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

MA 3972-MA-Book April 11, 2018 16:5

More Applications of Definite Integrals 339

10. y(t)=y 0 ekt

Half-life=4500 years⇒

1

2

=e^4500 k.

Take ln of both sides:

ln

(

1

2

)

=lne^4500 k

⇒−ln 2= 4500 k

ork=

−ln 2

4500

.

y(t)= 100 e

(−ln 2

4500

)

(5000)=25(2 2 / (^9) )
≈ 46. 293.
There are approximately 46.29 grams left.

11. Step 1: Separate the variables:
    dy=xcos(x^2 )dx.
    Step 2: Integrate both sides:
    ∫
    dy=

∫

xcos(x^2 )dx

∫

dy=y

∫

xcos (x^2 )dx: Letu=x^2 ;

du= 2 xdx,

du

2

=xdx

∫

xcos(x^2 )dx=

∫

cosu

du

2

=

sinu

2

+c=

sin (x^2 )

2

+C.

Thus,y=

sin (x^2 )

2

+C.

Step 3: Substitute given values.

y(0)=

sin (0)

2

+C=π⇒C=π

y=

sin (x^2 )

2

+π

Step 4: Verify result by differentiating:

dy

dx

=

cos (x^2 )(2x)

2

=xcos(x^2 ).

12. Step 1: Rewrite
    d^2 y
    dx^2
    as
    dy′
    dx
    dy′
    dx
    =x− 5.

Step 2: Separate variables:

dy′=(x−5)dx.

Step 3: Integrate both sides:

∫

dy′=

∫

(x−5)dx

y′=

x^2

2

− 5 x+C 1.

Step 4: Substitute given values:

Atx=0,y′=

0

2

−5(0)+C 1

=− 2 ⇒C 1 =− 2

y′=

x^2

2

− 5 x− 2.

Step 5: Rewrite:

y′=

dy

dx

;

dy

dx

=

x^2

2

− 5 x− 2.

Step 6: Separate the variables:

dy=

(

x^2

2

− 5 x− 2

)

dx.

Step 7: Integrate both sides:

∫

dy=

∫ (

x^2

2

− 5 x− 2

)

dx.

y=

x^3

6

−

5 x^2

2

− 2 x+C 2

Step 8: Substitute given values:

Atx=0,y= 0 − 0 − 0 +C 2

= 1 ⇒C 2 = 1

y=

x^3

6

−

5 x^2

2

− 2 x+ 1.