Math 1210
Midterm 3
April 4th, 2014
This exam consists of 2 sections, A and B. Section A is conceptual, whereas
section B is more computational. The value of every question is indicated at
the beginning of it. You may only use scratch paper and a small note card.
No cell phones, calculators, notes, books or music players are allowed during
the test.
Name:
UID:
Section A: Conceptual questions.
1. (5 points) If f is a function which is differentiable over the open interval (a, b) and
continuous over the closed interval [a, b], the mean value theorem for derivatives states
(a)
that there exists some c inside of [a, b] such that f 0 (c) = f (b)−f
.
b−a
Consider the function f (x) = x2/3 over the interval [−8, 27]. Show that the mean value
theorem for derivatives fails and explain why.
Solution: One the one hand f 0 (x) =
2 1
,
3 x1/3
and on the other hand
f (27) − f (−8)
9−4
1
=
=
27 − (−8)
35
7
3
1
> 43 = 64 so clearly c doesn’t lie
Imposing f 0 (c) = 23 x1/3
= 17 we find that c = 14
3
inside the interval [−8, 27] and the mean value theorem fails. The reason is that the
function f is not differentiable at x = 0.
2. (5 points) Recall that a function f is odd if f (−x) = −f (x) for all x. Show that if f is
an odd function, then for any real number a we have
Z a
f (x) dx = 0
−a
Hint: use the change of variables u = −x.
Solution: Write
Z
a
Z
0
f (x) dx =
−a
Z
f (x) dx +
−a
a
f (x) dx
0
We apply the suggested change of variable u = −x to the first one and we get
Z 0
Z 0
Z 0
Z a
Z a
[1]
[2]
[3]
[4]
f (x) dx =
f (−u)(− du) = −
f (−u) du =
f (−u) du = −
f (u) du
−a
a
a
0
0
Z a
[5]
= −
f (x) dx
0
where in [1] we applied the change of variable u = −x (note the change in the limits of
integration), in [2] we took the minus sign out if the integral, in [3] we reversed the limits
of integration by changing the sign, in [4] we used the fact the f is an odd function, so
f (−u) = −f (u), and in [4] we wrote x instead of u (these are just dummy variables).
Overall, going back to our original integral we get
Z a
Z 0
Z a
Z a
Z a
f (x) dx =
f (x) dx +
f (x) dx = −
f (x) dx +
f (x) dx = 0
−a
−a
0
0
0
as claimed.
Section B: Practical questions.
3. (15 points) Compute the following indefinite integrals:
R 2 2
(i) (5 points) (x √+1)
dx Solution: We start by expanding the numerator:
x
Z
(x2 + 1)2
√
dx =
x
(ii) (5 points)
R
Z
√ 3x
2x2 +5
Z
2
4
x7/2 + x−1/2 + 2x3/2 dx = x9/2 + 2x1/2 + x5/2 + C
9
5
dx Solution:
3x
3
√
dx =
4
2x2 + 5
Z
4x(2x2 + 5)−1/2 dx =
3√ 2
2x + 5 + C
2
R
(iii) (5 points) x2 (x3 + 5)8 cos [(x3 + 5)9 ] dx
Hint: Use the change of variables u = (x3 + 5)9 .
Solution:
Z
Z
3
du
2 3
8
9
2 3
8
x (x + 5) cos (x + 5) dx =
x (x + 5) cos(u)
27x2 (x3 + 5)8
Z
1
1
=
cos(u) du =
sin(u) + C
27
27
1
sin (x3 + 5)9 + C
=
27
Page 2
4. (15 points) Find the area under the curve y = 2x + 2 over the interval [−1, 1] as follows.
(i) (2 points) Subdivide the interval [−1, 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 =
2
n
xk = −1 +
2k
n
(ii) (4 points) Write down an expression for the area of the rectangle over [xk , xk+1 ]
which depends ONLY on k and n. Solution: The area of the rectangle over
[xk , xk+1 ] is given by
2
8k
2k
= 2
+2
f (xk )∆x = (2xk + 2)∆x = 2 −1 +
n
n
n
(iii) (7 points) Find the sum A(Rn ) of the areas of the n rectangles.
P
Hint: Remember that nk=1 k = n(n+1)
.
2
Solution: The sum of the areas of the n rectangles is given by
A(Rn ) =
n−1
8 (n − 1)n
8 X
1
k= 2
= 2
=4 1−
n2
n k=0
n
2
n
n−1
X
8k
k=0
(iv) (2 points) Find the limit A = limn→∞ A(Rn ). Solution: From (iii) we clearly have
1
lim A(Rn ) = lim 4 1 −
=4
n→∞
n→∞
n
Page 3
5. (15 points) Consider the function F (x) =
(i) (2 points) Compute F π4 and F (0).
R cos x
sin x
t5 dt
Solution:
f
π 4
Z
cos
π
4
=
sin
Z
t3 dt =
√
2
2
π
4
cos 0
Z
3
f (0) =
√
2
2
Z
t dt =
sin 0
0
1
t3 dt = 0
1
1
1 4 t dt = t =
4 0 4
3
(ii) (10 points) Compute F 0 (x)
Ra
R cos x
Hint: Write F (x) = sin x t3 dt + a t3 dt.
Solution:
Z a
Z cos x
Z sin x
Z cos x
d
d
d
d
3
3
3
f (x) =
t dt +
t dt = −
t dt +
t3 dt
dx sin x
dx a
dx a
dx a
= − sin3 x cos x − cos3 x(− sin x) = − sin3 x cos x − cos3 x sin x
0
6. (10 points) Consider the function f (x) =
Rx
0
√ u
1+u2
du
(i) (5 points) Find the intervals where f is increasing or decreasing.
Solution: This boils down to determining over which intervals is f 0 (x) positive or
negative. By the first fundamental theorem of calculus,
Z x
u
x
d
0
√
du = √
f (x) =
2
dx 0
1+u
1 + x2
so f is decreasing over (−∞, 0) and increasing over (0, ∞).
(ii) (5 points) Find the intervals where f is concave up or down.
This boils down to studying the sign of f 00 (x). Note that
2
0 √
2 − √x
1
+
x
1
x
2
1+x
=
=
f 00 (x) = √
2
2
1+x
(1 + x2 )3/2
1+x
which is always positive, so f is concave-up over the whole reals.
Page 4
7. (10 points) Compute the definite integral
Z
π/2
sin x sin(cos x) dx
0
Solution: We use the change of variable u = cos x, so that du = − sin xdx and hence
π/2
Z
0
u(π/2)
−du
sin x sin(cos x) dx =
sin x sin(u)
=−
sin x
u(0)
= 1 − cos 1
Z
Z
0
sin u du = cos u|01
1
8. (10 points) Use symmetry to compute the following integral
Z
π/2
−π/2
sin x
+ |x| sin5 x + x2 dx
1 + cos x
Be explicit in your justification.
Solution: We split the integral into 3
Z
π/2
sin x
+ |x| sin5 x + x2 dx =
1 + cos x
−π/2
Note that the functions f (x) =
Z
sin x
1+cos x
π/2
−π/2
sin x
dx +
1 + cos x
Z
π/2
5
Z
π/2
|x| sin x dx +
x2 dx
−π/2
−π/2
and g(x) = |x| sin5 x are odd. Indeed:
sin(−x)
− sin x
=
= −f (x)
1 + cos(−x)
1 + cos x
g(−x) = | − x| (sin(−x))5 = |x| (− sin x)5 = −|x| sin5 x = −g(x)
f (−x) =
The integrals of f and g over symmetric intervals are hence 0 and therefore
Z
π/2
−π/2
sin x
+ |x| sin5 x + x2 dx =
1 + cos x
=
Z
π/2
Z
2
x dx = 2
−π/2
2 π 3
3
Page 5
2
0
3
=
π
12
π/2
π/2
1 3 x dx = 2 x 3 0
2
9. (15 points) YOU DON’T HAVE TO COMPUTE ANY INTEGRAL IN THIS EXERCISE.
√
(i) (5 points) Let R be the region bounded by the graphs of y = x, y = −x + 6 and
y = 0. Write down the integral that would compute the area of R using vertical
slices.
Solution: We first need to determine the points of intersection of the 2 curves,
namely the points satisfying both equations. The x-value of such a point must
satisfy
√
x = −x + 6
Taking squares yields
x = x2 + 36 − 12x =⇒ x2 − 13x + 36 = 0
and this equations has the following solutions
√
13 ± 169 − 144
13 ± 5
9 =⇒ out of bound
x=
=
=
4 =⇒ y = 2
2
2
so the curves intersect at the point (4, 2).
Using vertical slices, the area will be given by
Z 4
Z 6
√
A=
x dx +
−x + 6 dx
0
4
(ii) (5 points) Write down the integral that would compute the area of R using horizontal slices.
Using horizontal slices, we need to integrate from y = 0 to y = 2 in order to cover
the region. The height of a thin rectangular slice will be dy and the base will be
given by 6 − y − y 2 .
You need to invert the equations defining the curves to do this: given √
any x-value
x = a,
√ the corresponding y-value of a point lying on the graph of y = x must be
y = a. Conversely, given√ a y-value y = b, the corresponding x-value of a point
lying on the graph of y = x must be x = b2 .
The area will then be given by
Z
A=
2
6 − y − y 2 dy
0
√
(iii) (5 points) Now consider the region bounded by the graphs of y = x, y = −x + 6
and x = 0. Write down the integral that would compute the area of R.
Solution: In this case we may use vertical slices and the area will be given by
Z 4
√
A=
−x + 6 − x dx
0
Page 6