{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

quiz3-sol - (2-2 A 2(2 5 ⇒-7 = 0 7 A 2 ⇒ A 2 =-1 2 Set...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Quiz 3 – Math 2153, Calculus II – Sept. 9, 2011 1. Evaluate the integral t - 9 t 2 + 3 t - 10 dt Solution We use partial fractions to solve this problem. Notice that the denominator can be factored as t 2 + 3 t - 10 = ( t + 5)( t - 2) . Therefore the integrand can be written as t - 9 t 2 + 3 t - 10 = A 1 t + 5 + A 2 t - 2 (1) Now we need to compute the values of A 1 and A 2 . Notice that A 1 t + 5 + A 2 t - 2 = A 1 ( t - 2) ( t + 5)( t - 2) + A 2 ( t + 5) ( t + 5)( t - 2) = A 1 ( t - 2) + A 2 ( t + 5) ( t + 5)( t - 2) So Equation (1) can be written as t - 9 t 2 + 3 t - 10 = A 1 ( t - 2) + A 2 ( t + 5) ( t + 5)( t - 2) Compare the numerators and denominators of the above equation, we have t - 9 = A 1 ( t - 2) + A 2 ( t + 5) (2) 1. Set
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (2-2) + A 2 (2 + 5) ⇒-7 = 0 + 7 A 2 ⇒ A 2 =-1 2. Set t =-5 in Equation (2), we have-5-9 = A 1 (-5-2) + A 2 (-5 + 5) ⇒-14 =-7 A 1 + 0 ⇒ A 1 = 2 Plug in the values of A 1 and A 2 back into Equation (1), we have t-9 t 2 + 3 t-10 = 2 t + 5-1 t-2 Hence the integral is Z t-9 t 2 + 3 t-10 dt = Z ± 2 t + 5-1 t-2 ² dt = 2 ln | t + 5 | -ln | t-2 | + C (Or if you like, you can further simplify the answer into ) = ln ³ ³ ³ ³ ( t + 5) 2 ( t-2) ³ ³ ³ ³ + C...
View Full 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