MAT1332W10A3S - MAT1332 B C Assignment#3 solutions Question...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
MAT1332 B & C Assignment #3 solutions Question 1 a ) Using the definition of an improper integral, we have Z 1 1 3 x dx = lim q →∞ Z q 1 x - 1 / 3 dx = lim q →∞ 3 2 x 2 / 3 | q 1 = + . Thus, this integral diverges. b ) Using the definition of tan, we have Z 3 π/ 4 0 sin( x ) tan( x ) dx = Z 3 π/ 4 0 cos( x ) dx = sin( x ) fl fl fl 3 π/ 4 0 = 2 2 . c ) Using integration by parts with f ( x ) = ln( x ) and g ( x ) = ln( x ), we have Z 1 0 ln( x ) 1 x dx = (ln( x ) ln( x ) | 1 0 - Z 1 0 1 x ln( x )) . Thus Z 1 0 ln( x ) 1 x dx = 1 2 lim q 0+ ([ln(1)] 2 - [ln( q )] 2 ) = -∞ . d ) (2 points) Integration by parts gives 2 Z 0 arctan x 1 1 + x 2 dx = 2 lim q →∞ [arctan( x )] 2 | q 0 - 2 Z 0 arctan x 1 + x 2 dx. Hence 2 Z 0 arctan x 1 + x 2 dx = lim q →∞ [arctan( x )] 2 | q 0 = π 2 4 . Question 2 a ) (3 points) For the integral I = R x 3 +1 x 2 +3 dx , the numerator x 3 + 1 has higher degree than the denominator x 2 + 3, so we need to use long division. If we do this, the result is x with remainder - 3 x + 1. Thus x 3 + 1 x 2 + 3 = x + - 3 x + 1 x 2 + 3 = x - 3 x x 2 + 3 + 1 x 2 + 3 so we can calculate I = Z x dx - Z 3 x x 2 + 3 dx + Z 1 x 2 + 3 dx .
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
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