5 Steps to a 5 AP Calculus BC 2019

More Applications of Definite Integrals 343

13.12 Solutions to Cumulative Review Problems

26. 3 ey=x^2 y

3 ey

dy

dx

= 2 xy+

dy

dx

(

x^2

)

3 ey

dy

dx

−

dy

dx

x^2 = 2 xy

dy

dx

(

3 ey−x^2

)

= 2 xy

dy

dx

=

2 xy

3 ey−x^2

27. Letu=x^3 +1;du= 3 x^2 dxor

du

3

=x^2 dx.

∫

x^2

x^3 + 1

dx=

∫

1

u

du

3

=

1

3

ln|u|+C

=

1

3

ln

∣∣

x^3 + 1

∣∣

+C

∫ 3

0

x^2

x^3 + 1

dx=

1

3

ln|x^3 + 1 |

] 3

0

=

1

3

(ln 2−ln 1)=

ln 2

3

28. (a) F(− 2 )=

∫− 2

− 2

f(t)dt= 0

F( 0 )=

∫ 0

− 2

f(t)dt=

1

2

( 4 + 2 ) 2 = 6

(b) F′(x)=f(x);F′(0)=2 andF′(2)= 4.

(c)Since f>0on[−2, 4],Fhas a

maximum value atx=4.

(d) The functionfis increasing on (1, 3),

which implies that f′>0 on (1, 3).

Thus,Fis concave upward on (1, 3).

(Note: f′is equivalent to the 2nd

derivative ofF.)

29. (a)
    dy
    dx

=

y

2 x+ 1

;f(0)= 2

dy

dx

∣∣

∣∣

x= 0

=

2

2 ( 0 )+ 1

= 2 ⇒m=2atx= 0.

y−y 1 =m

(

x−x 1

)

y− 2 = 2 (x− 0 )⇒y= 2 x+ 2

The equation of the tangent tof at

x=0isy= 2 x+2.

(b) f(0.1)=2(0.1)+ 2 = 2. 2

(c) Solve the differential equation:

dy

dx

=

y

2 x+ 1

.

Step 1. Separate variables:

dy

y

=

dx

2 x+ 1

Step 2. Integrate both sides:

∫

dy

y

=

∫

dx

2 x+ 1

ln|y|=

1

2

ln| 2 x+ 1 |+C.

Step 3. Substitute given values (0, 2):

ln 2=

1

2

ln 1+C⇒C=ln 2

ln|y|=

1

2

∣∣

2 x+ 1

∣∣

+ln 2

ln|y|−

1

2

∣∣

2 x+ 1

∣∣

=ln 2

ln

∣

∣∣

∣

y

( 2 x+ 1 )^1 /^2

∣

∣∣

∣=ln 2

eln

∣∣

∣∣( 2 x+y 1 ) 1 / 2

∣∣

∣∣

=eln 2

y

( 2 x+ 1 )^1 /^2

= 2

y= 2 ( 2 x+ 1 )^1 /^2.