080205quiz3solution

# 080205quiz3solution - Z 1 √ x 2 16 dx We use the...

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

MATH 126 QUIZ #3 Solution 02/07/08 Mathias Knape Last Name : Circle discussion section First Name: 11 am 12 pm 1 pm Note: In order to receive full credit, please show all your work (conceptual steps, definitions you use, ...) 1. Evaluate the integrals (a) 1 x 3 + 2 x 2 + x dx At first we have to find the partial fraction representation. Observe that 1 x 3 + 2 x 2 + x = 1 x ( x + 1) 2 = A x + B x + 1 + C ( x + 1) 2 Multiplying by x ( x + 1) 2 we obtain that 1 = A ( x + 1) 2 + Bx ( x + 1) + Cx. Plugging in x = 0 and x = - 1 gives us A = 1 and C = - 1 respectively. Now we can choose an arbitrary point x , e.g. x = 1, and insert it in the equation above to obtain B = - 1. Thus 1 x 3 + 2 x 2 + x dx = 1 x - 1 x + 1 - 1 ( x + 1) 2 dx = ln | x | - ln | x + 1 | + 1 x + 1 + D (b) x + 1 x 2 - 4 x + 6 dx x + 1 x 2 - 4 x + 6 dx = x + 1 ( x - 2) 2 + 2 dx = u + 3 u 2 + 2 du = u u 2 + 2 du + 3 u 2 + 2 du = 1 2 ln( u 2 + 2) + 3 2 arctan( u 2 ) + C = 1 2 ln( x 2 - 4 x + 6) +

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Z 1 √ x 2 + 16 dx We use the trigonometric substitution x = 4 tan θ , then dx = 4 sec 2 θdθ . Thus Z 1 √ x 2 + 16 dx = Z 4 sec 2 θ 4 sec θ dθ = Z sec θdθ = ln | sec θ + tan θ | + C = ln ± ± ± ± ± √ x 2 + 16 4 + x 4 ± ± ± ± ± + C 2. Use Simpson’s Rule to approximate Z 3 1 1 + y 5 dy for n = 6. You do not have to com-pute a ﬁnal numerical value. Deﬁne f ( y ) = 1 1+ y 5 and notice that Δ x = 3-6 = . 5. Using Simpson’s Rule we have that Z 3 1 1 + y 5 dy ≈ 1 3 1 2 [ f (0) + 4 f ( 1 2 ) + 2 f (1) + 4 f ( 3 2 ) + 2 f (2) + 4 f ( 5 2 ) + f (3)] ....
View Full Document

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern