10pm1-solutions

# 10pm1-solutions - MATH 3B CALCULUS WITH APPLICATIONS 2...

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MATH 3B CALCULUS WITH APPLICATIONS 2 SOLUTIONS TO PRACTICE MIDTERM #1 Exercise 1. Find the following integrals. (a) Z e s cos ( e s ) ds (b) Z 1 0 4 x 3 x 2 + 4 dx (c) Z 1 0 ( 4 u + 1) 2 du (d) Z e x x dx (a) Let u = e s . Then du = e s ds and the integral becomes Z cos ( u ) du. Since Z cos ( u ) du = sin ( u ) + c, it follows that Z e s cos ( e s ) ds = sin ( e s ) + c. (b) Let u = x 2 . Then du = 2 x dx and the integral becomes Z 1 0 2 3 u + 4 du. By the Second Part of the Fundamental Theorem of Calculus we have that Z 1 0 2 3 u + 4 du = 2 3 ln (3 u + 4) 1 0 = 2 3 [ln(3 + 4) - ln(4)] = 2 3 ln 7 4 . (c) We have that Z 1 0 ( 4 u + 1) 2 du = Z 1 0 ( 4 u + 1)( 4 u + 1) du = Z 1 0 u + 2 4 u + 1 du. 1

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2MATH 3B CALCULUS WITH APPLICATIONS 2 SOLUTIONS TO PRACTICE MIDTERM #1 By the Second Part of the Fundamental Theorem of Calculus we have that Z 1 0 u + 2 4 u + 1 du = 2 3 u 3 / 2 + 8 5 u 5 / 4 + u 1 0 = 2 3 + 8 5 + 1 - 0 = 49 15 . (d) Let u = x . Then du = 1 2 x dx and the integral becomes Z 2 e u du. Since Z 2 e u du = 2 e u + c, it follows that Z e x x dx = 2 e x + c.
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