|---|
| The area between $y = f(x)$ and $y = g(x)$ | ||||
| A volume of revolution, cut across the axis | ||||
| The length of a curve | ||||
| The work done by a force $F(x)$ |
A tank of square cross-section, side $5$ metres, is $9$ metres deep and full of water. For the slab of water at depth-height $y = 2$ metres above the bottom, fill in the three factors of its work.
| Value | |
|---|---|
| The slab's cross-sectional area | 25 |
| The distance it must be lifted | |
| Their product |
Put the steps of setting up any application of the integral into the order you do them.
Number the steps in order (write the number in the box):
Water enters a tank at $4t$ litres per minute at time $t$. How much enters during the first $4$ minutes?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A spring obeys $F(x) = 6x$ newtons at extension $x$ metres. How much work is done stretching it from $0$ to $3$ metres?
Answer:
You can set up an application you have never seen by describing one slice. Say in your own words why pumping a full tank to the top takes half of weight times depth. Next: what happens when an interval is infinite, or the integrand is not.
10. Your turn: the work to stretch a spring with $F(x) = 40x$ from $0.1$ m to $0.3$ m, step 3