Air is trapped under a piston in a metal cylinder. The cylinder stands in a large tub of water at room temperature, \(300\,\text{K}\). The piston rises very slowly, so heat flows in from the water and the air stays at \(300\,\text{K}\) all the time. As the volume grows, the pressure falls in proportion, \(p = nRT/V\). The work done by the air is the area under this curve, \(\int p\,dV\). That area needs the integral of \(1/V\), the one power that the rule of Lesson 1 cannot handle.
The one power the rule misses
The reverse power rule gives \(\int x^n\,dx = x^{n+1}/(n+1)\). At \(n = -1\) it asks us to divide by zero, so it gives nothing. Yet the area under \(y = 1/x\) plainly exists. The chapter on differentiation supplies the answer. The derivative of the natural logarithm \(\ln x\) is \(1/x\). So \(\ln x\) is the antiderivative we need.
The logarithm has a property that the figure below shows directly. Compare the area under \(1/x\) from 2 to 4 with the area from 1 to 2. Each strip of the second region is twice as wide as the matching strip of the first, and half as tall. So the two areas are equal. Every doubling of \(x\) adds the same area, \(\ln 2 = 0.693\). This is how a logarithm behaves: equal ratios give equal steps.
A short table of integrals
Each entry of the table below is a derivative read backwards, and each is checked by differentiating. The derivative of \(e^{kx}\) is \(ke^{kx}\), so \(e^{kx}/k\) integrates \(e^{kx}\). The derivative of \(\cos kx\) is \(-k\sin kx\), so \(-\cos kx/k\) integrates \(\sin kx\). The derivative of \(\sin kx\) is \(k\cos kx\), so \(\sin kx/k\) integrates \(\cos kx\). In \(\ln|x|\) the bars let the rule work for negative \(x\) too.
This is the worked number of The First Law and Thermodynamic Processes, lesson 4, done with the tool. One mole of an ideal gas at \(300\,\text{K}\) expands slowly at that temperature until its volume doubles. Find the work it does. Take \(R = 8.31\,\text{J/(mol K)}\).
At a steady temperature \(p = nRT/V\). The work is the area under the p-V curve: \(W = \int_{V_1}^{V_2} p\,dV = nRT\int_{V_1}^{V_2} \dfrac{dV}{V}\).
\(\int dV/V = \ln V\), so \(W = nRT\left(\ln V_2 - \ln V_1\right) = nRT\ln(V_2/V_1)\).
\(V_2/V_1 = 2\). So \(W = 1 \times 8.31 \times 300 \times \ln 2 = 2493 \times 0.693 = 1728\,\text{J}\).
The gas does \(1728\,\text{J}\) of work, as in the physics chapter. Only the ratio of the volumes enters, never the volumes themselves, because every doubling adds the same area \(\ln 2\) under \(1/V\).
One mole of an ideal gas at \(300\,\text{K}\) expands at that temperature from \(1\,\text{L}\) to \(8\,\text{L}\). Find the work it does, in joules. Take \(R = 8.31\,\text{J/(mol K)}\) and \(\ln 2 = 0.693\).
What is \(\int \sin 2t\,dt\)?
Substitution uses the chain rule in reverse
The chain rule differentiates a function of an inner expression. For example, the derivative of \(\sin(x^2)\) is \(\cos(x^2) \times 2x\). Read backwards, this says \(\int 2x\cos(x^2)\,dx = \sin(x^2) + C\). The integrand holds an inner expression, \(x^2\), together with its derivative, \(2x\). This pattern is the signal for a substitution.
To integrate a function of an inner expression, put \(u\) for the inner expression. Differentiate to get \(du = (du/dx)\,dx\), and replace every piece of the integrand by \(u\) and \(du\). Integrate in \(u\) with the table. Then put the inner expression back in place of \(u\). In a definite integral, change the limits to the values of \(u\) instead.
Find (a) \(\int (x + 2)^5\,dx\), (b) \(\int \cos(7\theta + 5)\,d\theta\), (c) \(\int_0^1 x(x^2 + 1)^3\,dx\).
(a) Put \(u = x + 2\), so \(du = dx\). The integral is \(\int u^5\,du = u^6/6\), which is \((x + 2)^6/6 + C\).
(b) Put \(u = 7\theta + 5\), so \(du = 7\,d\theta\) and \(d\theta = du/7\). The integral is \(\tfrac17\int \cos u\,du\), which is \(\tfrac17\sin(7\theta + 5) + C\).
(c) Put \(u = x^2 + 1\), so \(du = 2x\,dx\) and \(x\,dx = du/2\). The limits \(x = 0\) and \(x = 1\) become \(u = 1\) and \(u = 2\). The integral is \(\tfrac12\int_1^2 u^3\,du = \tfrac18(16 - 1) = \tfrac{15}{8}\).
A linear substitution is the one physics uses most. A body is cut into elements, and each element sits at a distance from a point. The distance is often an inner expression like \(4 - x\). In the figure below, a rod lies from \(x = 0\) to \(2\,\text{m}\), and a point P is on its line at \(x = 4\,\text{m}\). An element at \(x\) is a distance \(r = 4 - x\) from P. Its gravitational field at P varies as \(1/r^2\).
This is the rod of Gravitational Force and Field, lesson 4: mass \(12\,\text{kg}\), length \(2\,\text{m}\), and P on its line \(2\,\text{m}\) from the near end. With \(\lambda = 6\,\text{kg/m}\) and \(r = 4 - x\), the field is \(E = G\lambda\int_0^2 \dfrac{dx}{(4 - x)^2}\). Find \(E\) as a multiple of \(G\).
Products: integration by parts
Some integrands are a product of two different kinds of function, such as \(t\,e^{-\lambda t}\). Substitution does not help here. The product rule of differentiation, used in reverse, gives a second method. It is called integration by parts. Choose one factor \(u\) to differentiate and the other factor \(dv\) to integrate. The rule trades the integral for a simpler one.
\[ \dfrac{d}{dx}(uv) = u\,\dfrac{dv}{dx} + v\,\dfrac{du}{dx} \]
Why. the product rule of differentiation for two functions \(u\) and \(v\) of \(x\).
\[ uv = \int u\,dv + \int v\,du \]
Why. integrate both sides with respect to \(x\). The left side undoes its derivative, and \(\tfrac{dv}{dx}\,dx = dv\).
\[ \int u\,dv = uv - \int v\,du \]
Read it. move \(\int v\,du\) to the other side. The constant of integration is absorbed into the integral that remains.
Choose for \(u\) the factor that becomes simpler when differentiated. A power of \(t\) is the usual choice, because its derivative has a lower power. The mean life of a radioactive nucleus is found this way in Radioactivity, lesson 4. The integral there is \(\int_0^\infty t\,e^{-\lambda t}\,dt\). Take \(u = t\) and \(dv = e^{-\lambda t}\,dt\), so \(du = dt\) and \(v = -e^{-\lambda t}/\lambda\). The term \(uv\) is zero at both limits. What remains is \(\tfrac1\lambda\int_0^\infty e^{-\lambda t}\,dt = 1/\lambda^2\).
The mean life of a nucleus is \(\tau = \lambda\int_0^\infty t\,e^{-\lambda t}\,dt\). Using the integral found above, what is \(\tau\)?
Which substitution turns \(\int (x^2 + 1)^3\,2x\,dx\) into a standard integral?
The integral of \(1/x\) has two limits of its own. The logarithm takes only positive numbers, so write \(\ln|x|\) and never integrate across \(x = 0\), where \(1/x\) is infinite. The trigonometric entries need the angle in radians; in degrees every answer carries a stray factor of \(\pi/180\). And after a substitution in a definite integral, change the limits too, or put \(x\) back before using the old limits.
The slow gas in the tub is now solved. Its pressure falls as \(1/V\), so its work is \(nRT\) times the area under \(1/V\), which is a logarithm. One mole at \(300\,\text{K}\) that doubles its volume does \(1728\,\text{J}\) of work. Doubling it once more adds another \(1728\,\text{J}\), because each doubling adds the same area.
The short table and substitution work through the whole book. In The First Law and Thermodynamic Processes, \(\int dV/V\) gives the work of an isothermal gas. In Gravitational Force and Field and in Electric Charge, Coulomb's Law and the Electric Field, substitution adds the fields of rods, rings and discs. In Momentum and Its Conservation, \(\int dm/m\) gives the rocket equation. In Radioactivity, integration by parts gives the mean life. Heat flow through a cylinder and the mean value of a sinusoid use the same table.
- Using the power rule for \(\int dx/x\).At \(n = -1\) the rule divides by zero. The integral is \(\ln|x|\).
- Forgetting to divide by \(k\) in \(\int e^{kx}\,dx\) or \(\int \cos kx\,dx\).Differentiate your answer; if a factor \(k\) appears that is not in the integrand, divide it out.
- Keeping the old limits after a substitution.Change the limits to values of \(u\), or return to \(x\) before evaluating.