MA1102R14chap7 - CHAPTER 7 TECHNIQUES OF INTEGRATION 7.1 Integration by Substitution(Reference TC 8.3 Recall(see 5.4 that by letting u = g(x f(g(x)g(x

MA1102R14chap7 - CHAPTER 7 TECHNIQUES OF INTEGRATION 7.1...

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Unformatted text preview: CHAPTER 7: TECHNIQUES OF INTEGRATION § 7.1 Integration by Substitution (Reference: TC, § 8.3) Recall (see § 5.4) that by letting u = g ( x ), integraldisplay f ( g ( x )) g ′ ( x ) dx = integraldisplay f ( u ) du. This often converts the more complicated integral integraldisplay f ( g ( x )) g ′ ( x ) dx into a simpler integral integraldisplay f ( u ) du . On the other hand, sometimes we begin with an integral integraldisplay f ( u ) du that cannot be eval-uated directly. By setting u = g ( x ) for some suitable one-to-one function g (which has an inverse), the resulting integral integraldisplay f ( g ( x )) g ′ ( x ) dx becomes manageable. This is called inverse substitution . Substitution Rules: Theorem (Inverse Substitution): Let f be a continuous function. Suppose that x = g ( t ) is a differentiable function, and g ′ is continuous. Then integraldisplay f ( x ) dx = integraldisplay f ( g ( t )) g ′ ( t ) dt. [Change notations in integraldisplay f ( g ( x )) g ′ ( x ) dx = integraldisplay f ( u ) du : replace x by t , and u by x .] 166 Example Evaluate integraldisplay 1 x (1 + x 4 ) dx . Solution : Example Evaluate integraldisplay 1 1 + √ x dx . Solution : Let x = t 2 , t > 0. Then dx dt = 2 t = ⇒ dx = 2 t dt . Thus integraldisplay 1 1 + √ x dx = integraldisplay 2 t 1 + t dt = 2 integraldisplay (1 + t ) − 1 1 + t dt = 2 integraldisplay parenleftbigg 1 − 1 1 + t parenrightbigg dt = 2 ( t − ln(1 + t )) + C....
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