{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Integ_tut_sol1 - Integration I Solutions to Integration I...

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Integration I Solutions to Integration I Exercises 1. a) 4 7 3 3 3 x dx x C 7 = + b) 2 1 2 3 3 dx 3 x dx 3x C C x x - - = = - + = - + c) ( 29 2 3 2 2 1 1 x x dx x x 2 dx x 2x C x 3 - - + = + + = - + + 2. a) ( 29 3 2 3 2 2 1 2 1 1 2 d 2 d 2 C 2 1 1 2 C 2 - - - - + - θ = θ + θ - θ θ θ θ = - - θ - θ + = - - - θ+ θ θ b) ( 29 2 1 2 3 3 2 3 4 ax bx c dx ax bx cx dx x a b c x x x C 2 3 4 - - - + + = + + = + + + 3. a) ( 29 2 2 2 2 2 4 1 1 2 3 1 1 2 x 1 dx x 2x dx x x 2x 2 8 1 1 2 3 3 8 3 1 6 3 4 3 - -   - =-       = +     = +- +   +-- = = 3 4 3 6 1 3 8 2 3 1 1 3 8 x 2 3 x 2 1 1 3 = - - + = + - + = + = - b) ( 29 ( 29 ( 29 ( 29 [ ] 4 4 2 0 0 4 3 2 0 4 4 3 2 0 x x 1 x 2 dx x x 3x 2 dx x 3x 2x dx x x x 4 64 64 16 16 - - = - + = - + = - + = - + = 4. 1
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 I ( 29 ( 29 ( 29 2 2 2 2 2 2 2 2 2 1 dy 1 x 1 dx dy 1 dx 1 x dx y 1 x Let x tan 1 x 1 tan sec and dx sec d sec y d sec y d C tan x C - + = = + = + = θ ∴ + = + θ = θ = θ θ θ = θ θ = θ = θ+ = + 5. ( 29 2 2 dy 2x 1 dx y 2x 1 dx y x x C when x 1 and y 1 1 1 1 C C 1 y x x 1 = - = - = - + = = = - + = = - + 6. 2 2 2 1 3 2 1 2 d x 2 3t (acceleration) dt dx 3 2t t C (velocity) dt 2 t x t C t C (distance) 2 = + = + + = + + + when t = 0 x = 5 5 = 0 + 0 + 0 + C 2 C 2 = 5 When t = 1 dx 10 dt = 2
Image of page 2
Integration I 1 1 3 2 3 10 2 C 2 13 C 2 t 13t x t 5 2 2 When t 1 1 13 x 1 5 2 2 x 13 = + + = = + + + = = + + + = 7. a) ( 29 ( 29 ( 29 3 3 3 4 4 Evaluate 2 x dx Let 2 x u dx du 2 x dx u du u C 4 1 2 x C 4 - - = - = - = - = - + = - - + b) ( 29 ( 29 ( 29 ( 29 3 3 3 2 2 Evaluate 2x 1 dx Let 2x 1 u 2dx du 1 2x 1 dx u du 2 u C 4 1 2x 1 C 4 1 C 4 2x 1 - - - - - - - = = - = =- + =- - + =- + - = – 2 ) 1 x 2 ( 4 1 - + C 3
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
Integration I c) 1 2 1 2 dx Evaluate 5x 7 Let 5x 7 u 5dx du dx 1 u du 5 5x 7 2 u C 5 2 5x 7 C 5 - - - = = = - = + = - + d) ( 29 ( 29 ( 29 2 2 2 Evaluate sec 2x 1 dx Let 2x 1 u 2dx du 1 sec 2x 1 dx sec u du 2 1 tan u C 2 1 tan 2x 1 C 2 - - = = - = = + = - + 8. a) Evaluate θ θ - π π d 2 3 sin 6 / 0 Let θ - π 2 3 = u, when θ = 0, u = 3 π –2d θ = du θ = 6 π , u = 3 π 3 π = 0 [ ] 6 0 6 0 0 3 3 0 3 0 Evaluate sin 2 d 3 Let 2 u when 0 u 3 3 2d du u 0 6 3 3 1 sin 2 d sin u du 3 2 1 sin u du 2 1 cosu 2 1 1 1 2 2 1 4 π π π π π π - θ θ π π - θ= θ= = π π π - θ= θ= = - = π - θ θ=- = =- =- - = 4
Image of page 4
Integration I b) Evaluate + = + 3 / 2 0 2 3 / 2 0 2 u 4 9 1 du 4 1 u 9 4 du Let 4 9 u 2 = tan 2 θ , when u = 0, θ = 0 2 3 u = tan θ u = 3 2 , tan θ = 1 [ ] 2 3 2 3 2 2 0 0 2 2 2 2 3 4 2 2 2 0 0 4 2 2 0 4 0 4 0 du 1 du Evaluate 9 4 9u 4 1 u 4 9 Let u tan when u 0 0 4 3 2 u tan u tan 1 2 3 2 du sec d 3 4 du 1 sec d 4 9u 6 1 tan 1 sec d 6 sec 1 d 6 1 6 24 π π π π = + + = θ = θ= = θ = θ= π = θ θ θ= θ θ = + + θ θ θ = θ = θ = θ π = c) Evaluate - = - 2 0 2 2 0 2 4 x 1 dx 2 1 x 4 dx Let 4 x 2 = sin 2 θ , when x = 0, θ = 0 2 x = sin θ x = 2, sin θ = 1 dx = 2 cos θ d θ θ = 2 π [ ] 2 d cos d cos sin 1 d cos x 4 dx 2 / 0 2 / 0 2 / 0 2 / 0
Image of page 5

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

View Full Document Right Arrow Icon
Image of page 6
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