{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MATH test 2 KEY

# MATH test 2 KEY - f t = t 2 5 t 6 on 1,5 1 6 t 2 5 t 6 1...

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

Math 2015 Test 2 Spring 2007 Answers 1. Evaluate cos x sin 3 x dx ! . u = sin x ; du = cos xdx ans : 1 4 sin 4 x + C 2. Find x 1 + x 2 dx 1 2 ! . u = 1 + x 2 ; du = 2 xdx ; when x = 1, u = 2; when x = 2, u = 5 1 2 1 u du 2 5 ! = 1 2 ln u 2 5 = 1 2 ln 5 2 " # \$ % & 3. Solve ! F ( x ) = 3 x 2 + 1 x with F (1) = 5 . F ( x ) = x 3 + ln x + 4 4. ! T = k ( T " 60) C = 20; k = ln(0.75) 2 # " 0.144 T = 60 + 20 e " 0.144 t t = " 4.6 hours; 4.6 hours before midnight 5. Smallest: (5, ! 6) Largest: (0, 1) 6. Evaluate 3 e ! 2 x dx 0 " # = 3 2 . 7. Determine the average value of the following function in the given interval.
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: f ( t ) = t 2 ! 5 t + 6 on [ ! 1,5] 1 6 t 2 ! 5 t + 6 ! 1 5 " dt = 3 8. Find x 2 x 3 + 1 dx ! u = x 3 + 1 ! du = 3 x 2 dx ; ans : 2 9 x 3 + 1 ( ) 3 2 + C MULTIPLE CHOICE: Answers: 4, 3, 1, 2, 4, 2, 4, 1, 3, 2, 3, 4, 1, 2, 3...
View Full Document

{[ snackBarMessage ]}