Substitution with Definite Integrals

Main idea

When you substitute in a definite integral, the limits are $x$ -values, so they have to become $u$ -values too. Once they do, you can finish entirely in $u$ and never switch back to $x$ .

\int_a^b f\big(g(x)\big)\,g'(x)\,dx = \int_{{\htmlData{sym=0}{g(a)}}}^{{\htmlData{sym=1}{g(b)}}} f(u)\,du

What the symbols mean

$g(a)$
New lower limit
The u-value when x = a.
$g(b)$
New upper limit
The u-value when x = b.

Explanation

Substitution with definite integrals is simply the substitution rule plus one extra step: translating the endpoints into the new variable. All we're really saying is that if you rename $x$ as $u$ , you have to rename everything, including where the interval starts and stops. The formula has just that one new piece: $g(a)$ and $g(b)$ (the $u$ -values you get by plugging the old endpoints $a$ and $b$ into $u = g(x)$ ). Everything else is the substitution you already know.

Think about why the limits have to change. The limits $a$ and $b$ say "from $x = a$ to $x = b$ ." Once the integral is written in $u$ , it's asking about $u$ , and $u$ doesn't run from $a$ to $b$ ; it runs from $g(a)$ to $g(b)$ . Leaving the old numbers on would answer a different question.

Two correct ways to finish:

Change the limits to $u = g(a)$ and $u = g(b)$ , and evaluate in $u$ .
Find the antiderivative in $u$ , substitute back to $x$ , and use the original limits.

A common mistake. It's tempting to mix the two: find the antiderivative in $u$ , then plug in the original $x$ -limits, because those are the numbers written on the integral. That gives the wrong answer. Either change the limits or switch back to $x$ , but not half of each.

Symmetry shortcuts on $[-a, a]$ :

$f$ odd ( $f(-x) = -f(x)$ ): $\int_{-a}^a f(x)\,dx = 0$ .
$f$ even ( $f(-x) = f(x)$ ): $\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$ .

See it

This is the rate a drug is absorbed, r(t) = 50te^(−t²/4) mg per hour. With u = t²/4 (so du = t/2 dt), the total absorbed by time T becomes 100∫₀^(T²/4) e⁻ᵘ du = 100(1 − e^(−T²/4)). Notice that the limits changed from 0 and T to 0 and T²/4. As you can see, the absorption peaks at about 1.4 hours (t = √2), and by 6 hours almost the whole 100 mg dose is in.

Worked examples

Example 1

Evaluate $\displaystyle\int_0^1 2x\left(x^2 + 1\right)^3dx$ .

1
Let $u = x^2 + 1$ , so $du = 2x\,dx$ . Convert the limits: when $x = 0$ , $u = 1$ , and when $x = 1$ , $u = 2$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Evaluate $\displaystyle\int_0^2 x\left(x^2 + 1\right)^2dx$ .

Problem 2+5 XP

Evaluate $\displaystyle\int_0^{\pi/2}\cos x\sin^2 x\,dx$ .

Problem 3+5 XP

Evaluate $\displaystyle\int_0^1 xe^{x^2}\,dx$ .

Problem 4+5 XP

Evaluate $\displaystyle\int_1^e\frac{\ln x}{x}\,dx$ .

Problem 5+5 XP

Evaluate $\displaystyle\int_{-2}^2 x^3\cos x\,dx$ .

Problem 6+5 XP

Water flows into a tank at $r(t) = 6t\sqrt{t^2 + 9}$ liters per minute. How much enters between $t = 0$ and $t = 4$ ?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Evaluate $\displaystyle\int_0^1\frac{x}{1 + x^4}\,dx$ .

Where it's used

Symmetry arguments like the odd-function rule save enormous effort in physics and signal processing: half of a Fourier calculation vanishes before you start.

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