Transform definite-integral bounds
The ends travel with their points
The teal region lies under the curve \(y = 2x\,e^{-x^2}\), from \(x = 0.5\) to \(x = 2\), and two orange lines mark its ends.
Press the button to relabel the axis with \(u = x^2\). Each point slides to its \(u\)-value, and each orange line slides with its point.
The left end goes from \(x = 0.5\) to \(u = 0.5^2 = 0.25\), and the right end goes from \(x = 2\) to \(u = 2^2 = 4\).
The strips change width and height, but the area of each strip stays the same, so the region keeps its area.
The region now lies under \(y = e^{-u}\) from \(u = 0.25\) to \(u = 4\). As a check, its area is \(e^{-0.25} - e^{-4} \approx 0.760\).
Any limits
The region lies under a curve, from \(x = a\) to \(x = b\).
Drag the right end to a new place. Then move the Relabel slider to relabel the axis with \(u = g(x)\).
Each end stays attached to its point, so the end at \(a\) goes to \(g(a)\), and the end at \(b\) goes to \(g(b)\).
The region keeps its area, so the new integral has the same value. Its limits are where the two ends now sit, g(a) and g(b).
Notation
A definite integral has limits \(a\) and \(b\), which are values of \(x\). After the change of variable, each limit must be a value of \(u\).
The limit at \(a\) becomes \(g(a)\), and the limit at \(b\) becomes \(g(b)\). Keep them in the same order, even if \(g(a)\) is the larger number.
\[\begin{aligned} x = a &\to u = \cy{g(a)} \\ x = b &\to u = \cy{g(b)} \end{aligned}\]In words, the integral from \(a\) to \(b\) of \(f(g(x))\,g'(x)\) equals the integral of \(f(u)\) from \(g(a)\) to \(g(b)\).
\[\begin{aligned} &\int_{a}^{b} f(g(x))\,g'(x)\,dx \\ &\quad = \int_{\cy{g(a)}}^{\cy{g(b)}} f(u)\,du \end{aligned}\]Evaluate the new integral between the new limits. You do not need to return to \(x\).
The other route is to write the antiderivative in \(x\) again and use the limits \(a\) and \(b\). Both routes give the same number. Do not mix them: an antiderivative in \(u\) with limits in \(x\) is wrong.
The slope \(g'\) must be continuous on the interval from \(a\) to \(b\). The function \(f\) must be continuous on the values that \(g\) takes there.
When it fails
This is the opening region, with \(a = 0.5\) and \(b = 2\), values of \(x\). Press Keep the x limits to relabel the axis and leave the limits at 0.5 and 2.
The red region lies under \(e^{-u}\) from \(u = 0.5\) to \(u = 2\), not from 0.25 to 4. Its area is \(e^{-0.5} - e^{-2} \approx 0.471\), not 0.760. Each limit must be the value of \(u\) at the same point as the old limit.
Example 1: new limits in order
Evaluate \(\displaystyle\int_1^3 \frac{x}{x^2+1}\,dx\).
Let \(u\) be the denominator. Its slope is \(2x\), so \(x\,dx\) is one half of \(du\).
\[u = x^2 + 1, \qquad x\,dx = \dfrac{1}{2}\,du\]Find the new limits. The orange end lines slide to these values.
\[\begin{aligned} x = 1 &\to u = \cy{2} \\ x = 3 &\to u = \cy{10} \end{aligned}\]Rewrite the whole integral, limits included.
\[\dfrac{1}{2}\int_{2}^{10} \dfrac{1}{u}\,du\]The antiderivative of \(1/u\) is \(\ln u\), because \(u\) is positive here. Evaluate between the new limits.
\[\begin{aligned} \dfrac{1}{2}\Big[\ln u\Big]_{2}^{10} &= \dfrac{1}{2}(\ln 10 - \ln 2) \\ &= \cb{\dfrac{1}{2}\ln 5} \end{aligned}\]As a check, \(\tfrac12\ln 5 \approx 0.805\).
Example 2: a falling inside function
Evaluate \(\displaystyle\int_0^{\pi/2} \sin x\cos^4 x\,dx\).
The expression \(\cos x\) sits inside the fourth power. Its slope is \(-\sin x\), so \(\sin x\,dx\) is \(-du\).
\[u = \cos x, \qquad \sin x\,dx = -du\]Find the new limits. The value of \(\cos x\) falls from 1 to 0, so the \(u\) numbers on the figure fall from left to right. The lower limit is larger than the upper limit.
\[\begin{aligned} x = 0 &\to u = \cy{1} \\ x = \tfrac{\pi}{2} &\to u = \cy{0} \end{aligned}\]A tempting wrong move: Write the limits in increasing order, from 0 to 1, and keep the minus sign.
\[-\int_{0}^{1} u^4\,du = \cr{-\dfrac{1}{5}}\]The shaded region lies above the axis, so the integral is positive. The value \(-\tfrac15\) is negative, because the swap of the limits changed the sign of the answer.
Keep each limit with its point. The lower limit is the value at \(x = 0\), which is 1, and the upper limit is the value at \(x = \pi/2\), which is 0.
\[\int_{1}^{0} -u^4\,du = \left[-\dfrac{u^5}{5}\right]_{1}^{0} = \cb{\dfrac{1}{5}}\]
Practice
Evaluate \(\displaystyle\int_0^{\sqrt3} \frac{2x}{\sqrt{x^2+1}}\,dx\).
Move the slider and follow the two orange ends to their values of \(u\). Those values are the new limits.
Solution
Let \(u\) be the expression under the root. Its slope is \(2x\), which is the numerator.
\[u = x^2 + 1, \qquad du = 2x\,dx\]Find the new limits.
\[\begin{aligned} x = 0 &\to u = 1 \\ x = \sqrt3 &\to u = 4 \end{aligned}\]Rewrite the integral and evaluate. The antiderivative of \(u^{-1/2}\) is \(2u^{1/2}\).
\[\begin{aligned} \int_{1}^{4} u^{-1/2}\,du &= \Big[2\sqrt{u}\Big]_{1}^{4} \\ &= 2(2 - 1) = 2 \end{aligned}\]
Questions
The graph shows the inside function \(u = g(x) = (x-2)^2 + 1\) for \(x\) from 0 to 4. Each question takes an integral \(\int f(g(x))\,g'(x)\,dx\) between two values of \(x\).
The region runs from \(x = 0\) to \(x = 1\). What are the limits of the integral in \(u\), from the lower limit to the upper limit?
- From 5 to 2
- From 2 to 5
- From 0 to 1
- From 1 to 2
Answer
From 5 to 2
The region runs from \(x = 1\) to \(x = 3\). What is the value of \(\displaystyle\int_1^3 f(g(x))\,g'(x)\,dx\) for any continuous \(f\)?
- 0
- \(f(2)\)
- \(\displaystyle\int_1^3 f(u)\,du\)
Answer
0
Evaluate \(\displaystyle\int_0^1 x^2(x^3+1)^2\,dx\).
- \(\tfrac79\)
- \(\tfrac73\)
- \(\tfrac19\)
- \(\tfrac89\)
Answer
\(\tfrac79\)
Summary
You replace each limit with the value of u at the same point, and you keep the limits in the same order.
This works because each end stays attached to its point and the region keeps its area, so the new integral has the same value.
You change the limits whenever the problem gives limits in x and you change the variable.
This is one of four skills in topic 6.9 of Which & Why: AP Calculus.