SampleMT1-1solutions - Solutions to Practice Midterm #1...

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Unformatted text preview: Solutions to Practice Midterm #1 x1/3 ln x dx = 3 3 4/3 x = x4/3 ln x − 4 4 3 4/3 9 x1/3dx = x ln x − x4/3 + C ; 4 16 ln xd 2 3 − u−3 u−1 3 4/3 x d ln x 4 1. (a) 3 3 = x4/3 ln x − 4 4 (b) (c) u+3 du = (u − 1)(u − 3) du = 3 ln|u − 3| − 2 ln|u − 1| + c; dx (1 − sin x) (1 − sin x) = dx = dx 1 + sin x (1 − sin x)(1 + sin x) (1 − sin2 x) (1 − sin x) dx sin xdx = dx = − = tan x − sec x + C . 2x 2x cos cos cos2 x √ (d) Make the substitution u = (1 + x)1/2. Then x = (u2 − 1)2 , dx = 4u(u2 − 1); (1 + √ 1/2 x) dx = 3 √ √ 4 4 4 4 4u2(u2 − 1)du = u5 − u3 + C = (1 + x)5/2 − (1 + x)3/2 + C. 5 3 5 3 3 t 2. (a) 1 dx = lim x − 1 t→1+ dx = 1 + ex dx = lim (ln 2 − ln(t − 1)) = ∞. Integral diverges. x − 1 t→1+ du = u(u + 1) du − u (b) First make the substitution u = ex. Then du u+1 u ex = ln|u| − ln|u + 1| + C = ln + C. + C = ln u+1 1 + ex t dx ex e0 = lim = lim ln − ln t→∞ 0 1 + ex t→∞ 1 + et 1 + e0 et 1 = ln lim − ln = ln 1 + ln 2 = ln 2. t t→∞ 1 + e 2 ∞ 0 dx 1 + ex (b − a)5K (c) We use the formula |Sn | ≤ if |f (4)(x)| ≤ K for a ≤ x ≤ b. First let us 180n4 estimate K . (sin(x2 ))(4) = (16x4 − 12) sin(x2 ) − 48x2 cos(x2). One can choose K = 12 + 48 = 60 since |(16x4 − 12) sin(x2)| ≤ 12 and |48x2 cos(x2)| ≤ 48. 60 60 Then |Sn | ≤ . Apply the condition |Sn | ≤ 0.001. Since ≤ 0.001 implies that 4 180n 180n4 4 n ≥ 60000/180 ≈ 333. Thu n = 5 is sufficiently large but since n is even in Simpson rule take n = 6. The formula for calculation of the integral is S6 = 1 18 0 + 4 sin 1 1 1 4 25 + 2 sin + 4 sin + 2 sin + 4 sin + sin 1 . 36 9 4 9 36 4. The first integral is divergent because ex 1 ≥ for 0 ≤ x ≤ 1 x x and 0 1 dx x is divergent. The second integral converges because ex e √ ≤ √ for 0 ≤ x ≤ 1 x x and the integral 0 1 e √ dx = e x 1 0 dx √ x is convergent. ...
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SampleMT1-1solutions - Solutions to Practice Midterm #1...

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