Page 205 - Calculus Workbook For Dummies

189
Chapter 10: Integration: Reverse Differentiation

2 π
j # sinx dx = - cosx@ 2 0 π = -^ 1 - h 0 Do you see why the answer is zero?
1 =
0 3
2
3
k # _ x - 4 x + 5 x - 10i dx = –6.58.
2
3
3
2
# _ x - 4 x + 5 x - 10i dx
2
3
1 4 5
2
3
4
= x - x + x - 10 xE
4 3 2
2
1 4 5 1 4 5
=c $ 81 - $ 27 + $ 9 - 30 - cm $ 16 - $ 8 + $ 4 - 20m
4 3 2 4 3 2
. - . 6 58
2
l # e dx . . 7 02
x
- 1
x l x x
_ e i = e , so e is its own antiderivative as well as its own derivative. Thus,
2
# e dx e A 2 = e - e . . 7 02.
x
x
2
1
-
=
- 1
- 1
3 1 4
1 +
m #^ 4 x - 1h dx = ^ 4 x - h C
16
1 4
1. Guess your answer: ^ 4 x - 1h
4
3
1 $
4 ^
2. Differentiate: x - h 4 (by the chain rule). It’s 4 times too much.
1 4
3. Tweak guess: 4 ^ x - 1h
16
1 3 3
1 $
4. Differentiate to check: ^ 4 x - h 4 = ^ 4 x - 1h . Bingo.
4
n # sec x dx = 1 tan x + C
2
6
6
6
Your guess at the antiderivative, tan x, gives you tan x = l sec x 6 $ . Tweak the guess
2
6
6
^
6 h
1 1 l 1 2 2
6 $
6
6
6 m
to tan x. Check: c tan x = sec x 6 = sec x.
6 6 6
o # cos x - 1 dx 2 sin x - 1 + C
=
2 2
x - 1 x - 1 1
Your guess is sin . Differentiating that gives you cosc $ m .
2 2 2
x - 1
Tweaked guess is sin . That’s it.
2
2
p # 3 dt = 3 ln t 2 + 5 + C
t 2 + 5 2
1
2
ln t + 5 is your guess. Differentiating gives you: 2 $ .
t 2 + 5
You wanted a 3, but you got a 2, so tweak your guess by 3 over 2. (I’m a poet!)
This “poem” always works. Try it for the other problems. Often what you want is a 1. For
example, for problem 15, you’d have “You wanted a 1 but you got a ⁄2, so tweak your guess by
1
1
1 over ⁄2.” That’s 2, of course. It works!
3
Back to problem 16. Your tweaked guess is ln t + 5 . That’s it.
2
2