More Applications of Definite Integrals 339
Step 9. Verify result by differentiating:
dy
dx
=
x^2
2
− 5 x− 2
d^2 y
dx^2
=x− 5.
Part B Calculators are allowed.
- Average Value=
1
π/ 3 −π/ 4
∫π/ 3
π/ 4
tanxdx.
Enter=(1/(π/ 3 −π/4))
∫
(tanx,x,π/4,π/3)
and obtain
6 ln(2)
π
= 1. 32381.
- v(t)=
∫
a(t)dt
=
∫
3 e^2 t=
3
2
e^2 t+C
v(0)=
3
2
e^0 +C=
1
2
⇒
3
2
+C=
1
2
orC=− 1
v(t)=
3
2
e^2 t− 1
Displacement=
∫ 3
0
(
3
2
e^2 t− 1
)
dt.
Enter
∫
( 3 / 2 ∗e∧(2x)−1,x,0,3)
and obtain 298. 822.
Distance Traveled=
∫ 3
0
∣∣
v(t)
∣∣
dt.
Since
3
2
e^2 t− 1 >0 fort≥0,
∫ 3
0
∣
∣v(t)
∣
∣dt=
∫ 3
0
(
3
2
e^2 t− 1
)
dt= 298. 822.
- Step 1. y(t)=y 0 ekt
y(1)= 5000 ⇒ 5000 =y 0 ek⇒y 0
= 5000 e−k
y(3)= 4000 ⇒ 4000 =y 0 e^3 k
Substituting:
y(0)= 5000 e−k, 4000=(5000e−k)e^3 k
4000 = 5000 e^2 k
4
5
=e^2 k
ln
(
4
5
)
=ln
(
e^2 k
)
= 2 k
k=
1
2
ln
(
4
5
)
≈− 0. 111572.
Step 2. 5000 =y 0 e−^0.^111572
y(0)=(5000)/e−^0.^111572 ≈ 5590. 17
y(t)=( 5590. 17 )e−^0.^111572
Step 3. y(7)=(5590.17)e−^0 .111572(7)
≈ 2560
Thus, sales for the 7th month are
approximately 2560 units.
- Step 1. Separate variables:
dy
dx
=
2 y
x+ 1
dy
2 y
=
dy
x+ 1
.
Step 2. Integrate both sides:
∫
dy
2 y
=
∫
dx
x+ 1
1
2
ln|y|=ln
∣∣
x+ 1
∣∣
+C.
Step 3. Substitute given value (0, 4):
1
2
ln( 4 )=ln( 1 )+C
ln 2=C
1
2
ln|y|−ln|x+ 1 |=ln 2
ln
∣∣
∣∣y
1 / 2
x+ 1
∣∣
∣∣=ln 2
