Explanation Given: The intgeral is ∫ d x x 2 − 4 x + 8 . Key concept used: Steps of integration by substitution: If the integral is in the below form: ∫ f ( g ( x ) ) g ′ ( x ) d x Let u = g ( x ) and d u = g ′ ( x ) d x The intgeral becomes, ∫ f ( u ) d u . Thus, integrate f ( u ) . After that substitute g ( x ) back. Trigonometic formula: sec 2 x = tan 2 x + 1 Integration formula: ∫ sec x d x = ln | sec x + tan x | + c Calculation: The intgeral is ∫ d x x 2 − 4 x + 8 . Part a: We have to complete the square of x 2 − 4 x + 8 . x 2 − 4 x + 8 = x 2 − 4 x + 4 + 4 = ( x 2 − 4 x + 4 ) + 4 = ( x 2 − 2 ⋅ 2 ⋅ x + 2 2 ) + 4 = ( x − 2 ) 2 + 4 Hence, the complete the square of x 2 − 4 x + 8 is ( x − 2 ) 2 + 4 . Part b: We have to use the substitution u = x − 2 to show that I = ∫ d u u 2 + 2 2 . The intgeral is ∫ d x x 2 − 4 x + 8 . By part a, we can rewrite the integral as: ∫ d x x 2 − 4 x + 8 = ∫ d x ( x − 2 ) 2 + 4 ... (1) Let u = x − 2 . Now, differentiate u = x − 2 . d u = d x Substitute u for x − 2 and d u for d x in equation (1). ∫ d x ( x − 2 ) 2 + 4 = ∫ d u u 2 + 2 2 … (2) Now, solve the integral. Let u = 2 tan t . Now, differentiate u = 2 tan t . d u = 2 sec 2 t d t Substitute 2 tan t for u and 2 sec 2 t d t for d u in equation (2)...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 8.3, Problem 37E

To determine

a. Complete the square to show that $x^{2} - 4 x + 8 = {(x - 2)}^{2} + 4$ .

b. Use the substitution $u = x - 2$ to show that $I = \int \frac{d u}{\sqrt{u^{2} + 2^{2}}}$ . Also, evaluate the u-integral.

c. Show that $I = \ln | \sqrt{{(x - 2)}^{2} + 4} + x - 2 | + c$ .

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Chapter 8.3, Problem 36E

Chapter 8.3, Problem 38E

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Chapter 8 Solutions

CALCULUS (CLOTH)

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a. Complete the square to show that x 2 − 4 x + 8 = ( x − 2 ) 2 + 4 . b. Use the substitution u = x − 2 to show that I = ∫ d u u 2 + 2 2 . Also, evaluate the u -integral. c. Show that I = ln | ( x − 2 ) 2 + 4 + x − 2 | + c .

Solution Summary: The author explains the steps of integration by substitution: complete the square, evaluate the u -integral, integrate f, and substitute g.

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a. Complete the square to show that $x^{2} - 4 x + 8 = {(x - 2)}^{2} + 4$ .

b. Use the substitution $u = x - 2$ to show that $I = \int \frac{d u}{\sqrt{u^{2} + 2^{2}}}$ . Also, evaluate the u-integral.

c. Show that $I = \ln | \sqrt{{(x - 2)}^{2} + 4} + x - 2 | + c$ .

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Chapter 8 Solutions

a. Complete the square to show that x 2 − 4 x + 8 = ( x − 2 ) 2 + 4 . b. Use the substitution u = x − 2 to show that I = ∫ d u u 2 + 2 2 . Also, evaluate the u -integral. c. Show that I = ln | ( x − 2 ) 2 + 4 + x − 2 | + c .

Solution Summary: The author explains the steps of integration by substitution: complete the square, evaluate the u -integral, integrate f, and substitute g.

Videos

a. Complete the square to show that x2−4x+8=(x−2)2+4. b. Use the substitution u=x−2 to show that I=∫duu2+22. Also, evaluate the u-integral. c. Show that I=ln| (x−2)2+4+x−2 |+c.

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Chapter 8 Solutions

a. Complete the square to show that $x^{2} - 4 x + 8 = {(x - 2)}^{2} + 4$ .

b. Use the substitution $u = x - 2$ to show that $I = \int \frac{d u}{\sqrt{u^{2} + 2^{2}}}$ . Also, evaluate the u-integral.

c. Show that $I = \ln | \sqrt{{(x - 2)}^{2} + 4} + x - 2 | + c$ .