Carry the chosen method through correctly
The ends travel with the axis
The curve is \(f(x) = 3x\sqrt{x^2 + 2}\), and the teal region lies under it from \(x = 0\) to \(x = 2\). The inner function is \(u = x^2 + 2\).
Move the Axis slider to relabel the axis with \(u\). The two orange end lines move with the axis: the end \(x = 0\) goes to \(u = 2\), and the end \(x = 2\) goes to \(u = 6\).
Since \(du = 2x\,dx\), \(3x\,dx = \tfrac32\,du\), so on the \(u\) axis the curve is \(\tfrac32\sqrt{u}\). The region has the same area as before, and it now lies between \(u = 2\) and \(u = 6\).
An antiderivative of \(\tfrac32\sqrt{u}\) is \(u^{3/2}\), so the area is \(6^{3/2} - 2^{3/2} = 6\sqrt6 - 2\sqrt2\).
As a check, \(6\sqrt6 - 2\sqrt2 \approx 11.87\), the teal area in the readout.
One variable from start to end
The inner function is now a general \(u = g(x)\). Move the Axis slider. The end lines \(a\) and \(b\) move to \(g(a)\) and \(g(b)\), and the region and its area stay the same.
The integral must stay in one variable. Either the curve and the ends are both in \(u\), or both are in \(x\).
The same care applies to every method. From one line to the next, the integrand, the constant factors, the signs and the ends must all match.
Notation
Choose \(u = g(x)\) and write \(du = g'(x)\,dx\). Replace every \(x\) and \(dx\), including any \(x\) that is left over.
\[u = g(x), \qquad du = g'(x)\,dx\]Move the ends with the axis. The new ends are \(g(a)\) and \(g(b)\).
\[\begin{aligned} &\int_a^b f(g(x))\,g'(x)\,dx \\ &\quad = \int_{\cy{g(a)}}^{\cy{g(b)}} f(u)\,du \end{aligned}\]Evaluate in \(u\) between the new ends. You can also write the antiderivative in \(x\) and use the original ends \(a\) and \(b\). Do not mix the two.
Keep every sign. If \(du = -\sin x\,dx\), the minus sign stays in the integral. If \(g\) is decreasing, then \(g(a)\) is greater than \(g(b)\), and the integral still runs from \(g(a)\) to \(g(b)\).
Before you integrate, find any point where the integrand is unbounded, such as \(\tfrac1x\) at \(x = 0\). If it lies between \(a\) and \(b\), the integral is improper, and \(F(b) - F(a)\) for an antiderivative \(F\) can give a false value.
Check the answer by differentiating it. The result must be the integrand.
When it fails
Use the old ends \(a\) and \(b\) on the \(u\) axis. The red region runs from \(u = a\) to \(u = b\), so it is a different region from the teal one.
The area changed because the ends did not travel with the axis. Convert the ends, or return to \(x\) before you use them.
Example 1: new ends
Find \(\displaystyle\int_0^4 \frac{6}{2x + 1}\,dx\).
Take \(u = 2x + 1\), so \(du = 2\,dx\) and \(dx = \tfrac12\,du\). The factor 6 and the factor \(\tfrac12\) give 3.
\[\frac{6}{2x+1}\,dx = \frac{3}{u}\,du\]Move the ends. At \(x = 0\), \(u = 1\). At \(x = 4\), \(u = 9\).
\[\begin{aligned} &\int_0^4 \frac{6}{2x+1}\,dx \\ &\quad = \int_{\cy{1}}^{\cy{9}} \frac{3}{u}\,du \end{aligned}\]Integrate, and evaluate between the new ends. The term \(3\ln 1\) is 0.
\[\begin{aligned} &\Big[3\ln u\Big]_1^9 \\ &\quad = 3\ln 9 - 3\ln 1 \\ &\quad = \cb{3\ln 9} \end{aligned}\]As a check, the derivative of \(3\ln u\) is \(\tfrac3u\), the integrand in \(u\), and \(3\ln 9 \approx 6.59\).
Example 2: a minus sign
Find \(\displaystyle\int_0^{\pi/3} \sin x\cos^3 x\,dx\).
The inner function is \(\cos x\). Its derivative is \(-\sin x\), so \(du = -\sin x\,dx\) and \(\sin x\,dx = -du\). Move the ends.
\[\begin{gathered} u = \cos x \\ x = 0 \to u = 1 \\ x = \tfrac{\pi}{3} \to u = \tfrac12 \end{gathered}\]A tempting wrong move: Replace \(\sin x\,dx\) by \(du\) and keep the new ends.
\[\begin{aligned} &\int_{1}^{1/2} u^3\,du \\ &\quad = \left[\frac{u^4}{4}\right]_1^{1/2} \\ &\quad = -\frac{15}{64} \end{aligned}\]The teal region lies above the x-axis, so the integral is positive, but this answer is negative. The minus sign in \(\sin x\,dx = -du\) was lost.
Keep the minus sign: \(\sin x\,dx = -du\).
\[\begin{aligned} &-\int_{1}^{1/2} u^3\,du \\ &\quad = -\left[\frac{u^4}{4}\right]_1^{1/2} \\ &\quad = -\left(\frac{1}{64} - \frac14\right) \\ &\quad = \cb{\frac{15}{64}} \end{aligned}\]
Practice
Find \(\displaystyle\int_0^4 \frac{x}{\sqrt{2x + 1}}\,dx\).
Replace every \(x\), including the \(x\) in the numerator. Move both ends with \(u\).
Solution
Take \(u = 2x + 1\), so \(x = \tfrac{u-1}{2}\) and \(dx = \tfrac12\,du\). The ends are \(u = 1\) and \(u = 9\).
\[\begin{aligned} &\int_0^4 \frac{x\,dx}{\sqrt{2x+1}} \\ &\quad = \int_1^9 \frac{u-1}{2}\cdot\frac{1}{\sqrt{u}}\cdot\frac{du}{2} \end{aligned}\]The two factors \(\tfrac12\) give \(\tfrac14\), and \(\tfrac{u-1}{\sqrt u} = u^{1/2} - u^{-1/2}\). Integrate each power of \(u\).
\[\begin{aligned} &\frac14\int_1^9 \left(u^{1/2} - u^{-1/2}\right)du \\ &\quad = \frac14\left[\tfrac23 u^{3/2} - 2u^{1/2}\right]_1^9 \end{aligned}\]Evaluate between the ends. At 9, the bracket is \(18 - 6 = 12\). At 1, it is \(\tfrac23 - 2 = -\tfrac43\).
\[\begin{aligned} &\frac14\left(12 + \frac43\right) \\ &\quad = \cb{\frac{10}{3}} \end{aligned}\]
Questions
The figure shows the region under \(\dfrac{2x}{x^2+1}\) from \(x = 0\) to \(x = 2\), and relabels the axis with \(u = x^2 + 1\). On the \(u\) axis, the curve is \(\dfrac1u\).
The right end line is at \(x = 2\). Where is it on the \(u\) axis?
- \(u = 5\)
- \(u = 2\)
- \(u = 4\)
Answer
\(u = 5\)
Which integral has the same area as the teal region?
- \(\displaystyle\int_1^5 \frac{du}{u}\)
- \(\displaystyle\int_0^2 \frac{du}{u}\)
- \(\displaystyle\int_1^5 u\,du\)
Answer
\(\displaystyle\int_1^5 \frac{du}{u}\)
What is \(\displaystyle\int_0^3 \frac{x}{x^2 + 16}\,dx\)?
- \(\ln\dfrac54\)
- \(\ln\dfrac{25}{16}\)
- \(\ln 5\)
Answer
\(\ln\dfrac54\)
Summary
You replace every \(x\) and \(dx\), move the ends or return to \(x\), keep every sign, and check the result by differentiating.
This works because the region and its area do not change when the axis is relabeled, so the ends must travel with it.
You use these steps for every definite integral, and you look for points where the integrand is unbounded before you integrate.
This is one of five skills in topic 6.14 of Which & Why: AP Calculus.