Exam 4 - Math 151
R. Ketchersid1)[18pts, 2pts each] Evaluate the following indefinite integrals. a)
Z
t·sin(t2)dt
u=t2;du/2 =t dt Z
t·sin(t2)dt= 1 2
Z
sin(u)du= −cos(u) 2 +C=
−cos(t2)
2 +C
b)
Z y dy
4 + 9y2
u= 4−9y2;du/18 =y dy Z y dy
4 + 9y2 =
1 18
Z du
u =
ln|u|
18 +C=
ln(4 + 9y2)
18 +C
c)
Z dy
4 + 9y2
Z dy
4 + 9y2 =
1 4
Z dy
1 + 9y2/4 =
1 4
Z dy
1 + (3/2y)2
u= 3/2y; 2/3du=dy
1 4
Z dy
1 + (3/2y)2 =
1 6
Z du
1 +u2 =
arctan(u) 6 +C=
arctan(3/2y) 6 +C
d)
Z (u−2)2
u du
Z (u2−4u+ 4)
u du=
Z
(u−4 + 4/u)du= u
2
2 −4u+ 4 ln|u|+C
e)
Z 1 +x √
1−xdx
u= 1−x; −du=dx;x= 1−u
Z 1 +x √
1−xdx=−
Z 2−u √
u du=
Z
2
-f)
Z π/4
0
tan(x) sec2(x)dx
u= tan(x);du= sec2(x)dx
Z π/4
0
tan(x) sec2(x)dx=
Z 1
0
u du= u
2 2 1 0 = 1 2 g) Z 1 0
x+x3
√
1−x4dx
Correct: The function x+x
3
√
1−x4 has a vertical asymptote atx= 1so the correct answer isDNE.
However, if you went through the formalism, I also accepted the following:
Z 1
0
x+x3
√
1−x4dx= Z 1
0
x dx
√
1−x4 + Z 1
0
x3dx
√
1−x4
For the first integral,u=x2anddu/2 =x dx
Z 1
0
x dx
√
1−x4 =
1 2 Z 1 0 du √
1−u2 =
1
2arcsin(u)
1 0 =π 4
For the second integral,v= 1−x4 and−du/4 =x3dx Z 1
0
x3dx
√
1−x4 =−
1 4 Z 0 1 dv √ v = 1 4 Z 1 0
v−1/2dv= 1 42v
1/2 1 0= 1 2
So the answer is
1
2arcsin(u)
1 0 +1 42v
1/2 1 0 = π 4 + 1 2 h)1
Z −π/6
−π/3
cos(x) sin(x)dx
u= sin(x);du= cos(x)dx
Z −π/6
−π/3
cos(x) sin(x)dx=
Z −1/2
−√32
du
u = ln|u|
−1/2
−√3/2
= ln(1/2)−ln(√3/2) = −ln(3)
2 i) Z 2 1 d dx p
4−x2dx
Z 2
1
d dx
p
4−x2dx= p4−x2 2 1 =
√
3
1A number of students madeprecisely the mistakes I warned you about. For example in (h) many wrote:
Z −π/6
−π/3
cos(x) sin(x)dx=
Z −π/6
−π/3 du
u = ln|u|
−π/6
−π/3 and many wrote
Z −π/6
−π/3
cos(x) sin(x)dx=
Z −1/2
−√32 du
u = ln|u|
−1/2
−√3/2
= ln|sin(x)|
−1/2
−√3/2 .
3) [9pts] (a) [3pts] A particle moves along a straight path. The velocity is recorded 6 times in the interval fromt= 1tot= 6att= 1,2,3,4,5,6the data is given as follows (data is in meters/seconds and time is measured in seconds):
t 1 2 3 4 5 6 v(t) 8.4 9 1.4 -7.5 -9.5 -2.8
Write down the approximate net displacement of this particle over the interval beginning att= 1
and ending att = 6, that is s(6)−s(1) where s(t) is the displacement function, using the right hand endpoints of the intervals for test points. In other words write downR5using the given data.
s
(6)
−
s
(1)
≈
R
5= (1)(9 + 1
.
4 +
−
7
.
5 +
−
9
.
5 +
−
2
.
8) =
−
9
.
4
(b) [3pts] The data is entered into a computer and it is found thatv(t) = 10·sin(t)models the data well. Write down the approximate net displacement of the particle on the interval[1,6]using 100 subintervals and using the right endpoint of each subinterval as the test point, that is, write down R100.
s
(6)
−
s
(1)
≈
R
100=
100X
i=1
10 sin
1 +
5
i
100
!
5
100
=
−
4
.
478026668
(c) [3pts] Compute the net displacement of the particle on the interval starting att= 1and lasting tot= 6using the modelv(t) = 10·sin(t)for velocity.
s
(6)
−
s
(1) =
Z 6
1
10 sin(
t
)
dt
=
−
10 cos(
t
)
6
4
-4) [5pts] For whatxin[0, π/2]doesF(x) =R
√
2x 0 cos(t
2)dtobtain its maximum value? (You must
explain how you know that you have found the maximum! You may use the sign diagram forF0(x)
on[0, π/2], this is probably the simplest thing to do.) F0(x) = cos √2x2(1/2)(2x)−1/2(2) = cos(2x)√
2x
The only c.p.’s occuring will be whereF0(x) = 0and that happens when cos(2x) = 0so 2x=π/2
and hencex = π/4. Since F0(x) >0 on (0, π/4) and F0(x) <0 on (π/4, π/2) we know F(x) is increasing on(0, π/4)and then decreasing on(π/4, π/2)so the maximum ofF occurs atx=π/4. 5) [5pts] Compute the volume of the solid whose base is the triangle determined by (0,0), (1,2), and (3,−1) and whose cross sections parallel to the y-axis are semicircular. (This is not a solid formed by rotation.)
The cross sec-tional area A(x) depends on whether 0 ≤ x ≤ 1 or 1 ≤ x ≤ 3. For 0 ≤ x ≤ 1, the diam-eter of the semicircle is d(x) = 2x+ 1/3x = 7/3x and hence r(x) = 7/6x. For 1 ≤ x ≤ 3, d(x) =−3/2(x−1) + 2 + 1/3x= 7/2−7/6xand thusr(x) = (7/4)(1−x/3). SinceA(x) =πr2(x)/2
we have
A(x) =
(
(1/2)π(7/6)2x2 for0≤x≤1 (1/2)π(7/4)2(1−x/3)2 for1< x≤3 =
(
49/72πx2 for0≤x≤1 49/32π(1−2x/3 +x2/9) for1< x≤3
So
V =
Z
dV = 49π 72
Z 1
0
x2dx+49π 32
Z 3
1
1−2x
3 +
x2
9
dx
= 49π 8 " 1 9 x3 3 1 0 +1 4
x−x 2 3 + x3 27 3 1 #
= 49π 8 1 27+ 1 4
(3−3 + 1)−
1−1
3+ 1 27
= 49π 8 1 27+ 1 4 1 3 − 1 27
= 49π 8 1 27+ 1 4 8 27
= 49π 8
3 27
1) Write equation of tangent line to the curvesin(xy) = cos(xy)at (√π/2,√π/2).
cos(xy)(y+xy0) =−sin(xy)(y+xy0)
cos(π/4)(√π/2 +√π/2y0) =−sin(π/4)(√π/2 +√π/2y0) (√2/2)(√π/2)(1 +y0) =−(√2/2)(√π/2)(1 +y0)
1 +y0=−(1 +y0) 2y0=−2
y0=−1
So the equation of the tangent line is y−√π/2 =−(x−√π/2) . 2) Differentiatef(x) = (1 + 1/x)x. (You need not simplify.)
ln(y) =xln(1 + 1/x)
y0/y= ln(1 + 1/x) +x
−1/x2
1 + 1/x
y0/y= ln(1 + 1/x)− 1
x+ 1
y0= (1 + 1/x)x
ln(1 + 1/x)− 1
1 +x
3) Ifg0(x) = 1 1−x2 find
d
dxg(sin(x)).
d
dxg(sin(x)) =
1
1−[sin(x)]2
d dxsin(x)
= cos(x) 1−sin2(x) =
cos(x) cos2(x) =
1
cos(x) = sec(x)
4) Differentiate arctan(3√ x)
1 +x2 . (You need not simplify.)
√ 1+x2 1+(3x)2
−arctan(3x)(1/2)(1 +x2)−1/2(2x)
1 +x2 =
1+x2 1+9x2
Extra Credit for Exam 3
1) (a) Forf(x) =x(x+ 1)2/3 findf0(x)and simplify.
(x+ 1)2/3+x(2/3)(x+ 1)−1/3= 3(x+ 1) + 2x 3(x+ 1)1/3 =
5x+ 3 3(x+ 1)1/3
(b) What are the critical points off(x)?
The critical points are x=−3/5 (since f0(x) = 0) and x=−1 (since f0(x) is undefined and f(x)is defined).
(c) On what intervals isf(x)increasing/decreasing?
f0(x)<0on(−1,−3/5)and f0(x)>0on(−∞,−1)∪(−3/5,∞)so:
• f(x)is increasing on(−∞,−1]and[−3/5,∞)
• f(x)is decreasing on[−1,−3/5].
(d) Findf00(x)and simplify.
5(3)(x+ 1)1/3−(5x+ 3)(x+ 1)−2/3
9(x+ 1)2/3 =
15(x+ 1)−(5x+ 3) 9(x+ 1)4/3 =
10x+ 12 9(x+ 1)4/3
(e) On what intervals isf(x)concave up/down?
f00(x)<0 on(−∞,−12/10)andf00(x)>0 on(−12/10,−1)∪(−1,∞)hence:
• f(x)is concave down on (−∞,−12/10)and
2) Findlimx→∞(1 + 2/x)x. This has indeterminate form1∞.
lim
x→∞(1 + 2/x)
x= lim u→Le
u
whereL= limx→∞xln(1 + 2/x).
lim
x→∞xln(1 + 2/x) = limx→∞
ln(1 + 2/x)
1/x (form 0/0)
=L’H lim
x→∞
−2/x2 1+2/x
−1/x2
= lim
x→∞
2
1 + 2/x = 2 So
lim
x→∞(1 + 2/x)
x= lim u→2e
u=e2
3) An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 6 km east of the refinery. The cost of laying pipe is $400,000/km over land to a pointP on the north bank and $800,000/km under the river to the tanks. To minimize the cost of the pipeline, where shouldP be located?
Letxrepresent how far to the east of the refineryP is. Then the cost of pipe from the refinery to P is8(105)√x2+ 22 and the cost for the pipe fromP to the storage tanks is 4(105)(6−x)where
0≤x≤6. So we must minimize
C(x) = 8(105)px2+ 22+ 4(105)(6−x)on0≤x≤6
We have
C0(x) = 8(10
5)x √
x2+ 4−4(10 5)
The only critical point will occur whereC0(x) = 0this is when
8(105)x= 4(105)px2+ 4
4x2=x2+ 4 3x2= 4
So the critical point is atx= 2/√3. Now you can just computeC(x)atx= 0,2, and6. x C(x)
0 8√2(105) + 24(105) = (8√2 + 24)(105)
2/√3 4/√3(8)(105) + (6−4/√3)(4)(105)(24 + 16/√3)(105)
6 2√10(8)(105) = 16√10(105)
