Calculus Volume 1

834

Answer Key

⌠ ⌡ 1 4

4 x 2

169 . T 10 =3.058,

dx =3

171 . F ( x ) = x 3 173 . F ( x ) = − t 5 5 3 + 175 . F ( x ) = x 177 . F ( x ) = x 3 100 100 ,

3 x 2

x , F (3)− F (−2) = − 35 6

2 −5

3 3 −36

+ 13 t

t , F (3)− F (2) = 62 15

F (1)− F (0) = 1 100

1 x , F (4)− F ⎛

⎞ ⎠ = 1125 64

⎝ 1 4

3 +

179 . F ( x ) = x , F (4)− F (1) =1 181 . F ( x ) = 4 3 183 . F ( x ) =−cos x , F ⎛ ⎝ π 2 ⎞

t 3/4 , F (16)− F (1) = 28 3

⎠ − F (0) =1

185 . F ( x ) = sec x , F ⎛ ⎝ π 4 ⎞

⎠ − F (0) = 2−1

187 . F ( x ) =−cot( x ), F ⎛ ⎝ π 2 ⎞

⎛ ⎝ π 4

⎞ ⎠ =1

⎠ − F

189 . F ( x ) = − 1 x + 1 2 x 2

, F (−1)− F (−2) = 7 8

191 . F ( x ) = e x − e 193 . F ( x ) =0 195 . ∫ −2 −1 ⎛

3 ⎛

4 ⎛ ⎝ t 2 −2 t −3 ⎞

⎝ t 2 −2 t −3 ⎞

⎝ t 2 −2 t −3 ⎞

⎠ dt − ∫

⎠ dt + ∫

⎠ dt = 46 3

−1

3

⌡ − π /2 0

197 . − ⌠

π /2

sin tdt + ∫

sin tdt =2

0

199 . a. The average is 11.21×10 9 since cos ⎛ ⎝ πt 6 ⎞

⎠ has period 12 and integral 0 over any period. Consumption is equal to

the average when cos ⎛ ⎝ πt 6 ⎞

⎠ =0, when t =3, and when t =9. b. Total consumption is the average rate times duration: 11.21 × 12 × 10 9 =1.35×10 11 c. 10 9 ⎛ ⎝ ⎠ =11.84 x 10 9 201 . If f is not constant, then its average is strictly smaller than the maximum and larger than the minimum, which are attained over ⎡ ⎣ a , b ⎤ ⎦ by the extreme value theorem. 203 . a. d 2 θ = ( a cos θ + c ) 2 + b 2 sin 2 θ = a 2 + c 2 cos 2 θ +2 ac cos θ = ( a + c cos θ ) 2 ; b. ⎜ 11.21− 1 6 ⌠ ⌡ 3 9 cos ⎛ ⎝ πt 6 ⎞ ⎠ dt ⎞ ⎠ ⎟ =10 9 ⎛ ⎝ 11.21+ 2 π ⎞

2 π

d –

= 1 2 π ∫

( a +2 c cos θ ) dθ = a

0

⌠ ⌡ ⎮⎮ 0 2 π

205 . Mean gravitational force = GmM 2

1

dθ .

⎛ ⎝ a +2 a 2 − b 2 cos θ ⎞ ⎠

2

⎛ ⎝ x − 1 x

⎞ ⎠ dx = ∫ x 1/2 dx − ∫ x −1/2 dx = 2 3

207 . ∫ 209 . ⌠ ⌡ 211 . ⌠

x 3/2 + C

1/2 + C

x 3/2 −2 x 1/2 + C

1 −2 x

2 = 2 3

dx 2 x = 1 2 ln|

x | + C

⌡ 0 π

π cos xdx =−cos x | 0

sin xdx − ∫

π −(sin x ) |

0 π = ⎛

⎝ −(−1)+1 ⎞

⎠ −(0−0) =2

0

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