More Applications of Definite Integrals 343
13.12 Solutions to Cumulative Review Problems
- 3 ey=x^2 y
3 ey
dy
dx
= 2 xy+
dy
dx
(
x^2
)
3 ey
dy
dx
−
dy
dx
x^2 = 2 xy
dy
dx
(
3 ey−x^2
)
= 2 xy
dy
dx
=
2 xy
3 ey−x^2
- Letu=x^3 +1;du= 3 x^2 dxor
du
3
=x^2 dx.
∫
x^2
x^3 + 1
dx=
∫
1
u
du
3
=
1
3
ln|u|+C
=
1
3
ln
∣∣
x^3 + 1
∣∣
+C
∫ 3
0
x^2
x^3 + 1
dx=
1
3
ln|x^3 + 1 |
] 3
0
=
1
3
(ln 2−ln 1)=
ln 2
3
- (a) F(− 2 )=
∫− 2
− 2
f(t)dt= 0
F( 0 )=
∫ 0
− 2
f(t)dt=
1
2
( 4 + 2 ) 2 = 6
(b) F′(x)=f(x);F′(0)=2 andF′(2)= 4.
(c)Since f>0on[−2, 4],Fhas a
maximum value atx=4.
(d) The functionfis increasing on (1, 3),
which implies that f′>0 on (1, 3).
Thus,Fis concave upward on (1, 3).
(Note: f′is equivalent to the 2nd
derivative ofF.)
- (a)
dy
dx
=
y
2 x+ 1
;f(0)= 2
dy
dx
∣∣
∣∣
x= 0
=
2
2 ( 0 )+ 1
= 2 ⇒m=2atx= 0.
y−y 1 =m
(
x−x 1
)
y− 2 = 2 (x− 0 )⇒y= 2 x+ 2
The equation of the tangent tof at
x=0isy= 2 x+2.
(b) f(0.1)=2(0.1)+ 2 = 2. 2
(c) Solve the differential equation:
dy
dx
=
y
2 x+ 1
.
Step 1. Separate variables:
dy
y
=
dx
2 x+ 1
Step 2. Integrate both sides:
∫
dy
y
=
∫
dx
2 x+ 1
ln|y|=
1
2
ln| 2 x+ 1 |+C.
Step 3. Substitute given values (0, 2):
ln 2=
1
2
ln 1+C⇒C=ln 2
ln|y|=
1
2
∣∣
2 x+ 1
∣∣
+ln 2
ln|y|−
1
2
∣∣
2 x+ 1
∣∣
=ln 2
ln
∣
∣∣
∣
y
( 2 x+ 1 )^1 /^2
∣
∣∣
∣=ln 2
eln
∣∣
∣∣( 2 x+y 1 ) 1 / 2
∣∣
∣∣
=eln 2
y
( 2 x+ 1 )^1 /^2
= 2
y= 2 ( 2 x+ 1 )^1 /^2.