improper-ans - 1 Math1014 Calculus II Improper Integrals Evaluating improper integrals by taking suitable limits determining convergence or divergence

improper-ans - 1 Math1014 Calculus II Improper Integrals...

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Unformatted text preview: 1 Math1014 Calculus II Improper Integrals • Evaluating improper integrals by taking suitable limits; • determining convergence or divergence of improper integrals by comparison. 1. Determine whether each integral is convergent or divergent. Evaluate those that are convergent. 0 1 dx 2x − 5 (i) −∞ ∞ (iv) 0 8 (vii) 6 0 (i) −∞ ∞ (ii) (x2 0 6 1 dz z 2 + 3z + 2 (viii) 0 0 L→−∞ L→−∞ −∞ 6 6 −∞ 6 − ∞ e−2t dt (iv) L 0 L→∞ = lim L→∞ 3 3rder/3 6 6 3er/3 dr = 3rer/3 −∞ − 9er/3 1 0 ln x √ dx = x 1 (viii) 0 L→0+ L 1/2 2 +2=2 2 0 e1/2 dx x3 1 − e1/2 x−2 2 2 L 1/2 1 e = − e1/2 + lim =∞ 2 8 L→0+ 2L Divergent. 0 0 e1/x dx ≥ x3 = lim 1 2x−1/2 dx L→∞ (3 − x)−1/2 dx Convergent. 2 1 2 − 2(3 − x)1/2 L→3− 2 ln xdx1/2 = 2x1/2 ln x 2 ln x = − lim −1/2 − + x x→0 0 = lim L 1 dx = lim L→3− 3−x = − lim 2(3 − L) (x − 6)−3 dx 0 L −∞ 8 1 = lim − (x − 6)−2 2 L L→6+ 1 1 = − + lim =∞ 8 L→6+ 2(L − 6)2 Convergent. (ix) ln |z + 1| − ln |z + 2| L→3− 8 L 0 = lim Convergent. 6 √ (vi) 2 1 dx = lim (x − 6)3 L→6+ 0 1 dz (z + 1)(z + 2) ln |z + 1 |z + 2| L + 1| 1 = lim ln − ln = ln 1 + ln 2 = ln 2 L→∞ L+2 2 Convergent. r→−∞ 8 x dx (x2 + 2)2 1 1 − dz z+1 z+2 = lim L ∞ 1 dz = + 3z + 2 L = 9e2 − lim (3rer/3 − 9er/3 ) = 9e2 (vii) z2 1 −∞ 0 ln x √ dx x − L→∞ −∞ = 3rer/3 L x dx = lim L→∞ + 2)2 1 √ dx 3−x L 1 2 (x + 2)−1 2 0 1 1 1 = − lim + = L→∞ 2(L2 + 2) 4 4 Convergent. = lim 1 1 −2L = − e−2 + lim e =∞ L→−∞ 2 2 Divergent. rer/3 dr = (x2 0 1 e−2t dt = lim 6 0 (ii) 0 1 ln |2x − 5| 2 L 1 1 ln |2L − 5| = −∞ = ln 5 − lim L→−∞ 2 2 Divergent. 1 (ix) ∞ 1 dx 2x − 5 L 1 e dx x3 L→−∞ (v) 3 (vi) 1/x = lim = lim −∞ 2 2 1 − e−2t 2 e−2t dt −∞ 1 dx = lim L→−∞ 2x − 5 −∞ (iii) rer/3 dr (v) 1 dx (x − 6)3 (iii) 1 x dx + 2)2 1 − 0 L′ Hospital = 2x1/2 d ln x = − lim 2x1/2 ln x − x→0+ 2 x 1 x→0+ − x−3/2 2 − lim 1 0 2x1/2 · 1 − 4x1/2 0 = 0 − 4 = −4 1 dx x L 0 2 2. The average speed of molecules in an ideal gas is 4 M v= √ ¯ π 2RT ∞ 3/2 v 3 e−Mv 2 /(2RT ) dv 0 where M is the molecular weight of the gas, R is the gas constant, T is the temperature, and v is 8RT the molecular speed. Show that v = ¯ . πM Integration by parts! M 4 v= √ ¯ π 2RT ∞ 3/2 v 3 e−M v 2 /(2RT ) 0 M 4 = −√ π 2RT 3/2 RT M 4 M = lim − √ v→∞ π 2RT ∞ v 2 e−M v 2 /(2RT ) 0 v2 3/2 RT v2 v→∞ eM v2 /(2RT ) 3/2 · − 0 ∞ 3/2 0 3/2 RT 2RT = M M 2v = lim ∞ 3/2 4 M +√ π 2RT M 4 +√ π 2RT M eM v2 /(2RT ) 4 M = √ π 2RT (By L’Hospital’s rule, lim M 4 dv = √ π 2RT − RT 2 −M v2 /(2RT ) v de M RT −M v2 /(2RT ) 2 e dv M RT 2RT −M v2 /(2RT ) e M M ∞ 0 8RT πM = 0.) v→∞ 2M v eM v 2 /(2RT ) 2RT 3. (§7.8, Q 78.) Find the value of the constant C for which the integral ∞ C x − x2 + 1 3x + 1 0 dx converges. Evaluate the integral for this value of C. L x C x C − dx = lim − dx L→∞ 0 x2 + 1 3x + 1 x2 + 1 3x + 1 0 L 1 C 1 C = lim ln(x2 + 1) − ln |3x + 1| = lim ln(L2 + 1) − ln |3L + 1| L→∞ 2 L→∞ 2 3 3 0 (L2 + 1)1/2 L(1 + 1/L2 )1/2 = lim ln = lim ln C/3 L→∞ L→∞ (3L + 1)C/3 L (3 + 1/LC/3 )C/3 So the limits exist if and only if C/3 = 1, i.e. ,C = 3. Moreover, for C = 3, ∞ ∞ x2 0 x C − +1 3x + 1 dx = lim ln L→∞ L(1 + 1/L2 )1/2 1 = ln L(3 + 1/L) 3 4. (§7.8, Q80.) Show that if a > −1, and b > a + 1, then the following integral is convergent. ∞ 0 a Note that for x ≥ 1, 0 ≤ a x x ≤ b = xa−b . Hence for a − b + 1 < 0, 1 + xb x ∞ 1 = lim L→∞ Note that xa dx 1 + xb xa dx ≤ 1 + xb 1 xa−b+1 a−b+1 ∞ 0 L 1 xa−b dx = lim L→∞ 1 = lim L→∞ xa dx = 1 + xb 1 0 xa dx + 1 + xb ∞ 1 + but we have to be careful with x = 0, since x −→ ∞ as x → 0 0 a x dx = lim 1 + xb L→0+ 1 L a x dx ≤ lim 1 + xb L→0+ xa−b dx 1 1 1 1 La−b+1 − =− a−b+1 a−b+1 a−b+1 a 1 L ∞ 1 L xa dx = lim if a < 0. Now, if a + 1 > 0, L→0+ Hence the improper integral converges if a > −1 and b > a + 1. xa dx 1 + xb 1 a+1 x a 1 L = 1 1 1 − lim La+1 = a L→0+ a a ...
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