Math 3B Midterm Practice
Written by Victoria Kala
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SH 6432u Office Hours: R 11:00 am - 12:00 pm
Last updated 1/22/2015
1. Evaluate
Z
0
(1 +
p
9 − x2 )dx
−3
by interpreting it in terms of areas.
Solution. This is like unto a quiz we did the second week. Split up the integral
Z 0
Z 0
Z 0p
p
2
1dx +
(1 + 9 − x )dx =
9 − x2 dx
−3
−3
−3
R0
Let’s look at the first integral: −3 1dx. The graph of f (x) = 1 is a horizontal line through
the point y = 1. This graph forms a rectangle on [−3, 0] (draw a picture!). The area of a
rectangle is A = lw = 3 · 1 = 3 (notice that the area is positive because we are above the
R0
x-axis). Thus −3 1dx = 3.
√
R0 √
Now let’s look at the second integral: −3 9 − x2 dx. The graph of f (x) = 9 − x2 is a
semi-circle with radius r = 3. This graph forms a quarter of a circle on [−3, 0] (draw a
πr2
π · 32
9π
picture!). The area of a quarter circle is A =
=
=
(notice that the area is
4
4
4
R0 √
9π
.
positive because we are above the x-axis. Thus −3 9 − x2 dx =
4
Now we want to plug into our formula above:
Z 0
Z 0
Z 0p
p
9π
(1 + 9 − x2 )dx =
9 − x2 dx = 3 +
1dx +
4
−3
−3
−3
2. Find f if f 0 (x) = cos x − (1 − x2 )−1/2 .
Solution. We really just need to find the anti-derivative of f 0 (x). Let’s rewrite it so it looks
a little nicer:
1
f 0 (x) = cos x − (1 − x2 )−1/2 = cos x − √
1 − x2
We know how to find the anti-derivative of this:
Z
Z 1
0
f (x) = f (x)dx =
cos x − √
dx = sin x − sin−1 x + C
1 − x2
1
3. Express the Riemann sum
lim
n→∞
n
X
sin xi ∆x
i=1
as a definite integral on the interval [0, π] and then evaluate the integral.
Solution. Recall that the definite integral and Riemann sum are related to each other by the
following formula:
Z b
n
X
lim
f (xi )∆x =
f (x)dx
n→∞
a
i=1
Let’s compare our Riemann sum with a general Riemann sum:
lim
n→∞
n
X
f (xi )∆x = lim
n→∞
i=1
n
X
sin xi ∆x
i=1
By comparing these two, we see that f (xi ) = sin xi ⇒ f (x) = sin x. We can then write out
our integral on the interval [0, π]:
lim
n→∞
n
X
Z
sin xi ∆x =
0
i=1
π
π
sin xdx = − cos x = − cos π − (− cos 0) = 1 − (−1) = 2
0
4. Evaluate the upper and lower sums for f (x) = 1 + x2 , −1 ≤ x ≤ 1 with n = 3 and n = 4.
Illustrate with diagrams. (Leave your answers as fractions, no need to use a calculator!)
Solution. (a) n = 3: First let’s find ∆x:
∆x =
b−a
1 − (−1)
2
=
=
n
3
3
We will have three intervals: −1 ≤ x ≤ − 13 , − 13 ≤ x ≤ 31 , 13 ≤ x ≤ 1.
Upper sum: On the first interval, the highest point occurs at f (−1) = 2. On the second
interval, the highest point occurs at f ( 31 ) = f (− 13 ) = 1 + ( 13 )2 = 1 + 19 = 10
9 . On the
third interval, the highest point occurs at f (1) = 2. So
A ≈ Aupper =
3
X
10
2
92
1
+ f (1) ∆x = 2 +
+2 · =
f (xi )∆x = f (−1) + f
3
9
3
27
i=1
Lower sum: On the first interval, the lowest point occurs at f (− 13 ) = 10
9 . On the second
interval, the lowest point occurs at f (0) = 1. On the third interval, the lowest point
occurs at f ( 13 ) = 10
9 . So
A ≈ Alower
3
X
1
1
10
10 2
58
=
f (xi )∆x = f (− ) + f (0) + f ( ) ∆x =
+1+
· =
3
3
9
9
3
27
i=1
2
(b) n = 4: First let’s find ∆x:
∆x =
b−a
1 − (−1)
2
1
=
= =
n
4
4
2
We will have four intervals: −1 ≤ x ≤ − 21 , − 12 ≤ x ≤ 0, 0 ≤ x ≤ 21 , 12 ≤ x ≤ 1.
Upper sum: On the first interval, the highest point occurs at f (−1) = 2. On the second
interval, the highest point occurs at f (− 21 ) = 1 + (− 12 )2 = 1 + 14 = 54 . On the third
interval, the highest point occurs at f ( 12 ) = 54 . On the fourth interval, the highest point
occurs at f (1) = 2. So
4
X
1
1
+f
+ f (1) ∆x
A ≈ Aupper =
f (xi )∆x = f (−1) + f −
2
2
i=1
5 5
1
13
= 2+ + +2 · =
4 4
2
4
Lower sum: On the first interval, the lowest point occurs at f (− 21 ) = 54 . On the second
interval, the lowest point occurs at f (0) = 1. On the third interval, the lowest point
occurs at f (0) = 1. On the fourth interval, the lowest point occurs at f ( 12 ) = 54 . So
4
X
1
1
f (xi )∆x = f −
+ f (0) + f (0) + f
∆x
2
2
i=1
5
5 1
9
+1+1+
· =
=
4
4 2
4
A ≈ Alower =
5. If
R6
0
R4
f (x)dx = 10 and
0
R6
f (x)dx = 7, find
Solution. Recall that
Z
4
4
6
Z
f (x)dx +
0
f (x)dx.
Z
6
f (x)dx =
4
f (x)dx
0
(this is a nice property of integrals). We want to solve for
Z
6
Z
Z
f (x)dx −
f (x)dx =
4
6
0
√
Z
h(x) =
ex
3
x
4
f (x)dx:
4
f (x)dx = 10 − 7 = 3
0
6. Find the derivative of
R6
z2
dz
sin z + 2
Solution. Use the Fundamental Theorem of Calculus, part 1:
√
√
( x)2
(ex )2
1
e2x
x
0
x 0
√
√
√
h (x) =
· ( x)0 −
·
−
·
(e
· ex
)
=
sin ex + 2
sin x + 2
sin x + 2 2 x sin ex + 2
√
e3x
x
√
−
=
2(sin x + 2) sin ex + 2
7. A particle is moving with the given data, find the position of the particle:
a(t) = 10 sin t + 3 cos t, s(0) = 0, s(2π) = 12
Solution. Find the anti-derivatives:
a(t) = 10 sin t + 3 cos t
v(t) = −10 cos t + 3 sin t + C
s(t) = −10 sin t − 3 cos t + Ct + D
Need to use s(0) = 0 and s(2π) = 12 to solve for C and D:
s(0) = 0 ⇒ −10 sin 0 − 3 cos 0 + C · 0 + D = 0 ⇒ −3 + D = 0 ⇒ D = 3.
s(2π) = 12 ⇒ −10 sin 2π − 3 cos 2π + C · 2π + D = 12 ⇒ −3 + 2πC + 3 = 12 ⇒ C =
Thus s(t) = −10 sin t − 3 cos t + π6 t + 3.
Evaluate the integrals:
Z
8.
tan xdx
Solution. Solve using substitution.
Z
Z
tan xdx =
sin x
dx
cos x
Let u = cos x, du = − sin xdx:
Z
Z
sin x
1
dx = −
du = − ln |u| + C = − ln | cos x| + C
cos x
u
We can rewrite this answer using log properties:
1 + C = ln | sec x| + C
− ln | cos x| + C = ln cos x Either answer is acceptable.
4
6
π.
Z
9.
1
(xe + ex )dx
0
Solution.
e+1
e+1
e+1
Z 1
x
1
0
1
1
(xe + ex )dx =
+ ex =
+ e1 −
+ e0 =
+e−1
e
+
1
e
+
1
e
+
1
e
+
1
0
0
Z
10.
sec t(sec t + tan t)dt
Solution.
Z
Z
sec t(sec t + tan t)dt =
(sec2 t + sec t tan t)dt = tan t + sec t + C
√
Z
11.
e x
√ dx
x
Solution. Solve using substitution. Let u =
√
Z
√
Z
12.
3
√
1/ 3
√
1/ 3
Z
e
e4
Z
x, du =
1
√
dx:
2 x
√
eu du = 2eu + C = 2e
x
+C
8
dx
1 + x2
Solution.
Z √3
13.
e x
√ dx = 2
x
√
√3
π π 4π
√ √
8
−1 dx
=
8
tan
x
−
=
√ = 8 tan−1 ( 3) − tan−1 (1/ 3) = 8
2
1+x
3
6
3
1/ 3
dx
√
x ln x
Solution. Solve using substitution. Let u = ln x, du = x1 dx; when x = e, u = 1, when
x = e4 , u = 4:
Z e4
Z 4
Z 4
4
p
p
dx
1
√
√ du =
=
u−1/2 du = 2u1/2 = 2( (4) − (1)) = 2(2 − 1) = 2
u
1
x ln x
e
1
1
5