{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1 - Solution set for MATH 155 of Assignment 1 Prob 1 Given...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution set for MATH 155, of Assignment 1 Prob. 1 Given the function as, f ( x ) = x - 2 x 2 + 3 x 3 - 4 x 4 Integrating, we get f ( x ) = F ( x ) = x 2 - 2 x 3 3 + 3 x 4 4 - 4 x 5 5 + C . where C is a constant. Prob. 2 Given the function as, f ( x ) = x 1 + x We can rewrite it as, f ( x ) = 1 + x - 1 1 + x = 1 - 1 1 + x Integrating, we get f ( x ) = F ( x ) = x - ln ( 1 + x ) + C . where C is a constant. Prob. 3 Given the function as, f ( x ) = x 7 + 1 x 7 Integrating, we get f ( x ) = F ( x ) = x 8 8 - 1 6 x 6 + C . where C is a constant. Prob. 4 Given the function as, f ( x ) = e x 2 + e - x 2 Integrating, we get f ( x ) = F ( x ) = 2 e x 2 - 2 e - x 2 = 2 ( e x 2 - e - x 2 ). + C where C is a constant. 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
Prob. 5 Given the equation as, dy dx = 1 + sec x 4 Thus, we have the general solution by finding the antiderivative of the above function. Hence, y = x + 4 tan x 4 + C where C is a constant. Prob. 6 Given the equation as, dT dt = cos t ) and given that, W ( 0 ) = 3 Substituting x = π t we have dx = π dt Thus, we have the new function as, dT dx = 1 π sin x and when t = 0 , x = 0 Thus, we have the general solution by finding the antiderivative of the above function.
Image of page 2
Image of page 3
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