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Calculus 2-A2

# Calculus 2-A2 - Calculus II Calculus Early Transcendental...

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Calculus II Calculus: Early Transcendental Functions , 4th ed., by Ron Larson, Robert Hostetler, and Bruce H. Edwards (Boston: Houghton Mifflin, 2007; ISBN-10: 0-618-60624-6). Written Assignment 2 Complete the following textbook exercises for sections 8.2, 8.3, 8.4, and 8.5, and submit them to your mentor for correction and grading. Show all calculations. SECTION 8.2 Exercises Do exercises 8, 10, 14, 16, 18, and 26 on page 531 of the textbook. In Exercises 5– 10, identify and for finding the integral using integration by parts. (Do not evaluate the integral.) #8 u = ln3x and dv = dx #10 u = x 2 and dv = cosxdx In Exercises 11– 36, find the integral. (Note: Solve by the simplest method— not all require integration by parts.) #14 ƒ t -2 e t -1 d t Regular u substitution u = t -1 => du = -t -2 => -du = t -2 dt -ƒ e u du = -e u + C = -e t-1 + C = -e 1/t + C #16 u = lnx => du = 1/x dx dv = x 4 dx => v= ƒ x 4 dx = x 5 /5

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Integration using the parts formula: uv – ƒ v d u 1/5x 5 lnx – ƒ x 5 /5 * 1/x dx = 1/5x 5 lnx – 1/5 ƒ x 4 dx =1/5x 5 lnx – 1/5 (x 5 /5) +C =1/5x 5 lnx – 1/25x 5 +C #18 t = ln x => dt = 1/x dx Rewrite as ƒ1/t 3 dt = ƒ t - 3 dt = -1/2t 2 + C =-1/(2(lnx) 2 ) + C #26 Rewrite as: ƒ x (2+3x) -1/2 dx u = x => du = d x d v = (2 + 3 x ) -1/2 dx = ƒ (2 + 3 x ) -1/2 dx = 2/3(2 + 3 x ) u =x du= dx x/√(2 + 3 x ) dx = (2x√(2 + 3 x ))/3 -2/3ƒ (2 + 3 x ) dx = (2x√(2 + 3 x ))/3 -4/27 * (2 + 3 x ) 3/2 + C = (2√(2 + 3 x ))/27 [9x - 2(2 + 3 x )] +C = (2√(2 + 3 x ))/27 (3x – 4) + C SECTION 8.3 Exercises Do exercises 6, 8, 14, 20, 26, 30, 66, and 72 on pages 540–541 of the textbook. In Exercises 5– 18, find the integral.
#6 =ƒ cosx(1 – sin 2 x) sin 4 x dx u = sinx, du = cosx ƒ(1-u 2 )u 4 du = ƒu 4 – u 6 du = ƒu 5 /5 – u 7 /7 du = (sin 5 x)/5 – (sin 7 x)/7 +C #8 sin 3 x = sin 2 x * sinx cos 2 x + sin 2 x = 1 sin 2 x = 1 - cos 2 x (sinx) sin 3 x = sin 2 x (sinx) 1 - cos 2 x (sinx) = sinx – cos 2 x sin x ƒsinx dx = - cosx

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