An auto-rickshaw crawls through Bengaluru traffic. Its speed rises, falls and rises again, and the needle never rests for long. The fare meter charges by the distance covered. Suppose you had only a video of the speedometer needle, with a clock beside it. Could you find the distance from that video alone? Distance equals speed times time only while the speed is steady. Here the speed changes every second.
Displacement is an area
At a steady velocity \(v\), the displacement in a time \(t\) is \(vt\). On a velocity-time graph this is a rectangle of height \(v\) and width \(t\). So the displacement is the area under the graph. The idea survives when the velocity changes. Cut the time into short intervals. In each short interval the velocity hardly changes, so the displacement in it is nearly \(v\,\Delta t\), a thin rectangle. Add the thin rectangles, and the total is the area under the curve.
Thin strips sum to the area
For a curve, the strips are not exact. Each strip has a flat top, while the curve above it rises or falls. So a sum of strips is only an estimate of the area. Take more strips, each thinner, and the flat tops follow the curve more closely. The error shrinks with every step. The limit of the sum, as the width of each strip tends to zero, is the exact area.
Divide the interval from \(a\) to \(b\) into \(n\) strips of width \(\Delta x = (b - a)/n\). Let \(x_i\) be a point in the \(i\)-th strip. The definite integral of \(f\) from \(a\) to \(b\) is the limit of the sum \(\sum f(x_i)\,\Delta x\) as \(n\) grows without limit. It is written \(\int_a^b f(x)\,dx\), and \(a\) and \(b\) are the limits of integration.
The notation carries the method inside it. The long S is a stretched "S" for sum. The product \(f(x)\,dx\) is one thin strip: a height times a very small width. So \(\int_a^b f(x)\,dx\) reads "add the strips \(f(x)\,dx\) from \(x = a\) to \(x = b\)". Physics uses the same reading everywhere. A velocity times a small time is a small displacement. A force times a small displacement is a small amount of work.
One antiderivative does all the adding
Adding thousands of strips by hand is slow. A single result, the fundamental theorem of calculus, replaces the whole sum with one subtraction. Find any antiderivative \(F\) of \(f\). Then the area from \(a\) to \(b\) is \(F(b) - F(a)\). The difference \(F(b) - F(a)\) is often written \(\left[F(x)\right]_a^b\).
Find the area under \(y = x^2\) from \(x = 1\) to \(x = 3\).
An antiderivative of \(x^2\) is \(x^3/3\).
At the upper limit, \(27/3 = 9\). At the lower limit, \(1/3\).
So the area is \(9 - \tfrac13 = \tfrac{26}{3} \approx 8.67\). The constant \(C\) would appear in both values and cancel, so it is not written.
A cyclist's velocity is \(v = 2t\), in m/s with \(t\) in s. Find the displacement from \(t = 0\) to \(t = 3\,\text{s}\), in metres.
Why an antiderivative measures an area
Let \(A(x)\) be the area under \(f\) from \(a\) up to a moving right edge at \(x\). Move the edge a little further, by \(h\). The area grows by one thin strip, of width \(h\) and height very nearly \(f(x)\). So the rate at which the area grows is the height of the curve at the edge. The derivation below makes this picture into the theorem.
\[ A(x) = \text{area under } f \text{ from } a \text{ to } x \]
Why. define the area function. It is zero at \(x = a\), where the region has no width.
\[ A(x + h) - A(x) \approx f(x)\,h \]
Why. moving the edge by \(h\) adds one thin strip, of width \(h\) and height nearly \(f(x)\).
\[ \dfrac{dA}{dx} = \lim_{h \to 0} \dfrac{A(x + h) - A(x)}{h} = f(x) \]
Why. divide by \(h\) and let \(h\) shrink. The strip becomes exact in the limit, so the area grows at the rate \(f(x)\).
\[ A(x) = F(x) - F(a) \]
Why. \(A\) and \(F\) are both antiderivatives of \(f\), so they differ by a constant. The condition \(A(a) = 0\) fixes that constant as \(-F(a)\).
\[ \int_a^b f(x)\,dx = A(b) = F(b) - F(a) \]
Read it. the area from \(a\) to \(b\) is the area function at the right edge \(x = b\).
This is not a formal proof, because step 2 is an approximation. A full proof bounds the strip between the lowest and the highest value of \(f\) in it. The picture is the true reason, though. Differentiation measures how fast a quantity grows. Integration adds the growth together. Each operation undoes the other.
This is the worked example "Work from a formula for the force" of Work, Kinetic Energy and Power, lesson 2, done with the tool. A force along the \(x\) axis is \(F(x) = (3x^2 - 2x + 4)\,\text{N}\), with \(x\) in metres. Find the work it does as a particle moves from \(x = 1\,\text{m}\) to \(x = 3\,\text{m}\).
The work is the area under the force-position graph: \(W = \int_1^3 F(x)\,dx\). This is the area that the strips of the figure above were estimating.
An antiderivative of \(3x^2 - 2x + 4\) is \(x^3 - x^2 + 4x\).
At \(x = 3\): \(27 - 9 + 12 = 30\). At \(x = 1\): \(1 - 1 + 4 = 4\). So \(W = 30 - 4 = 26\,\text{J}\).
The force does \(26\,\text{J}\) of work, the same answer as the physics chapter. The 32 strips of the figure gave \(25.38\,\text{J}\), close but low.
A lift's acceleration is plotted against time. What does the area under the a-t graph from \(t_1\) to \(t_2\) give?
Area below the axis is negative
When \(f\) is negative, each strip \(f(x)\,dx\) is negative too. So the integral counts area below the axis with a minus sign. For a velocity graph this is exactly right. A negative velocity is motion backwards, and it reduces the displacement. The distance travelled is different. The distance adds every stretch as positive, so it is the integral of the speed \(|v|\).
Two rules follow from the idea of adding strips. Reversing the limits reverses every strip, so \(\int_b^a f\,dx = -\int_a^b f\,dx\). An interval split at a point \(c\) gives two sums that add: \(\int_a^b f\,dx = \int_a^c f\,dx + \int_c^b f\,dx\). The second rule is how a distance is found. Split the time where the velocity changes sign, and add the sizes of the parts.
The same subtraction finds the area between two curves. If \(f\) lies above \(g\) from \(a\) to \(b\), the area between them is \(\int_a^b (f - g)\,dx\). For example, the line \(y = x + 2\) lies above the parabola \(y = x^2\) between their crossings at \(x = -1\) and \(x = 2\). The area between them is \(\left[x^2/2 + 2x - x^3/3\right]_{-1}^{2} = \tfrac{10}{3} + \tfrac76 = 4.5\).
A ball rolls up a slope and back with \(v = 6 - 2t\), in m/s with \(t\) in s. What are its displacement and the distance it travels from \(t = 0\) to \(t = 5\,\text{s}\)?
A footballer kicks a ball of mass \(0.4\,\text{kg}\) that is at rest. The force of the boot rises steadily from \(0\) to \(1200\,\text{N}\) and falls steadily back to \(0\), over \(0.01\,\text{s}\) in all. The impulse is the area under the force-time graph. Find the speed of the ball as it leaves the boot, in m/s.
Why does the sum of strips in Fig 2.2 approach \(26\,\text{J}\) as the strips become more numerous? Why is it always below \(26\,\text{J}\)?
Each strip has a flat top at the height of the force at its left edge. The force rises across every strip, so the flat top lies below the curve, and each strip misses a thin sliver of area. The sum is therefore always too small. Thinner strips leave thinner slivers, so the missing area shrinks towards zero, and the sum approaches the exact area of \(26\,\text{J}\).
A definite integral gives a signed total, not a distance. If the velocity changes sign inside the interval, \(\int v\,dt\) is the displacement, and the distance is larger. Split the interval where \(v = 0\) and add the sizes. The rule \(F(b) - F(a)\) also needs \(f\) finite on the whole interval. The area under \(1/x^2\) from \(-1\) to \(1\) is not the \(-2\) that the rule would give, because the curve rises without limit at \(x = 0\).
Archimedes found the area of a piece of a parabola more than two thousand years ago. In a letter now called The Method, he imagined the region cut into a great many thin parallel slices. He weighed the slices, in his mind, against the slices of a shape he already knew, on an imaginary lever. Then he proved the result again by a stricter method, because he did not trust the slices alone. The letter was lost for centuries. It was found in 1906, written underneath the text of a prayer book. The thin slices of Archimedes are the thin strips of this lesson.
The auto-rickshaw video can now be read. Note the speed from the needle every second, draw the velocity-time graph, and find the area under it. Each second contributes a thin strip, and the strips together make the distance. The fare meter does the same sum, one turn of the wheel at a time.
The area under a curve appears in every unit of the book. In Motion in One Dimension it gives the second equation of motion, \(s = ut + \tfrac12 at^2\). In Graphical Understanding of Motion the area under a v-t graph is the displacement, and under an a-t graph the change of velocity. In Accelerated Objects the area under a force-time graph is the impulse. In Work, Kinetic Energy and Power the area under a force-position graph is the work. The same area gives the work \(\int p\,dV\) of a gas and the energy stored in a capacitor.
- Reading the area under an a-t graph as a displacement.The area under an a-t graph is a change of velocity. Only the v-t graph gives displacement.
- Adding the area below the axis as positive when the displacement is asked.The integral counts it as negative. Use \(|v|\) only when the question asks for the distance.
- Subtracting the limits in the wrong order.Always take the value at the upper limit minus the value at the lower limit.