MA 3972-MA-Book April 11, 2018 16:5
More Applications of Definite Integrals 341
eln
∣∣y 1 / 2
x+ 1
∣∣
=eln 2
y^1 /^2
x+ 1
= 2
y^1 /^2 = 2 (x+ 1 )
y=( 2 )^2 (x+ 1 )^2
y= 4 (x+ 1 )^2.
Step 4: Verify result by differentiating:
dy
dx
= 4 ( 2 )(x+ 1 )=8(x+1).
Compare with
dy
dx
=
2 y
x+ 1
=
2(4(x+1)^2 )
(x+1)
=8(x+1).
- y(t)=y 0 ekt
y 0 =750, 000
y( 22 )=(750, 000)e(^0.^03 )(^22 )
≈
{
1. 45109 E 6 ≈1, 451, 090 using a TI-89,
1, 451, 094 using a TI-85.
- Step 1: Separate the variables:
4 ey=
dy
dx
− 3 xey
4 ey+ 3 xey=
dy
dx
ey(4+ 3 x)=
dy
dx
(4+ 3 x)dx=
dy
ey
=e−ydy.
Step 2: Integrate both sides:
∫
( 4 + 3 x)dx=
∫
e−ydy
4 x+
3 x^2
2
=−e−y+C
Switch sides:e−y=−
3 x^2
2
− 4 x+C.
Step 3: Substitute given value:y( 0 )= 0
⇒e^0 = 0 − 0 +c⇒c=1.
Step 4: Take ln of both sides:
e−y=−
3 x^2
2
− 4 x+ 1
ln(e−y)=ln
(
−
3 x^2
2
− 4 x+ 1
)
y=−ln
(
1 − 4 x−
3 x^2
2
)
.
Step 5: Verify result by differentiating:
Enterd(−ln(1− 4 x− 3
(x−∧^2 )/2),x) and obtain
− 2 ( 3 x+ 4 )
3 x^2 + 8 x− 2
, which is equivalent
toey(4+ 3 x).
- y(t)=y 0 ekt
k= 0 .0625,y( 10 )=50, 000
50, 000=y 0 e^10 (^0.^0625 )
y 0 =
50, 000
e^0.^625
⎧
⎪⎨
⎪⎩
$26763.1 using a TI-89,
$26763.071426≈$26763. 07
using a TI-85.
- Setv(t)= 2 − 6 e−t=0. Using the [Zero]
function on your calculator, compute
t= 1 .09861.
Distance traveled=
∫ 10
0
|v(t)|dt
∣∣
2 − 6 e−t
∣∣
=
{
−(2− 6 e−t)if0≤t< 1. 09861
2 − 6 e−t if t≥ 1. 09861
∫ 10
0
| 2 − 6 e−t|dt=
∫ 1. 09861
0
−(2− 6 e−t)dt
+
∫ 10
1. 09861
( 2 − 6 e−t)dt
= 1. 80278 + 15. 803 = 17. 606.
Alternatively, use the [nInt] function on
the calculator.
Enter nInt(abs(2− 6 e∧(−x)), x, 0, 10) and
obtain the same result.
