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Integration Techniques, L’Hopital’s Rule, and Improper Integrals

8.1 Basic Integration Rules

Focus on match standard integral rules, recognize an inner derivative, inverse-trigonometric forms.

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Ideas in this lesson

01Match standard integral rulesWorked steps · five questions02Recognize an inner derivativeWorked steps · five questions03Inverse-trigonometric formsWorked steps · five questions
Idea 1 of 3

Match standard integral rules

Simplify an integrand before choosing a rule. Definite evaluation still requires the correct antiderivative.

Worked example 1

Evaluate ∫02x dx\int_{0}^{2} x\,dx.
  1. F(x)=x22F(x) = \dfrac{x^{2}}{2}
  2. F(2)−F(0)=222−0=2F(2) - F(0) = \dfrac{2^{2}}{2} - 0 = 2

Answer: 22

Worked example 2

Evaluate ∫03x2 dx\int_{0}^{3} x^{2}\,dx.
  1. F(x)=x33F(x) = \dfrac{x^{3}}{3}
  2. F(3)−F(0)=333−0=9F(3) - F(0) = \dfrac{3^{3}}{3} - 0 = 9

Answer: 99

Idea 2 of 3

Recognize an inner derivative

An available u′u' factor often makes substitution the shortest method.

Worked example 1

Evaluate ∫012x(x2+2) dx\int_{0}^{1} 2x(x^{2} + 2)\,dx.
  1. u=x2+2u = x^{2} + 2; du=2xdxdu = 2x dx; x=0  ⟹  u=2,x=1  ⟹  u=3x = 0 \implies u = 2, x = 1 \implies u = 3
  2. ∫012x(x2+2) dx=∫23u du\int_{0}^{1} 2x(x^{2} + 2)\,dx = \int_{2}^{3} u\,du
  3. 32−222=52≈2.5\dfrac{3^{2} - 2^{2}}{2} = \dfrac{5}{2} \approx 2.5

Answer: 52≈2.5\dfrac{5}{2} \approx 2.5

Worked example 2

Evaluate ∫012x(x2+3)2 dx\int_{0}^{1} 2x(x^{2} + 3)^{2}\,dx.
  1. u=x2+3u = x^{2} + 3; du=2xdxdu = 2x dx; x=0  ⟹  u=3,x=1  ⟹  u=4x = 0 \implies u = 3, x = 1 \implies u = 4
  2. ∫012x(x2+3)2 dx=∫34u2 du\int_{0}^{1} 2x(x^{2} + 3)^{2}\,dx = \int_{3}^{4} u^{2}\,du
  3. 43−333=373≈12.33\dfrac{4^{3} - 3^{3}}{3} = \dfrac{37}{3} \approx 12.33

Answer: 373≈12.33\dfrac{37}{3} \approx 12.33

Idea 3 of 3

Inverse-trigonometric forms

Distinguish an a2+x2a^{2} + x^{2} denominator from an a2−x2a^{2} - x^{2} radical; they lead to different antiderivatives.

Worked example 1

Evaluate ∫0214+x2 dx\int_{0}^{2} \dfrac{1}{4 + x^{2}}\,dx.
  1. F(x)=arctan⁡(x2)2F(x) = \dfrac{\arctan (\dfrac{x}{2})}{2}
  2. F(2)−F(0)=π8≈0.39F(2) - F(0) = \dfrac{\pi }{8} \approx 0.39

Answer: π8≈0.39\dfrac{\pi }{8} \approx 0.39

Worked example 2

Evaluate ∫01.519−x2 dx\int_{0}^{1.5} \dfrac{1}{\sqrt{9 - x^{2}}}\,dx.
  1. F(x)=arcsin⁡(x3)F(x) = \arcsin (\dfrac{x}{3})
  2. F(1.5)−F(0)=π6≈0.52F(1.5) - F(0) = \dfrac{\pi }{6} \approx 0.52

Answer: π6≈0.52\dfrac{\pi }{6} \approx 0.52

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §8.1,p8.1, p. 508508. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.