Math 1210
Quiz 8
March 20th, 2014
This is a take-home quiz. It contains 2 ordinary questions (35 points) and
1 extra credit question (15 points). The value of every question is
indicated at the beginning of it. Please print this sheet and write your
answers in the space provided.
Name:
UID:
1. (15 points) Compute the following indefinite integrals.
R 2 2
dx
(i) (5 points) (x √+1)
x
Hint: Expand the numerator first.
Solution:
Z
Z 4
Z
(x2 + 1)2
x + 1 + 2x2
√
√
dx =
= x7/2 + x−1/2 + 2x3/2 dx
x
x
√
4
2 9/2
=
x + 2 x + x5/2 + C
9
5
(ii) (5 points)
R
Solution:
Z
√ 3y
2y 2 +5
3y
dy
3
p
dx =
2
4
2y + 5
Z
3
4y(2y 2 + 5)−1/2 dx = (2y 2 + 5)1/2 + C
2
R
(iii) (5 points) sin x(1 + cos x)4 dx
Hint: Use the generalized power rule.
Solution:
Z
Z
1
4
sin x(1 + cos x) dx = − − sin x(1 + cos x)4 dx = − (1 + cos x)5 + C
5
2. (20 points) Find the area under the curve y = x3 + x between x = 0 and x = 1 by
following the steps below:
Hint: The answer is 43 .
(i) (2 points) Subdivide the interval [0, 1] into n equal subintervals [x0 , x1 ], [x1 , x2 ], . . . , [xn−1 , xn ].
What is the length ∆x of every subinterval? For every k, write an expression for
xk .
Solution: ∆x = n1 and xk = nk .
(ii) (5 points) Sketch the rectangles that you will be using in the graph below and write
down an expression for the area of the rectangle over [xk , xk+1 ].
Figure 1: Question 2
Solution: The area of the rectangle over the interval [xk , xk+1 ] is given by
!
3
k
k
1
k3
k
f (xk )∆x = (x3k + xk )∆x =
+
= 4+ 2
n
n n
n
n
(iii) (5 points) Find the sum A(Rn ) of the areas of the n rectangles.
2
P
.
Hint: Remember that nk=1 k 3 = n(n+1)
2
Solution: The sum of the areas of the n rectangles is
A(Rn ) = f (x0 )∆x + · · · f (xn−1 )∆x =
n−1
X
f (xk )∆x =
k=0
=
1
n4
4
=
n−1
X
k3 +
k=0
3
1
n2
n−1
X
k=0
2 2
k=
k=0
2
k
+
n4 n2
1 (n − 1) n
1 (n − 1)n
+
n4
4
n2
2
n − 2n + n
n2 − n
+
4n4
2n2
Page 2
n−1 3
X
k
(iv) (4 points) The area under the curve will be the limit A = limn→∞ A(Rn ). Find this
limit.
Solution:
2
1
1
1
n4 − 2n3 + n2 n2 − n
1
1− + 2 +
1−
lim A(Rn ) = lim
+
= lim
n→∞
n→∞
n→∞ 4
4n4
2n2
n n
2
n
1 1
3
=
+ =
4 2
4
R
(v) (4 points) Compute the indefinite integral F (x) = x3 + x dx (so that F 0 (x) =
x3 + x) and find F (1) − F (0). You should be obtaining the same result as in (iv).
You will understand why in section 4.4.
Solution: The indefinite integral is
Z
1
1
F (x) = x3 + x dx = x4 + x2
4
2
and
1 1
3
+ =
4 2
4
We see that computations yield the same result: the reason for this resides in the
Second Fundamental Theorem of Calculus. We will study this in Section 4.4.
F (1) − F (0) =
Page 3
3. EXTRA CREDIT (15 points)
(i) (10 points) Consider the differential equation
dy
= −y 2 x(x2 + 2)4
dx
Use separation of variables in order to find the general solution of this equation.
Solution: Separating variables we obtain
1
dy = −x(x2 + 2)4 dx
2
y
Integrating both sides we obtain
Z
Z
1
dy = −x(x2 + 2)4 dx
y2
Computing both integrals we obtain
Z
Z
1
11
1
dy = − + C1 ,
−x(x2 + 2)4 dx = − (x2 + 2)5 + C2
2
y
y
25
and therefore
C
z }| {
1
1
1
− = − (x2 + 2)5 + (C2 − C1 ) = − (x2 + 2)5 + C
y
10
10
so solving for y we conclude
1
y= 1 2
(x + 2)5 − C
10
(ii) (5 points) Consider the differential equation
x + 3x2
dy
=
dx
y2
Show that y(x) =
q
3
3 2
x
2
+ 3x3 + 216 is a solution of this equation.
Solution: We just need to verify that the given function y(x) is a solution of the
equation
dy
x + 3x2
=
dx
y2
But note that
r
dy
d 3 3 2
1
1
2
=
x + 3x3 + 216 =
(3x + 9x )
dx
dx 2
3 3 x2 + 3x3 + 216 2/3
2
x + 3x
=
r
3
|
2
!2 =
3 2
x + 3x3 + 216
2
{z
}
y2
Page 4
x + 3x2
y2