A Delhi Metro train runs deep under the city, where no satellite signal reaches a phone. Yet the phone still holds a small sensor called an accelerometer. The sensor measures one thing only: the acceleration of the phone at each instant. A navigation program can calculate the speed of the train from this record, and then the distance it has covered. To do so, the program must run differentiation backwards. Differentiation takes you from position to velocity to acceleration. The program needs the opposite journey.
Running the derivative backwards
In the chapter on differentiation you learnt to go from a function to its rate of change. From \(x = t^3\) you found \(v = dx/dt = 3t^2\). Physics often gives the rate and asks for the function. A spring gives the force at each stretch, and we want the work. An accelerometer gives the acceleration, and we want the velocity. The question is always the same: which function has this derivative?
A function \(F(x)\) is an antiderivative of \(f(x)\) when \(dF/dx = f(x)\) at every \(x\) of the interval. Finding an antiderivative is called integration. It is written \(\int f(x)\,dx = F(x) + C\), and the result is called the indefinite integral of \(f\).
The long S in \(\int f(x)\,dx\) is the integral sign. The function \(f(x)\) inside it is the integrand. The \(dx\) names the variable of integration. It says that \(x\) is the quantity that changes. Read the whole symbol aloud as "the integral of \(f\) with respect to \(x\)".
The reverse power rule
The derivative of \(x^{n+1}\) is \((n+1)x^n\). This is the power rule of differentiation, read from left to right. Divide both sides by \(n + 1\). The derivative of \(x^{n+1}/(n+1)\) is then exactly \(x^n\). So \(x^{n+1}/(n+1)\) is an antiderivative of \(x^n\). In words: raise the power by one, then divide by the new power.
\[ \dfrac{d}{dx}\,x^{n+1} = (n+1)\,x^{n} \]
Why. the power rule of differentiation, used with the power \(n + 1\) in place of \(n\).
\[ \dfrac{d}{dx}\left(\dfrac{x^{n+1}}{n+1}\right) = x^{n} \]
Why. divide both sides by the number \(n + 1\). A constant factor passes through a derivative unchanged. This needs \(n \ne -1\).
\[ \dfrac{d}{dx}\left(\dfrac{x^{n+1}}{n+1} + C\right) = x^{n} \]
Why. the derivative of a constant \(C\) is zero, so adding \(C\) changes nothing.
\[ \int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C \]
Read it. by the definition of an antiderivative, any function whose derivative is \(x^n\) is an integral of \(x^n\).
Two more rules come straight from the rules of differentiation. A constant factor can be taken outside the integral: \(\int k f\,dx = k\int f\,dx\). A sum is integrated term by term: \(\int (f + g)\,dx = \int f\,dx + \int g\,dx\). A constant alone is the power \(x^0\), so \(\int 5\,dx = 5x + C\). Together these rules integrate any polynomial.
Find (a) \(\int x^3\,dx\), (b) \(\int (3x^2 + 4x - 5)\,dx\), (c) \(\int \sqrt{t}\,dt\). Check each answer by differentiating it.
(a) Raise the power to 4 and divide by 4: \(\int x^3\,dx = x^4/4 + C\). Check: the derivative of \(x^4/4\) is \(x^3\).
(b) Term by term: \(3 \times x^3/3 + 4 \times x^2/2 - 5x + C = x^3 + 2x^2 - 5x + C\). One constant at the end is enough. Check: the derivative is \(3x^2 + 4x - 5\).
(c) Write \(\sqrt t = t^{1/2}\). The new power is \(3/2\), so \(\int t^{1/2}\,dt = \tfrac23 t^{3/2} + C\). Check: \(\tfrac23 \times \tfrac32\,t^{1/2} = t^{1/2}\).
What is \(\int x^3\,dx\)?
Why every answer carries a constant
The functions \(x = t^2\), \(x = t^2 + 3\) and \(x = t^2 + 6\) all have the same derivative, \(2t\). The derivative of the added constant is zero. So a velocity \(v = 2t\) does not tell us which of them is the position. On a graph the three curves are the same curve, shifted up by different amounts. Their slopes agree at every instant.
The constant \(C\) is called the constant of integration. One more fact about the motion fixes it. That fact is usually the value at the start, called the initial condition. Suppose \(v = 2t\) and the body is at \(x = 3\,\text{m}\) when \(t = 0\). Then \(x = t^2 + C\) and \(3 = 0 + C\), so \(C = 3\,\text{m}\). The position is \(x = t^2 + 3\).
The constant of integration is not zero by habit. It is zero only when the starting value makes it zero. A body that starts at \(x = 3\,\text{m}\) has \(C = 3\,\text{m}\) in its position. Dropping \(C\) puts the body in the wrong place for the whole motion.
A body has velocity \(v = 2t\) in m/s. Student A writes \(x = t^2\). Student B writes \(x = t^2 + 5\). Which statement is correct?
Two steps up the ladder
Differentiation goes down a ladder: position, then velocity, then acceleration. Integration climbs the ladder again, one rung at a time. Integrate the acceleration and you get the velocity, plus a constant. Integrate the velocity and you get the position, plus a second constant. The first constant is fixed by the starting velocity. The second is fixed by the starting position.
A body starts from rest at the origin. Its acceleration is \(a = 6t\), in m/s² with \(t\) in s. Find its velocity and its position as functions of time, and its displacement in the first \(3\,\text{s}\).
Integrate the acceleration: \(v = \int 6t\,dt = 3t^2 + C_1\). The body starts from rest, so \(v = 0\) at \(t = 0\), and \(C_1 = 0\). So \(v = 3t^2\).
Integrate the velocity: \(x = \int 3t^2\,dt = t^3 + C_2\). The body starts at the origin, so \(C_2 = 0\), and \(x = t^3\).
At \(t = 3\,\text{s}\), \(x = 27\,\text{m}\). Check by differentiating: \(dx/dt = 3t^2\) and \(dv/dt = 6t\), as given.
The same two steps work for turning. An angular acceleration \(\alpha\) integrates to an angular velocity \(\omega\), and \(\omega\) integrates to the angle \(\theta\). The next example is taken from the physics of circular motion. Watch the mistake it warns against. A student who knows only the equations for constant acceleration multiplies the final angular acceleration by the time. That treats a growing \(\alpha\) as if it had been large all along.
This is the worked example "An angular acceleration that grows with time" of Kinematics of Circular Motion, lesson 5, done with the tool. A disc starts at \(t = 0\) with \(\omega_0 = 1\,\text{rad/s}\) and \(\theta_0 = 0\). Its angular acceleration is \(\alpha = 4t\), in rad/s² with \(t\) in s. Find \(\omega\) and \(\theta\) at \(t = 3\,\text{s}\).
\(\alpha = d\omega/dt = 4t\) and \(\omega = d\theta/dt\). The starting values are \(\omega = 1\) and \(\theta = 0\) at \(t = 0\).
\(\omega = \int 4t\,dt = 2t^2 + C_1\), and \(\theta = \int \omega\,dt\).
At \(t = 0\), \(\omega = 1\), so \(C_1 = 1\) and \(\omega = 1 + 2t^2\). Then \(\theta = t + \tfrac23 t^3 + C_2\), and \(\theta = 0\) at \(t = 0\) gives \(C_2 = 0\).
At \(t = 3\,\text{s}\): \(\omega = 1 + 18 = 19\,\text{rad/s}\) and \(\theta = 3 + 18 = 21\,\text{rad}\). The constant-acceleration equation with \(\alpha = 12\) would give \(37\,\text{rad/s}\), which is wrong.
The phone in the Metro does the same two steps, many times each second. It sums the measured acceleration to update the velocity, and sums the velocity to update the position. It must also be told two starting facts: the speed and the position when the record began. Without them, the constants \(C_1\) and \(C_2\) stay unknown, and the phone cannot say where the train is.
A car moves along a straight road with \(v = 3t^2 + 2\), in m/s with \(t\) in s. It passes the origin at \(t = 0\). How far from the origin is the car at \(t = 2\,\text{s}\), in metres?
The force needed to hold a spring stretched by \(x\) is \(F = kx\), with \(k = 800\,\text{N/m}\). The work to stretch it is \(\int F\,dx\), and \(\tfrac12 kx^2\) is an antiderivative of \(kx\). Find the work, in joules, to stretch the spring from \(5\,\text{cm}\) to \(10\,\text{cm}\).
What is \(\int 5\,dt\)?
The reverse power rule fails at one power, \(n = -1\). For \(\int x^{-1}\,dx\) it would give \(x^0/0\), a division by zero. The integral of \(1/x\) exists, but it is a new function, the natural logarithm \(\ln|x|\). Lesson 3 of this chapter meets it in the work done by a gas.
The antiderivative climbs the ladder from a rate back to a quantity. In Motion in One Dimension it converts a constant acceleration into \(v = u + at\). In Kinematics of Circular Motion it handles an angular acceleration that changes with time. In Work, Kinetic Energy and Power it gives the work \(\tfrac12 kx^2\) stored in a stretched spring. The same step gives pressure at a depth in a liquid and the energy stored in an inductor.
- Differentiating when the question asks for an integral.Integration raises the power by one. Check every answer by differentiating it back to the integrand.
- Leaving out the constant, or setting it to zero without a reason.Write \(+ C\) every time, then fix \(C\) from a starting value.
- Using \(\omega = \omega_0 + \alpha t\) when \(\alpha\) changes with time.That equation holds only for constant \(\alpha\). Integrate \(\alpha(t)\) instead.