5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 12:11

Take a Diagnostic Exam 33

45. Width of a rectangle=

6 − 0

3

=2.

Midpoints arex=1, 3, and 5 and f(1)=2,

f(3)=10, andf(5)=26.

∫ 6

0

f(x)dx≈2(2+ 10 +26)≈2(38)= 76

Chapter 14

46. Average value=

1

ln 2−(−ln 2)

∫ln 2

−ln 2

e−^4 xdx.

Letu=− 4 x;du=− 4 dx,or

−du

4

=dx.

∫

e−^4 xdx=

∫

eu

(

−du

4

)

=

− 1

4

eu+C

=

− 1

4

e−^4 x+C

Average value=

1

2ln 2

[

e−^4 x

− 4

]ln 2

−ln 2

=

1

2ln 2

[(

e−4ln2

− 4

)

−

(

e−4(−ln 2)

− 4

)]

=

1

2ln 2

[

(eln 2)−^4

− 4

+

(eln 2)^4

4

]

=

1

2ln 2

[

2 −^4

− 4

+

24

4

]

=

1

2ln 2

(

1

− 64

+ 4

)

=

1

2ln 2

(

255

64

)

=

255

128 ln 2

47.

dy

dx

=2 sinx⇒dy=2 sinxdx

∫

dy=

∫

2 sinxdx⇒y=−2 cosx+C

Atx=π,y= 2 ⇒ 2 =−2 cosπ+C

⇒ 2 =(−2)(−1)+C

⇒ 2 = 2 +C= 0.

Thus,y=−2 cosx.

48. Amount of water leaked

=

∫ 5

0

10 ln (t+1)dt.

Using your calculator, you obtain

10(6 ln 6−5) which is approximately

57.506 gallons.

49.

dy

dx

=ky⇒y=y 0 ekt

Half-life= 5730 ⇒y=

1

2

y 0

whent= 5730.

Thus,

1

2

y 0 =y 0 ek(5730)⇒

1

2

=e^5730 k.

ln

(

1

2

)

=ln

(

e^5730 k

)

⇒ln

(

1

2

)

= 5730 k

ln 1−ln 2= 5730 k⇒−ln 2= 5730 k

k=

−ln 2

5730

50. See the figure below.

− 1210

x

y

y = x + 2

y = x^2

To find points of intersection, setx^2 =x+ 2

⇒x^2 −x− 2 = 0 ⇒x=2orx=−1.

Area of cross section=((x+2)−x^2 )^2.

Volume of solid,V=

∫ 2

− 1

(

x+ 2 −x^2

) 2

dx.

Using your calculator, you obtain:V=

81

10

.