prob6sol - Math 151 Spring 2008 Problem Set # 6 Improper...

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Problem Set # 6 Improper Integrals: Part 2 Solutions 1. If x> 0 0 cos 2 (4 x ) x 3 / 2 1 x 3 / 2 . Since 3 / 2 > 1 , the improper integral Z π 1 x 3 / 2 dx converges. By the comparison theorem, Z π cos 2 (4 x ) x 3 / 2 dx converges. Thus, Z π cos 2 (4 x ) x 3 / 2 dx converges as well. 2. If x 1 then x 1 so that 0 < e x x e x . The improper integral R 1 e x dx converges. Indeed, Z 1 e x dx = lim b + Z b 1 e x dx = lim b + ³ e x ¯ ¯ b 1 ´ = lim b + ¡ e b + e 1 ¢ = e 1 . By the comparison test, the given integral converges as well. 3. If x 9 then x 3 so that x x +4 3 x +4 > 0 . The improper integral Z 9 3 x +4 dx diverges. Indeed, Z b 9 3 x +4 dx =3ln( | x +4 | ) | b 9 =3ln( b +9) 3 ln (13) , so that lim b →∞ Z b 9 3 x +4 dx =l im b →∞ (3 ln ( b +9) 3 ln (13)) = + .
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prob6sol - Math 151 Spring 2008 Problem Set # 6 Improper...

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