{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 2
#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
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern