Single-Variable-Calculus · Unit 15 · Video 4 · Interactive Practice

Substitution in Definite Integrals: Change the Limits

IKey Formulas

Formula	Name	When it applies
$\displaystyle\int_{x_1}^{x_2} g(u(x))\,u'(x)\,dx = \int_{u_1}^{u_2} g(u)\,du$	Substitution with limits	$u_1 = u(x_1)$ and $u_2 = u(x_2)$
$\displaystyle\int_1^2 (x^3+2)^5\,x^2\,dx = \int_3^{10} \frac{u^5}{3}\,du = \frac{10^6 - 3^6}{18}$	Worked example	$u = x^3 + 2$ , $\;du = 3x^2\,dx$
$\displaystyle \left.\frac{(x^3+2)^6}{18}\right\|_1^2 = \left.\frac{u^6}{18}\right\|_3^{10}$	Two correct routes	Back to $x$ with $x$ -limits, or stay in $u$ with $u$ -limits
$\displaystyle\int_{-1}^{1} x^2\,dx = \int_{-1}^{0} x^2\,dx + \int_{0}^{1} x^2\,dx = \tfrac{1}{3} + \tfrac{1}{3}$	Split where $u'$ changes sign	$u = x^2$ with $x = -\sqrt{u}$ on the left, $x = +\sqrt{u}$ on the right

Key Insight: The limits belong to whichever variable the differential names. Substitute only where $u'$ keeps one sign, so that each $u$ comes from exactly one $x$ .

IIVisualization 1 — The Limits Travel With the Variable

As $x$ runs from $1$ to $2$ , $u = x^3 + 2$ runs from $3$ to $10$ , and the two integrals agree all the way.

Upper limit in x 2

IIIVisualization 2 — Match the Antiderivative to Its Limits

For $\int_1^2 (x^3+2)^5 x^2\,dx$ , only two of the four antiderivative-and-limits pairings give the true value.

Antiderivative Limits

IVVisualization 3 — When the Substitution Turns Around

On $[-1, 1]$ , $u = x^2$ falls and then climbs, so $x = \sqrt{u}$ holds only on the right half.

x −0.5 Left half uses

VQuiz Questions

Problem 1 · Convert the Limits

Given: $\displaystyle\int_0^1 2x\,(x^2+1)^3\,dx$ with $u = x^2 + 1$ — find the integral in $u$ , limits included.

✅ Correct!

du = 2x\,dx

absorbs the whole factor

2x\,dx

, and the limits become

u(0) = 1

and

u(1) = 2

❌ Not quite. The limits

0

and

1

are

x

-values. Once the integrand is in

u

, the limits must be

u(0)

and

u(1)

❌ Close, but… the upper limit is right, but the lower one was not converted: at

x = 0

u = 0^2 + 1 = 1

❌ Close, but… there is no leftover

\tfrac12

here. The integrand already contains

2x\,dx

, which is exactly

du

Show solution

With $u = x^2 + 1$ we have $du = 2x\,dx$ , so $(x^2+1)^3 \cdot 2x\,dx = u^3\,du$ .

The limits change with the variable: $u_1 = 0^2 + 1 = 1$ and $u_2 = 1^2 + 1 = 2$ .

\int_0^1 2x\,(x^2+1)^3\,dx = \int_1^2 u^3\,du = \left.\frac{u^4}{4}\right|_1^2 = \frac{16 - 1}{4} = \frac{15}{4}

Problem 2 · The Old-Limits Trap

Given: $\displaystyle\int_0^2 x\,(x^2+1)^2\,dx$ — find its value using $u = x^2 + 1$ .

✅ Correct! The

u

-antiderivative

\tfrac{u^3}{6}

is evaluated at the

u

-limits

1

and

5

❌ Not quite. This is

\tfrac{u^3}{6}

evaluated at

u = 0

and

u = 2

, the old

x

-limits. As

x

runs from

0

2

u

runs from

1

5

❌ Close, but…

du = 2x\,dx

, so

x\,dx = \tfrac{du}{2}

. The factor

\tfrac12

was dropped, which doubles the answer.

❌ Close, but… only the upper limit was converted. At

x = 0

u = 1

, so the lower term is

\tfrac{1^3}{6}

, not

0

Show solution

Let $u = x^2 + 1$ , so $du = 2x\,dx$ and $x\,dx = \tfrac{du}{2}$ . The limits: $u(0) = 1$ , $u(2) = 5$ .

\int_0^2 x\,(x^2+1)^2\,dx = \int_1^5 \frac{u^2}{2}\,du = \left.\frac{u^3}{6}\right|_1^5 = \frac{125 - 1}{6} = \frac{62}{3}

Check by changing back to $x$ : $\left.\tfrac{(x^2+1)^3}{6}\right|_0^2 = \tfrac{125 - 1}{6} = \tfrac{62}{3}$ , using the old limits with the $x$ -antiderivative.

Problem 3 · A Decreasing Substitution

Given: $\displaystyle\int_0^1 x^2\,(1 - x^3)^4\,dx$ with $u = 1 - x^3$ — find the new limits and the value.

As x runs from 0 to 1, u runs…

What is the value of the integral?

✅ Correct! The minus sign in

du = -3x^2\,dx

and the reversed limits

1 \to 0

cancel, leaving

\int_0^1 \tfrac{u^4}{3}\,du = \tfrac{1}{15}

❌ Not quite. Those are the

x

-limits copied over. Compute

u(0) = 1 - 0 = 1

and

u(1) = 1 - 1 = 0

❌ Check the limits. Put each

x

-limit into

u = 1 - x^3

u(0) = 1

is the lower limit and

u(1) = 0

is the upper one.

❌ Close, but… one of the two sign changes was lost. Keep both the minus sign in

x^2\,dx = -\tfrac{du}{3}

and the reversed limits

\int_1^0

❌ Not quite.

x^2\,dx = -\tfrac{du}{3}

, so the factor

\tfrac13

must stay in the integral.

❌ Check the value. Rewrite

x^2\,dx = -\tfrac{du}{3}

and integrate

u^4

between the converted limits.

Show solution

Let $u = 1 - x^3$ . Then $du = -3x^2\,dx$ , so $x^2\,dx = -\tfrac{du}{3}$ . Here $u' = -3x^2 \le 0$ , so $u$ moves steadily down and the substitution is safe.

The limits: $u(0) = 1$ and $u(1) = 0$ , so $u$ runs from $1$ down to $0$ .

\int_0^1 x^2 (1 - x^3)^4\,dx = \int_1^0 u^4 \left(-\frac{du}{3}\right) = \int_0^1 \frac{u^4}{3}\,du = \left.\frac{u^5}{15}\right|_0^1 = \frac{1}{15}

The integrand $x^2(1-x^3)^4$ is never negative on $[0, 1]$ , so a positive answer is expected.

Problem 4 · Where u Turns Around

Given: $\displaystyle\int_{-2}^{1} x^2\,dx$ , computed with $u = x^2$ . Find what the careless route gives and what splitting at $x = 0$ gives.

Using x = √u on the whole interval gives…

Splitting at x = 0 with the correct sign on each piece gives…

✅ Correct! The careless route gives

-\tfrac{7}{3}

, a negative area for a nonnegative integrand. Splitting gives

\tfrac{8}{3} + \tfrac{1}{3} = 3

, matching

\left.\tfrac{x^3}{3}\right|_{-2}^{1}

❌ Not quite.

3

is the true value, but

x = \sqrt{u}

is wrong where

x

is negative, so the careless route cannot produce it.

❌ Not quite. The limits only match when the interval is symmetric. Here

u(-2) = 4

and

u(1) = 1

, so the new integral runs from

4

1

❌ Check the careless route. It becomes

\int_4^1 \tfrac{u}{2\sqrt{u}}\,du = \int_4^1 \tfrac12\sqrt{u}\,du

; mind the order of the limits.

❌ Close, but…

\tfrac13

is only the right piece,

\int_0^1 x^2\,dx

. The left piece

\int_{-2}^0 x^2\,dx = \tfrac83

must be added.

❌ Check the split. With the correct sign each piece is positive: compute

\int_{-2}^{0}

and

\int_{0}^{1}

separately and add.

Show solution

Careless route. With $u = x^2$ , $du = 2x\,dx$ , the limits become $u(-2) = 4$ and $u(1) = 1$ . Writing $x = \sqrt{u}$ everywhere:

\int_4^1 u \cdot \frac{du}{2\sqrt{u}} = \int_4^1 \tfrac12\sqrt{u}\,du = \left.\frac{u^{3/2}}{3}\right|_4^1 = \frac{1 - 8}{3} = -\frac{7}{3}

This is wrong because $u' = 2x$ changes sign at $x = 0$ ; on $[-2, 0]$ , $x = -\sqrt{u}$ .

Split. On $[-2, 0]$ : $x = -\sqrt{u}$ , $dx = -\tfrac{du}{2\sqrt{u}}$ , $u$ runs from $4$ to $0$ :

\int_4^0 u\left(-\frac{du}{2\sqrt{u}}\right) = \int_0^4 \tfrac12\sqrt{u}\,du = \left.\frac{u^{3/2}}{3}\right|_0^4 = \frac{8}{3}

On $[0, 1]$ : $x = \sqrt{u}$ , $u$ runs from $0$ to $1$ , giving $\int_0^1 \tfrac12\sqrt{u}\,du = \tfrac13$ .

Total: $\tfrac83 + \tfrac13 = 3$ . Directly, $\left.\tfrac{x^3}{3}\right|_{-2}^{1} = \tfrac13 - \left(-\tfrac83\right) = 3$ .

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