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 ̄.
- 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.
- 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