5 Steps to a 5 AP Calculus BC 2019

More Applications of Definite Integrals 345

s ̄(0)=

(

−sin 0−

1

2

(0)

)

̄ı

+

(√

3

2

(0)+cos 0

)

j ̄+C= 0

0 ̄ı+ 1 j ̄+C= 0

C=−j ̄

s ̄(θ)=

(

−sinθ−

1

2

θ

)

̄ı

+

(√

3

2

θ+cosθ− 1

)

j ̄.

Substituting inθ=π:

s ̄(π)=

(

−sinπ−

1

2

π

)

̄ı

+

(√

3

2

π+cosπ− 1

)

j ̄

s ̄(π)=

(

0 −

1

2

π

)

̄ı+

(√

3

2

π− 1 − 1

)

j ̄

s ̄(π)=−

(

π

2

)

̄ı+

(√

3

2

π− 2

)

j ̄

s ̄(π)=−

π

2

ı ̄+

(√

3 − 4

2

π

)

j ̄.

32. Integrate

∫

x^2 e^5 x−^2 dxby parts. Let

u=x^2 ,du= 2 xdx,dv=e^5 x−^2 dx, and

v=

e^5 x−^2

5

. Then,

∫

x^2 e^5 x−^2 dx=

x^2 e^5 x−^2

5

−

2

5

∫

xe^5 x−^2 dx.

The remaining integral can be evaluated by

parts, usingu=x,du=dx,dv=e^5 x−^2 dx,

andv=

e^5 x−^2

5

·

∫

x^2 e^5 x−^2 dx

=

x^2 e^5 x−^2

5

−

2

5

[

xe^5 x−^2

5

−

1

5

∫

e^5 x−^2 dx

]

=

x^2 e^5 x−^2

5

−

2

5

[

xe^5 x−^2

5

−

1

5

·

e^5 x−^2

5

]

=

x^2 e^5 x−^2

5

−

2 xe^5 x−^2

25

+

2 e^5 x−^2

125

+C.

33. The path is defined byx=t−2,y=sin^2 t.
    Since
    dx
    dt
    =1 and
    dy
    dt
    =2 sintcost,
    dy
    dx

=

2 sintcost

1

. This is the slope of a
tangent line to the curve, and when the
slope is zero, the curve will reach either a
maximum or a minimum. The slope
2 sintcost=0 when sint=0 or when
cost=0. The first equation gives ust= 0
ort=πand the second,t=
π
2
. The
second derivative,
d^2 y
dx^2

=−2 sin^2 t+2 cos^2 t=2 cos 2t.

Evaluating at each of the possible values of

t, we find 2 cos 2t

∣∣

t= 0 =2, 2 cos 2t

∣∣

t=π=2,

and 2 cos 2t|t=π 2 =−2. The maximum value

will occur when the second derivative is

negative, so the maximumy-value is

achieved whent=

π

2

,x=

π

2

− 2 =

π− 4

2

,

andy=sin^2

π

2

=1.