Unformatted text preview: Solutions to Quiz 10A
www.math.uﬂ.edu/˜harringt
November 27, 2006
1. (2 pts.) Find f (x) by solving the intial value problem. f (x) = 1+ex +3x2 ; f (1) = e.
Solution: f (x) = f (x)dx = 1 + ex + 3x2 dx = x + ex + x3 + c where c is a
constant. Now, we have to ﬁnd the value for c. So, f (1) = 1 + e1 + 13 + c = 2 + e + c.
Recall: f (1) = e. Hence, 2+ e + c = e implies that c = −2. So f (x) = x + ex + x3 − 2.
ln x
dx.
x 2. (2 pts.) Find the indeﬁnite integral. 1
Solution: Here we have to use u subsitution. Let u = ln x, so du = x dx.
So we have: ln x
dx =
x 1
ln(x) ∗ dx
x
u = du udu 12
u +C
2
1
(ln(x))2 + C
=
2
= 1
Thus, ln x dx = 2 (ln(x))2 + C where C is a constant.
x
As a reminder, make sure you keep this fact straight: ln x2 = (ln x)2 . 3. (1 pt.) True or False. If f and g are integrable, then f (x)g (x)dx = f (x)dx g (x)dx. Solution: This is a false statement. Let f (x) = 1 and g (x) = 1.
f (x)g (x)dx =
f (x)dx g (x)dx = 1dx = x + c0
1dx 1dx = (x + c1 )(x + c2 ) = x2 + c1 x + c2 x + c1 c2 So, f (x)g (x)dx = f (x)dx g (x)dx when f (x) = 1 and g (x) = 1.
Note: each time we integrate, we may obtain diﬀerent constant terms. In other
words, c0 , c1 and c2 may be a diﬀerent numbers. 1 ...
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This note was uploaded on 07/18/2011 for the course MAC 2233 taught by Professor Smith during the Spring '08 term at University of Florida.
 Spring '08
 Smith
 Math, Calculus

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