Substitution in definite integrals

About this lesson

Calculus BC (AP-aligned) › Techniques: substitution

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Substitution in definite integrals

Warm-up

Three quick questions from earlier lessons.

Discover

In ∫022x(x2+1)3 dx\displaystyle\int_0^2 2x(x^2 + 1)^3\,dx you put u=x2+1u = x^2 + 1. When xx goes from 0 to 2, where does uu go?

  1. From 1 to 5
  2. From 0 to 2

Key idea

Change the limits with the variable.

∫abf(g(x)) g′(x) dx=∫g(a)g(b)f(u) du\displaystyle\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du. Then there is no need to go back to xx.

Worked example

∫022x(x2+1)3 dx\displaystyle\int_0^2 2x(x^2 + 1)^3\,dx.

∫15u3 du\displaystyle\int_1^5 u^3\,du
u = x² + 1 from 1 to 5.
625−14\dfrac{625 - 1}{4}
u⁴/4 from 1 to 5.
=156= 156
Compute.

Your turn, guided

Your turn: ∫0πcos⁡x sin⁡2x dx\displaystyle\int_0^{\pi} \cos x\,\sin^2 x\,dx with u=sin⁡xu = \sin x.

∫00u2 du=0\displaystyle\int_0^0 u^2\,du = 0
u runs from sin 0 = 0 to sin π = 0.

Practice ★

  1. Evaluate ∫012x(x2+1)4 dx\displaystyle\int_0^1 2x(x^2 + 1)^{ 4 }\,dx.
  2. Evaluate ∫012x(x2+1)3 dx\displaystyle\int_0^1 2x(x^2 + 1)^{ 3 }\,dx.
  3. Evaluate ∫012x(x2+1)2 dx\displaystyle\int_0^1 2x(x^2 + 1)^{ 2 }\,dx.

Watch out!

Practice ★★

  1. Evaluate ∫0π/2sin⁡6xcos⁡x dx\displaystyle\int_0^{\pi/2} \sin^{ 6 } x \cos x\,dx.
  2. Evaluate ∫0π/2sin⁡3xcos⁡x dx\displaystyle\int_0^{\pi/2} \sin^{ 3 } x \cos x\,dx.
  1. Evaluate ∫1e(ln⁡x)1x dx\displaystyle\int_1^{e} \dfrac{(\ln x)^{ 1 } }{x}\,dx.
  1. Boss problem. Evaluate ∫08xx2+36 dx\displaystyle\int_0^{ 8 } x\sqrt{x^2 + 36}\,dx with u=x2+36u = x^2 + 36.

Exit ticket

Three questions to check you have got it.

  1. Evaluate ∫012x(x2+1)4 dx\displaystyle\int_0^1 2x(x^2 + 1)^{ 4 }\,dx.
  2. Evaluate ∫012x(x2+1)3 dx\displaystyle\int_0^1 2x(x^2 + 1)^{ 3 }\,dx.
  3. Evaluate ∫012x(x2+1)2 dx\displaystyle\int_0^1 2x(x^2 + 1)^{ 2 }\,dx.

Summary card

Definite integrals by substitution

  • New limits: u(a)u(a) and u(b)u(b).
  • Integrate in uu between them.

Watch out: never mix xx-limits with a uu-integral.

End of the lesson

Next lesson: Integrals giving logarithms; exponentials and tangent Back to the chapter

Answers: Substitution in definite integrals

A

  1. 315\frac{31}{5}
  2. 154\frac{15}{4}
  3. 73\frac{7}{3}

B

  1. 17\frac{1}{7}
  2. 14\frac{1}{4}

C

  1. 12\frac{1}{2}

D

  1. 7843\frac{784}{3}

E

  1. 315\frac{31}{5}
  2. 154\frac{15}{4}
  3. 73\frac{7}{3}

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