U-Substitution with Definite Integrals

To do u-substitution with definite integrals, change the limits along with the variable. If u = g(x), the new limits are g(a) and g(b): \int_a^b f\big(g(x)\big)\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du. Then evaluate the u-integral with those limits. You never have to substitute back.

Before this lesson: U-Substitution and Fundamental Theorem of Calculus, Part 2

Why the limits change

The limits of a definite integral are values of the variable of integration. In \int_0^2 \ldots\,dx, the 0 and 2 are x-values. After you switch to u, the integral needs u-values instead.

So run each x-limit through the substitution. If u = x² + 1, then x = 0 gives u = 1 and x = 2 gives u = 5. The u-integral goes from 1 to 5.

U-substitution with definite integrals formula

If u = g(x) and g′ is continuous on [a, b], then:

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

Steps:

1. Choose u and find du, as in an indefinite integral.
2. Change the limits: put a and b into u = g(x).
3. Rewrite the integrand and dx in terms of u.
4. Integrate and evaluate with the new limits. Do not substitute back.

U-substitution with definite integrals examples

Example 1: A polynomial inside a power

Evaluate \int_0^2 2x(x^2 + 1)^2\,dx.

u = x^2 + 1

du = 2x\,dx

x = 0 \to u = 0^2 + 1 = 1

x = 2 \to u = 2^2 + 1 = 5

\int_0^2 2x(x^2 + 1)^2\,dx = \int_1^5 u^2\,du

= \Big[\dfrac{u^3}{3}\Big]_1^5

= \dfrac{125}{3} - \dfrac{1}{3} = \dfrac{124}{3}

Answer: \dfrac{124}{3} \approx 41.33

Example 2: A root in the denominator

Evaluate \int_0^1 \dfrac{x}{\sqrt{x^2 + 3}}\,dx.

u = x^2 + 3

du = 2x\,dx

x = 0 \to u = 3, \quad x = 1 \to u = 4

\int_0^1 \dfrac{x}{\sqrt{x^2 + 3}}\,dx = \int_3^4 \dfrac{1}{2}u^{-1/2}\,du

= \Big[u^{1/2}\Big]_3^4

= \sqrt{4} - \sqrt{3} = 2 - \sqrt{3}

Answer: 2 - \sqrt{3} \approx 0.268

Example 3: A trig substitution

Evaluate \int_0^{\pi/2} \sin^3 x\cos x\,dx.

u = \sin x

du = \cos x\,dx

x = 0 \to u = \sin 0 = 0

x = \dfrac{\pi}{2} \to u = \sin\dfrac{\pi}{2} = 1

\int_0^{\pi/2} \sin^3 x\cos x\,dx = \int_0^1 u^3\,du

= \Big[\dfrac{u^4}{4}\Big]_0^1 = \dfrac{1}{4} - 0 = \dfrac{1}{4}

Answer: \dfrac{1}{4}

Example 4: When the new limits flip

Evaluate \int_1^2 \dfrac{e^{1/x}}{x^2}\,dx.

u = \dfrac{1}{x}

du = -\dfrac{1}{x^2}\,dx

x = 1 \to u = 1, \quad x = 2 \to u = \dfrac{1}{2}

\int_1^2 \dfrac{e^{1/x}}{x^2}\,dx = \int_1^{1/2} -e^u\,du

= \int_{1/2}^1 e^u\,du

= \Big[e^u\Big]_{1/2}^1 = e - e^{1/2}

\approx 1.070

Answer: e - \sqrt{e} \approx 1.070. A flipped pair of limits is fine. Keep them in order, then use the minus sign to swap them.

Example 5: Water flowing into a tank

Water flows into a tank at a rate of r(t) = \dfrac{8t}{(t^2 + 1)^2} liters per minute. How much water flows in from t = 0 to t = 3 minutes?

u = t^2 + 1

du = 2t\,dt

t = 0 \to u = 1, \quad t = 3 \to u = 10

\int_0^3 \dfrac{8t}{(t^2 + 1)^2}\,dt = \int_1^{10} 4u^{-2}\,du

= \Big[-\dfrac{4}{u}\Big]_1^{10}

= -\dfrac{4}{10} - \left(-\dfrac{4}{1}\right)

= -0.4 + 4 = 3.6

Answer: 3.6 liters flow in during the first 3 minutes.

Common mistakes

Related methods

Substitute back and keep the x-limits

You can also find the antiderivative in x first, then use the original limits. In Example 1, \int 2x(x^2 + 1)^2\,dx = \dfrac{(x^2 + 1)^3}{3} + C, and \Big[\dfrac{(x^2 + 1)^3}{3}\Big]_0^2 = \dfrac{125}{3} - \dfrac{1}{3}. Both methods give 124/3.

Symmetric limits

If f is odd, \int_{-a}^{a} f(x)\,dx = 0. If f is even, it equals 2\int_0^a f(x)\,dx. Checking for symmetry first can save a substitution (see even and odd functions).

When u-substitution is not enough

Some integrals need other methods, such as integration by parts, in Calculus 2. If no inside function has its derivative nearby, u-substitution won’t help.

Practice problems

1. Evaluate \int_0^1 3x^2(x^3 + 1)^4\,dx.
2. Evaluate \int_0^2 (2x + 1)^3\,dx.
3. Evaluate \int_0^4 \dfrac{1}{\sqrt{2x + 1}}\,dx.
4. Evaluate \int_0^{\pi/4} \sin(2x)\,dx.
5. Evaluate \int_0^{\pi/3} \sin x\cos^2 x\,dx.
6. Evaluate \int_0^1 x e^{x^2}\,dx.
7. Evaluate \int_1^2 x(x^2 - 1)^3\,dx.
8. Evaluate \int_0^{\pi/4} \sec^2 x\tan^2 x\,dx.
9. Evaluate \int_{-1}^{1} x^3(x^4 + 2)^5\,dx.
10. A factory makes parts at a rate of r(t) = 6t\sqrt{t^2 + 16} parts per hour. How many parts does it make from t = 0 to t = 3 hours?

u = x^3 + 1,\; du = 3x^2\,dx

x = 0 \to u = 1, \quad x = 1 \to u = 2

\int_1^2 u^4\,du = \Big[\dfrac{u^5}{5}\Big]_1^2 = \dfrac{32}{5} - \dfrac{1}{5} = \dfrac{31}{5}

u = 2x + 1,\; du = 2\,dx

x = 0 \to u = 1, \quad x = 2 \to u = 5

\int_1^5 \dfrac{1}{2}u^3\,du = \Big[\dfrac{u^4}{8}\Big]_1^5

= \dfrac{625}{8} - \dfrac{1}{8} = 78

u = 2x + 1,\; du = 2\,dx

x = 0 \to u = 1, \quad x = 4 \to u = 9

\int_1^9 \dfrac{1}{2}u^{-1/2}\,du = \Big[u^{1/2}\Big]_1^9

= 3 - 1 = 2

u = 2x,\; du = 2\,dx

x = 0 \to u = 0, \quad x = \dfrac{\pi}{4} \to u = \dfrac{\pi}{2}

\int_0^{\pi/2} \dfrac{1}{2}\sin u\,du = \Big[-\dfrac{1}{2}\cos u\Big]_0^{\pi/2}

= 0 - \left(-\dfrac{1}{2}\right) = \dfrac{1}{2}

u = \cos x,\; du = -\sin x\,dx

x = 0 \to u = 1, \quad x = \dfrac{\pi}{3} \to u = \dfrac{1}{2}

\int_1^{1/2} -u^2\,du = \int_{1/2}^1 u^2\,du

= \Big[\dfrac{u^3}{3}\Big]_{1/2}^1 = \dfrac{1}{3} - \dfrac{1}{24} = \dfrac{7}{24}

u = x^2,\; du = 2x\,dx

x = 0 \to u = 0, \quad x = 1 \to u = 1

\int_0^1 \dfrac{1}{2}e^u\,du = \Big[\dfrac{1}{2}e^u\Big]_0^1 = \dfrac{e - 1}{2}

u = x^2 - 1,\; du = 2x\,dx

x = 1 \to u = 0, \quad x = 2 \to u = 3

\int_0^3 \dfrac{1}{2}u^3\,du = \Big[\dfrac{u^4}{8}\Big]_0^3 = \dfrac{81}{8}

u = \tan x,\; du = \sec^2 x\,dx

x = 0 \to u = 0, \quad x = \dfrac{\pi}{4} \to u = 1

\int_0^1 u^2\,du = \Big[\dfrac{u^3}{3}\Big]_0^1 = \dfrac{1}{3}

u = x^4 + 2,\; du = 4x^3\,dx

x = -1 \to u = 3, \quad x = 1 \to u = 3

\int_3^3 \dfrac{1}{4}u^5\,du = 0

u = t^2 + 16,\; du = 2t\,dt

t = 0 \to u = 16, \quad t = 3 \to u = 25

\int_{16}^{25} 3u^{1/2}\,du = \Big[2u^{3/2}\Big]_{16}^{25}

= 2(125) - 2(64) = 122

U-substitution with definite integrals FAQ

Do you have to change the limits in u-substitution?

You must either change the limits to u-values or substitute back to x and use the original limits. Changing the limits is usually faster.

How do you find the new limits?

Put each original limit into u = g(x). The lower limit a becomes g(a) and the upper limit b becomes g(b).

What if the new upper limit is smaller than the lower limit?

That is fine. Evaluate it as written, or swap the limits and change the sign of the integral.

Why don’t you substitute back with definite integrals?

After you change the limits, the u-integral already has the right value. Substituting back would need the old limits again.

Summary

• With u = g(x), \int_a^b f\big(g(x)\big)\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du.
• Change both limits by putting them into u = g(x).
• Evaluate in u with the new limits; don’t substitute back.
• Or substitute back and use the original x-limits. Never mix the two.
• Flipped limits are fine; a minus sign swaps them.
• Odd functions on symmetric limits give 0.

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