MATH 141 - Calculus II
Tutorial
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Contents
Tutorial 1
Tutorial 2
Tutorial 3
Tutorial 4
Strategy for Integration
Tutorial 5
Tutorial 6
Tutorial 7
Tutorial 9
2
3
5
8
10
20
22
23
25
1
Tutorial 1
√
(1) Quiz Problem: Express the area under the graph of f (x) = x2 + 1 + 2x, 4 ≤ x ≤ 7 as a limit.
(2) Find the area under f (x) = x + 1 using Riemann sums over [0, 2].
R2
(3) Express 0 x2 dx as a Riemann sum
Pn
1
(4) Express limn→∞ i=1 n+i
as a definite integral
Solutions:
3
3i
(1) Here ∆x = 7−4
n = n so xi = 4 + n . Hence the area under the curve is given by
"
2 s
#
n
3X
3i
3i
lim
4+
+ 1+2 4+
n→∞ n
n
n
i=1
No need to simplify this expression.
2
2i
(2) Let ∆x = 2−0
n = n and xi = 0 + i∆x = n so that
n
n X
X
2i
2
4 n(n + 1)
2
A = lim
f (xi )∆x = lim
+1
= 2·
+ ·n=2+2=4
n→∞
n→∞
n
n
n
2
n
i=1
i=1
(3) Recall that
Z
b
f (x)dx = lim
n→∞
a
where ∆x =
b−a
n
and xi = a + i∆x. Hence,
Z 2
n
X
2
(xi )2 · ,
x2 dx = lim
n→∞
n
0
i=1
= lim
n→∞
n
X
f (xi )∆x
i=1
xi = 0 + i ·
2
2i
=
n
n
n
X
8i2
i=1
n3
n
8 X 2
i
n→∞ n3
i=1
= lim
and we could evaluate this further, but we were only asked to express this as a Riemann sum.
1
(4) Here, we must write n+i
in the form
b−a
b−a
i ·
f (xi )∆x = f a +
n
n
for some a, b. So, write
!
1
1
1
=
n+i
n 1 + ni
1
i
which suggests that a = 1 and b = 2 so that ∆x = 2−1
n = n and xi = 1 + n . Now,
!
n
n
n
X
X
X
1
1
1
1
=
=
∆x.
i
n+i
x
1+ n n
i=1
i=1
i=1 i
This implies that f (x) =
1
x
and the definite integral here is
Z 2
1
dx.
1 x
2
Tutorial 2
R6
R6
(1) Quiz 2: If 0 f (x)dx = 12 find 0 (3f (x) − 2) dx.
R x3
(2) If f (x) = sin x 3t2 dt, find its derivative.
Rx
R t2 √ 4
du, find F 00 (2).
(3) If F (x) = 0 f (t)dt and f (t) = 1 1+u
u
R1 √
√
3
4
(4) Evaluate 0 x( x + x)dx
(5) Challenge Problem: Suppose f (x) is continuos, increasing and non-negative function on [a, b]
(where 0 < a < b). Demonstrate pictorially that the following identity is true:
Z b
Z f (b)
f (x)dx +
f −1 (y)dy = bf (b) − af (a).
a
f (a)
(6) (a)
(d)
Z
x cos(2x)dx
Z
(b)
Z
2x2
p
x3 + 1dx
√
x 1 + xdx
(e)
(c)
Z
Z
1
dx
x2 + 25
x3x dx
Solutions:
(1) Recall, the lineararity of integration
Z
Z
Z
(af (x) + bg(x)) dx = a f (x)dx + b g(x)dx.
With this, we find that
Z 6
Z
(3f (x) − 2)dx = 3
0
6
Z
6
dx = 3(12) − 2[x]60 = 36 − 12 = 24.
f (x)dx − 2
0
0
(2) We can write
Z
f (x) = −
sin x
3t2 dt +
0
Z
x3
3t2 dt
0
so that (by FTCI)
f 0 (x) = −3 sin2 x cos x + 3x6 (3x2 ) = −3 sin2 x cos x + 9x8
(3) By FTCI, we know that F 0 (x) = f (x) and F 00 (x) = f 0 (x). So,
p
√
1 + (x2 )4
2 1 + x8
00
0
F (x) = f (x) =
(2x) =
.
x2
x
Now, evaluating at 2 this becomes
√
√
2 1 + 28
00
F (2) =
= 257
2
(4)
Z 1
Z 1
√
√
3
4
x( x + x)dx =
(x4/3 + x5/4 )dx
0
0
3 7/3 4 9/4
x + x
7
9
55
= 3/7 + 4/9 =
63
1
=
3
0
(5) For this, simply sketching an arbitrary increasing function in the positive quadrant should geometrically reveal that bf (b) − af (a) represents the difference in area between the rectangle formed by the
points (0, 0), (0, b), (b, f (b)), (0, f (b)) and the rectangle whose corners are (0, 0) and (a, f (a)). Then
Rb
the graph of the function splits this area into two regions one, having area equal to a f (x)dx and the
R f (b)
other having area f (a) f −1 (y)dy. Putting this information together yields the identity we intended
to prove.
(6) (a) This is done with integration by parts where u = x, dv = cos 2xdx so that du = dx and
v = sin22x .
Z
Z
x
sin 2x
x cos(2x)dx = sin 2x −
dx
2
2
− cos 2x
x
+C
= sin 2x −
2
4
x
cos 2x
= sin 2x +
+C
2
4
(b) Here, use the substitution u = x3 + 1 and du = 3x2 .
Z
Z
p
2 √
2x2 x3 + 1dx =
udu
3
2 2
= · u3/2 + C
3 3
4
= u3/2 + C
9
(c)
Z
Z
1
1
1
1
x
dx
=
let u = , du = dx
2 dx,
2
x
x + 25
25
5
5
1+ 5
Z
1
1
=
du
25
1 + u2
1
= arctan(u) + C
5
x
1
= arctan
+C
5
5
(d)
Z
Z
√
√
x 1 + xdx = (u − 1) udu, with u = x + 1, x = u − 1, du = dx
Z
= (u3/2 − u1/2 )du
2 5/2 2 3/2
u − u +C
5
3
2
2
5/2
= (x + 1) − (x + 1)3/2 + C
5
3
=
(e) Here we use integration by parts with u = x, dv = 3x so that du = dx, v =
Z
Z x
x3x
3
x
x3 dx =
−
dx
ln 3
ln 3
x3x
3x
=
−
+C
ln 3 (ln 3)2
4
3x
ln 3 .
Tutorial 3
(1) Quiz: Evaluate the indefinite integral (where a 6= 0)
Z
1
dx.
ax + b
(2) Evaluate the following integrals
(c)
(a)
Z
2
1
x+2
dx
4x + x2
Z
√
e
2x
dx
0
(b)
Z
1
2
2 (ln 3)
(d)
ex
dx
9 + ex
Z
π/4
sin3 2t cos 2tdt
0
(3) Find the area of the region enclosed by y = ln x, the x-axis and x = e2 . Also show that integrating
on the y-axis gives the same area as integrating on the x-axis.
(4) Find the volume of a tetrahedron with base triangle equilateral of side length a and height h.
Recall that a tetrahedron looks like...
h
a
a
a
(5) Challenge Problem:
Compute the volume of a torus, having large
radius R and small radius r. Recall that a torus
looks as depicted to the right.
[Hint: cylindrical shells and trig substitution]
r
•
R
5
Solutions:
(1) Let u = ax + b so du = adx and the integral becomes
Z
Z
1
1
1
1 du
dx =
= ln |u| + C = ln |ax + b| + C
ax + b
u a
a
a
√
√
(c) First, make the substitution u = 2x so,
(2) (a) Substitution with u = 4x + x2 so
1
1
du = 2u
du = 2u
(4 + 2x)dx which is
2du ↔ udu = dx then (IBP) with
udu = (2 + x)dx
r = u, ds = eu du so that dr = du, s = eu
Z ln 3
Z 1 (ln 3)2 √
2
2x
dx
=
ueu du
e
Z x=2
Z 2
0
0
u
x+2
Z ln 3
√
dx =
du,
2
u
4x
+
x
u ln 3
x=1
1
= [ue ]0 −
eu du
x=2
0
= [u]x=1
√
√
= 3 ln 3 − (eln 3 − e0 )
=2 3− 5
= 3 ln 3 − 2
(b) Substitute u = ex , du = ex dx
Z
ex
dx =
9 + ex
(d) Substitution with u = sin 2t, du = 2 cos 2tdt
Z
1
du
9 + u2
Z
1
1
=
2 du
9
1 + u3
x
e
1
+C
= arctan
3
3
Z
0
π/4
1
sin 2t cos 2tdt =
2
3
Z
t=π/4
u3 du
t=0
π/4
1 sin4 2t
2
4
0
1
1 1
=
= ·
2 4
8
=
(3) With respect to x;
Z
Ax =
e2
u = ln x, dv = dx ⇒ du =
ln xdx,
1
e2
Z
e2
= [x ln x]1 −
1
dx, v = x
x
dx
1
e2
= 2e2 − [x]1
= 2e2 − e2 + 1 = e2 + 1
With respect to y;
Z
Ay =
2
(e2 − ey )dy
0
2
= e2 y − ey 0
= 2e2 − e2 + e0 = e2 + 1
√
(4) First, note that the height of an equilateral triangle of side length a is given by h = 23a , so then
the area is
√
√
1
3a
3 2
1
base · height = a ·
=
a .
2
2
2
4
Now, the area of a triangular slice at a distance y < h from the top of the tetrahedron remains
equilateral, so its area is expressed, as above, in terms of the side length a(y) as
√
3
A(y) =
(a(y))2 .
4
6
y
a
By similar triangles, we find the relationship a(y)
y = h so that a(y) = h a and thus,
√ √ 2
3 y 2
3a 2
A(y) =
y
a =
4 h
4h2
Finally, the volume is given by the definite integral
√ 2Z h
√ 2 3 h
√ 2 3
√
Z h
Z h√ 2
h
3a 2
3a
3a y
3a h
3 2
2
y
dy
=
y
dy
=
=
=
·
a
A(y)dy =
2
2
2
2 3
4h
4h
4h
3
4h
3
4
0
0
0
0
Rb
(5) Recall the formula for cylindrical shells looks like V = a 2πxf (x)dx. Here our bounds of integration
are a = R − r and b = R + r and the height (f (x)) is found from the implicit equation of a circle
having radius r, centred p
at (R, 0). This circle is given the relation (x − R)2 + y 2 = r2 , so the height
at an arbitrary x is y = r2 − (x − R)2 . So, finally, half of our desired volume should equal
Z
R+r
p
x r2 − (x − R)2 dx
2π
R−r
Consider the trig substitution x − R = r sin θ so that dx = r cos θdθ, we will also want to rearrange
to write that x = R + r sin θ and the new bound of integration are θ ranges from −π/2 to π/2. That
is
Z R+r p
Z π/2
p
2π
x r2 − (x − R)2 dx = 2π
(R + r sin θ) r2 − r2 sin2 θ ·r cos θdθ
{z
}
|
R−r
−π/2
=r cos θ
Z
π/2
(R + r sin θ)r2 cos2 θdθ
= 2π
−π/2
Z
π/2
Rr2 cos2 θ + r3 sin θ cos2 θ dθ
= 2π
−π/2
The second term in this integrand is an odd function, so the integral will evaluate to zero! So it
remains only to compute
Z π/2
Z π/2
2
2
2
2π
Rr cos θdθ = 2πRr
cos2 θdθ.
−π/2
−π/2
Recall, the half angle identity,
cos(2θ) = 2 cos2 θ − 1
which is equivalently expressed as
cos2 θ =
cos(2θ) + 1
2
So,
2πRr2
Z
π/2
−π/2
cos2 θdθ = 2πRr2
Z
π/2
−π/2
π/2
1 sin(2θ)
cos(2θ) + 1
dθ = 2πRr2
+θ
= π 2 Rr2 .
2
2
2
−π/2
|
{z
}
=π
Recall that we have only computed half the volume because we were only working with the top half
of the torus. Hence, the total volume of the torus given is
V = 2π 2 Rr2 = 2πR · πr2
and is indeed, simply the area of the circle slice drawn in red times the circumference of the solid at
the centroid of the circle.
Remark 1. More generally, the volume of the solid obtained by revolving any closed curve about an
axis is equal to the area enclosed by the curve times the circumference defined by the centroid (center
of mass) of that enclosure. This is known as Pappus’s Centroid Theorem.
7
Tutorial 4
(1) Quiz: Find the area enclosed by the parabola y 2 − 2y = x and the line y = 21 x.
(2) Given that
(
sin x, x ∈ [0, π/4]
f (x) =
cos x, x ∈ [π/4, π/2]
(a) set up the integration on x and y for the volume of solid obtained by revolving this region about
the x-axis.
(b) repeat part (a) for the solid generated by revolving this region about the y-axis
(3) Evaluate
Z
1
dx
x3/2 + x1/2
8
Solutions:
(1) equating the curves for points of intersection we find
2y = y 2 − 2y
has solutions y = 0 and y = 4. Now integrating with respect to y gives the desired area
2
4
Z 4
Z 4
y
y3
42
43
32
2
2
(4y − y )dy = 4 −
[2y − (y − 2y)]dy =
=4 −
=
2
3
2
3
3
0
0
0
(2) (a) For integration wrt x, we use the cross-sectional method;
Z π/4
Z π/2
V =
πf 2 (x)dx +
πf 2 (x)dx
0
π/4
π/4
Z
Z
2
π/4
0
Z
2
cos xdx
sin xdx +
=π
!
π/2
π/4
sin2 xdx
= 2π
0
π/4
1 − cos 2x
dx
2
0
π/4 1
sin 2x
1
2π π/8 −
= 2π x −
2
4
4
0
Z
= 2π
For integration wrt y, we use cylindrical shells;
Z 1/√2
2πy(arccos y − arcsin y)dy
V =
0
(b) Now, for integration wrt x, we use cylindrical shells;
Z π/4
Z π/2
V =
2πx sin xdx +
2πx cos xdx
0
π/4
and for integration wrt y, use cross-sectional method;
Z 1/√2
π(arccos2 y − arcsin2 y)dy
(3) with the substitution u =
√
0
1
x, so that x = u2 and du = 2u
dx we find,
Z
Z
1
2u
dx =
du
u(u2 + 1)
x3/2 + x1/2
Z
1
=2
du
u2 + 1
= 2 arctan u + C
√
= 2 arctan x + C
9
Strategy for Integration
(1) Simplify the integrand if possible
(2) Look for an obvious substitution
(3) Classify the integrand according to its form
(4) Try again
Problems We solve the odd numbered problems from §7.5
1.
Z
2
29.
Z
57.
ln(x +
cos x(1 + sin x)dx
p
x2
− 1)dx
31.
3.
Z
59.
sin x + sec x
dx
tan x
5.
Z r
t
dt
4
t +2
7.
arctan y
e
dy
1 + y2
−1
3 − 2x −
Z
0
x−1
dx
x2 − 4x + 5
5
Z
sec θ tan θ
dθ
sec2 θ − sec θ
√
69.
Z
4
3
√
θ tan θdθ
1
43.
Z
dx
(1 − x2 )3/2
17.
73.
Z
5 −x3
x e
t cos tdt
Z
x + arcsin x
√
dx
1 − x2
dx
0
47.
Z
Z
x
ex+e dx
x3 (x − 1)−4 dx
49.
21.
Z
arctan
√
x
1
√
dx
x 4x + 1
51.
Z
1
(1 +
√
x)8 dx
0
3x − 2
dx
x2 − 2x − 8
2
x sinh mxdx
55.
27.
Z
dx
1 + ex
√
xex
dx
1 + ex
Z
x sin2 x cos x
Z
√
81.
Z
2
Z
1
√
dx
x 4x2 + 1
53.
Z
1
dx
(x − 2)(x2 + 4)
79.
Z
25.
75.
Z
77.
Z
23.
e2x
dx
1 + ex
Z
x
dx
1 + x3
45.
2
19.
1 + x2
dx
x2
71.
√
Z
π
Z
dx
1
√ dx
x+1+ x
√
Z
2
sin t cos tdt
15.
x
67.
Z
41.
Z
√
xe
sin 2x
dx
1 + cos4 x
tan3 θ sec2 θdθ
39.
13.
√
65.
π/4
Z
1
Z
Z
cos 2x cos 6xdx
r4 ln rdr
11.
dθ
1 + cos θ
63.
Z
3
Z
Z
x2 dx
37.
9.
cos x cos3 (sin x)dx
61.
Z p
35.
1
Z
Z
1+x
dx
1−x
33.
Z
√
x 3 x + cdx
Z
1 − sin xdx
83.
Z
dx
√
x+x x
10
Z
sec3 tdt
Solutions: Note that the constant terms in the indefinite integrals here have been suppressed.
1.
Z
Z
2
Z
cos xdx + sin2 x cos xdx
Z
= sin x + u2 du
cos x(1 + sin x)dx =
= sin x + u3 /3
= sin x +
sin3 x
3
3.
Z
sin x + sec x
dx =
tan x
Z
1
cos x
sin x
cos x
sin x +
dx
Z 1
cos x +
dx
sin x
Z
= sin x + csc xdx
=
= sin x + ln | csc x − cot x|
5.
Z
Z
1
1
t
dt =
du
t4 + 2
2
u2 + 2
Z
1
1
=
du
√ 2
4
u/ 2 + 1
Z
1 √
1
2dy
=
2
4
y +1
√
√ 2
=
arctan t2 / 2
4
7.
Z
1
−1
earctan y
dy =
1 + y2
Z
π/4
eu du = eπ/4 − e−π/4
−π/4
9. By parts u = ln r, dv = r4 dr
Z
Z 4
r5
r
r5
r5
r4 ln rdr =
ln r −
dr =
ln r −
5
5
5
25
11. This is our first exciting one!
Z
Z
x−1
x−1
dx =
dx
2
2
x − 4x + 5
(x − 4x + 4) + 1
Z
x−1
=
dx
(x − 2)2 + 1
Z
x−2+1
=
dx
(x − 2)2 + 1
Z
Z
x−2
1
=
+
2
(x − 2) + 1
(x − 2)2 + 1
= ln |(x − 2)2 + 1| + arctan(x − 2)
11
13.
Z
sin5 t cos4 tdt =
Z
Z
=
Z
=
Z
z=
(1 − cos2 t)2 cos2 t sin tdt
(1 − u2 )2 u2 du
(1 − 2u2 + u4 )u2 du
(1 − 2u4 + u6 )du
2
u7
= u − u5 −
5
7
15.
Z
dx
=
(1 − x2 )3/2
Z
cos t
dt
(1 − sin2 t)3/2
1
=
dt
2
Zcos t
sec2 tdt = tan t
=
17. IBP with u = t, dv = cos2 tdt so that du = dv and v = 21 t + sin42t . Note that we have used the
2t
identity cos2 = 1+cos
. Now,
2
Z π
Z t sin 2t 1
sin 2t
1
−
t+
dt
t cos2 tdt = t2 +
2
2
2
2
0
1
t sin 2t t2
cos 2t
= t2 +
− +
2
2
4
8
1 2 t sin 2t cos 2t
= t +
+
4
2
8
19.
Z
x
ex+e dx = ee
x
21.
Z
arctan
23. With u = 1 +
√
x so that du =
Z
1
(1 +
1
√
dx
2 x
√
√
x
and rearranging is dx = 2(u − 1)du
Z
8
x) dx = 2
u8 (u − 1)du
0
2u10
2u9
−
10 √ 9
√
(1 + x)10
2(1 + x)9
=
−
5
9
25. This one will involve partial fractions. Factor the denominator as
=
x2 − 2x − 8 = (x − 4)(x + 2)
then write
1
A
B
=
+
(x − 4)(x + 2)
x−4 x+2
12
and solve to find A = −B = 16 . Now,
Z
Z 2
3x2 − 2
1
3x − 2 3x2 − 2
dx
=
−
dx.
x2 − 2x − 8
6
x−4
x+2
Dividing these polynomials yields that
46
3x2 − 2
= 3x + 12 +
x−4
x−4
and
3x2 − 2
10
= 3x − 6 +
.
x+2
x+2
Our integral then becomes
Z 46
10
1
3x + 12 +
− 3x + 6 −
dx
6
x−4
x+2
Z 1
46
10
=
18 +
−
dx
6
x−4 x+2
46
10
= 3x −
ln |x − 4| −
ln |x + 2|
6
6
27. Let u = 1 + ex so that du = ex dx = (u − 1)dx and
Z
Z
dx
1
=
du
1 + ex
u(u − 1)
Z 1
1
=
−
du
u−1 u
= ln |u − 1| − ln |u|
= x − ln(1 + ex )
√
x2 − 1), dv = dx so that
√
1
x
x + x2 − 1
1
√
√
√
du =
· 1+ √
dx =
dx = √
dx
2
2
2
2
2
x+ x −1
x −1
(x + x − 1) x − 1
x −1
29. IBP with u = ln(x +
and v = x. Now,
Z
Z
p
p
x
2
2
√
ln(x + x − 1)dx = x ln(x + x − 1) −
dx
2
x −1
and make the substitution u = x2 − 1 so that du = 2xdx. So, the integral becomes
Z
p
p
p
p
√
1
1
2
√ du = x ln(x + x2 − 1) − u = x ln(x + x2 − 1) − x2 − 1
x ln(x + x − 1) −
2
u
31.
Z r
1+x
dx
1−x
33. Here lets complete the square and write
p
p
3 − 2x − x2 = 4 − (1 + x)2 !
Now, letting u = x + 1, this is
Z p
3 − 2x −
x2 dx
13
=
Z p
4 − u2 du
and with the trig substitution u = 2 sin θ so that du = 2 cos θdθ and
Z
Z p
√
2
4 − u du =
4 − 4 sin θ(2 cos θ)dθ
Z
= 4 cos2 θdθ
Z
1 + cos 2θ
dθ
=4
2
θ sin 2θ
=4
+
2
4
= 2 arcsin(u/2) + 2 sin θ cos θ
√
u
2 − u2
= 2 arcsin(u/2) + 2 ·
2√
2
u 2 − u2
= 2 arcsin(u/2) +
2
35. Here, we use the identity
cos(mx) cos(nx) =
1
(cos(m − n)x + cos(m + n)x)
2
Z
1
cos(4x) + cos(8x)
2
sin 4x sin 8x
=
+
8
16
Z
cos 2x cos 6xdx =
37. Let u = tan θ so du = sec2 θdθ and
Z π/4
Z
3
2
tan θ sec θdθ =
0
π/4
u3 du
θ=0
=
tan4 θ
4
π/4
= 1/4
0
39. This one I’d like to solve in two ways!
Z
Z
sec θ tan θ
dθ
=
sec2 θ − sec θ
Z
=
Z
=
Z
=
Z
=
Z
=
tan θ
dθ
sec θ − 1
tan θ
sec θ + 1
·
dθ
sec θ − 1
sec θ + 1
tan θ(sec θ + 1)
dθ
sec2 θ − 1
Z
sec θ
1
dθ +
dθ
tan θ
tan θ
Z
cos θ
cos θ
dθ +
dθ
sin θ cos θ
sin θ
Z
1
cos θ
dθ +
dθ
sin θ
sin θ
= − ln | csc θ + cot θ| + ln | sin θ|
Alternatively, we could have simply made the substitution u = sec θ, so that du = tan θ sec θdθ
and then this is simply the integral
Z
1
du
u2 − u
14
and can be solved using partial fractions... That is,
u2
1
A
B
(A + B)u − A
= +
=
−u
u
u−1
u(u − 1)
implying that
A = −B = −1
Now,
Z
sec θ tan θ
dθ =
sec2 θ − sec θ
Z
1
du
−u
Z
Z
−1
1
=
du +
du
u
u−1
= ln |u − 1| − ln |u|
u2
= ln | sec θ − 1| − ln | sec θ|
Verify that these two solutions agree!
41. Here use IBP with u = θ, dv = tan2 θdθ so that du = dθ and
Z
Z
2
v = tan θdθ = (sec2 θ − 1)dθ = tan θ − θ.
Now,
Z
θ tan2 θdθ = θ(tan θ − θ) −
Z
(tan θ − θ)dθ
= θ(tan θ − θ) + ln | cos θ| +
= θ tan θ + ln | cos θ| −
θ2
2
θ2
2
√
43. Here, substitute u = x3/2 so dx = 32 xdx. Now,
Z √
Z
2
2
x
1
dx =
du = arctan(x3/2 )
1 + x3
3
1 + u2
3
45. Substitute u = x3 and du = 3x2 dx so
Z
Z
1
5 −x3
x e
dx =
ue−u du
3
Z
1
=
−ue−u + e−u du
IBP
3
−e−u
=
(u + 1)
3
3
−e−x
=
(x3 + 1)
3
47. Let u = x − 1 (a translation)
Z
Z
3
−4
x (x − 1) dx = (u + 1)3 u−4 du
Z 3
u + 2u2 + 3u + 1
=
du
u4
Z 1
1
1
1
=
+ 3 2 + 3 3 + 4 du
u
u
u
u
49. [INCORRECT - I think] Make the initial substitution u = 4x + 1 so that du = 4dx and x =
Now,
15
u−1
4 .
Z
1
dx =
x 4x + 1
√
Z
1
√ du
(u − 1) u
√
Let v = u so that dv = 2√1 u du and this becomes
Z
Z
Z
Z
Z
1
1
sec θ tan θ
sec θ
√ du = 2
dv
=
dθ
=
dθ
=
csc θdθ
v2 − 1
sec2 θ − 1
tan θ
(u − 1) u
θ+cot θ
2
Multiply by a phantom factor of 1 = csc
csc θ+cot θ , let v = csc θ + cot θ so dv = −(csc θ + csc θ cot θ)dθ.
Z
Z
csc θ + cot θ
1
csc θ
dθ = −
dv = − ln | csc θ + cot θ|
csc θ + cot θ
v
Recall that v = sec θ so we have the following right-angled triangle
csc θ = v =
v
4x + 1
so
1
cot θ =
θ√
√
√
v2 − 1 =
√
4x
v2 − 1
Finally,
Z
√
4x + 1
√
1
1 √
dx = − ln +
= ln |4x| − ln | 4x + 1 + 1|
4x
4x
x 4x + 1
51. That 4 with the x2 scares me, so we will start by letting u = 2x and
Z
Z
1
1
√
√
dx =
du
x 4x2 + 1
u u2 + 1
Now this is a nice looking trig substitution u = tan θ so that du = sec2 θdθ and
Z
Z
Z
sec2 θ
1
√
du =
dθ = csc θdθ
tan θ · sec θ
u u2 + 1
θ+cot θ
2
Multiply by a phantom factor of 1 = csc
csc θ+cot θ , let v = csc θ + cot θ so dv = −(csc θ + csc θ cot θ)dθ.
Z
Z
csc θ + cot θ
1
dθ = −
dv = − ln | csc θ + cot θ|
csc θ
csc θ + cot θ
v
Recall that u = tan θ so we have the following right-angled triangle
√
√
u2
csc θ =
+1
u
√
u2 +1
u
=
1
u
=
4x2 +1
2x
so
cot θ =
1
2x
θ
1
and, finally,
Z
√
4x2 + 1
p
1
1 √
dx = − ln +
= ln |2x| − ln | 4x2 + 1 + 1|
2x
2x x 4x2 + 1
16
x
−x
x
−x
53. Recall that sinh x := e −e
and cosh x = e +e
. So, we use integration by parts with u = x2 , dv =
2
2
1
sinh(mx)dx and du = 2xdx, v = m cosh(mx). Now,
Z
Z
x2
2
x2 sinh mxdx =
cosh(mx) −
x cosh(mx)dx
m
m
Integrate by parts again with u = x, dv = cosh(mx)dx so that du = dx, v =
then
x2
2
cosh(mx) −
m
m
Z
sinh(mx)
m
so the above is
Z
x2
x sinh(mx)
1
x cosh(mx)dx =
cosh(mx) −
−
sinh(mx)dx
m
m
m
2
x sinh(mx)
x
1
=
+ 2 cosh(mx) −
m
m
m
√
1
dx or dx = 2udu. Now
55. Let u = x so that du = 2√
x
Z
Z
Z √
√
dx
2udu
1
1
1
√ = 2
=2
du = 2
−
du = 2(ln x − ln( x + 1))
u + u3
u + u2
u 1+u
x+x x
57. Substitute u = x + c to get
Z
Z
Z
√
3c
3
3
1/3
x x + cdx = (u − c)u du = (u4/3 − cu1/3 )du = (x + c)7/3 − (x + c)4/3 =
7
4
59. Let u = sin x so
Z
Z
Z
Z
v3
cos x cos3 (sin x)dx = cos3 (u)du = (1 − sin2 u) cos udu = (1 − v 2 )dv = v −
3
3
sin (sin x)
= sin(sin(x)) −
3
1−cos θ
61. Multiply by the phantom factor of 1 = 1−cos
θ so then
Z
Z
Z
Z
Z
dθ
1 − cos θ
1 − cos θ
cos θ
2
=
dθ
=
dθ
=
csc
θdθ
−
dθ = − cot θ + csc θ
2
2
1 + cos θ
1 − cos θ
sin θ
sin2 θ
√
1
63. Let u = x so du = 2√
dx or dx = 2udu. Then
x
Z
Z
√ √x
xe dx = u2 eu du
Z
2 u
= u e − 2 ueu du IBP
Z
= u2 eu − 2 ueu − eu du
IBP
= u2 − 2ue2 + 2eu
2x
65. Write cos4 x = (cos2 x)2 and substitute cos2 x = 1+cos
, then make the substitution u = cos 2x so
2
that du = −2 sin 2xdx and
Z
Z
sin 2x
sin 2x
dx =
dx
4
2x 2
1 + cos x
1 + 1+cos
2
Z
1
1
=−
2 du
2
1 + 1+u
2 1+u
= − arctan
2
= − arctan cos2 x
17
67. Multiply by conjugate and
Z
Z
1
√
√ dx =
x+1+ x
Z
=
Z
=
√
√ 1
x+1− x
√
dx
√ · √
√
x+1+ x
x+1− x
√
√
x+1− x
dx
x+1−x
√
√
( x + 1 − x)dx
i
2h
(x + 1)3/2 − x3/2
=
3
69.
√
Z
1
x
3
√
1 + x2
dx
x2
x
71. Let u = 1 + e so that du = e dx and
Z
Z
e2x
u−1
dx
=
du = u − ln |u| = 1 + ex − ln(1 + ex )
x
1+e
u
73. Let x = sin t so that dx = cos tdt and
Z
Z
sin t + t
x + arcsin x
√
p
dx =
cos tdt
1 − x2
1 − sin2 t
Z
= (sin t + t)dt
= − cos t +
t2
2
p
arcsin2 x
= − 1 − x2 +
2
75. Here we use a special kind of partial fractions;
1
A
Bx + C
(A + B)x2 + (C − 2B)x + (4A − 2C)
=
+
=
(x − 2)(x2 + 4)
x−2
x2 + 4
(x − 2)(x2 + 4)
implies that A = −B = 81 and C = − 41 . So,
Z
Z
Z
1
1
1
x+2
dx =
dx −
dx
(x − 2)(x2 + 4)
8
x−2
x2 + 4
Z
Z
1
1
1
x
=
dx −
dx
ln |x − 2| −
8
x2 + 4
2
( x2 )2 + 1
Z
1
1
1
=
ln |x − 2| −
du − arctan (x/2)
8
2
u
1
1
2
=
ln |x − 2| − ln(x + 4) − arctan (x/2)
8
2
77. [INCOMPLETE] Let u = 1 + ex so that du = ex dx and x = ln(u − 1). Then
Z
Z
xex
ln(u − 1)
√
√
dx
=
du
x
u
1+e
√
1
Now try IBP with a = ln(u − 1), db = u−1/2 du so that da = u−1
and b = 2 u. Now,
Z
Z √
√
ln(u − 1)
u
√
du = 2 u ln(u − 1) − 2
du
u−1
u
18
79.
Z
x sin2 x cos x
Z
√
81.
1 − sin xdx
83. IBP with u = sec t, dv = sec2 dt so that du = sec t tan t and v = tan t so that
Z
Z
sec3 = sec t tan t − sec t tan2 tdt
Z
= sec t tan t − sec t(1 − sec2 t)dt
Z
Z
= sec t tan t + sec3 − sec tdt
Z
= sec t tan t + sec3 − ln | sec t + tan t|
Rearranging, this implies that
Z
2 sec3 dt = sec t tan t − ln | sec t + tan t|
so that
Z
sec3 dt =
1
(sec t tan t − ln | sec t + tan t|) .
2
19
Tutorial 5
(1) Consider the infinite series
∞
X
1
.
n
e
n=1
Determine whether is converges or diverges. In the case that it converges, find the sum.
(2) Determine convergence or divergence of the following series
(e)
(a)
∞
X
3n2
n2 + 1
n=1
∞
X
sin(nπ)
n=1
(b)
∞
X
2n−1
3n
n=1
(f)
∞
X
(c)
∞
X
e1/n
n2
n=1
n=1
(−1)
n=1
n+1
1
n
(g)
(d)
∞
X
ln
∞
X
1
n(ln n)3
n=1
3
2n
Solutions:
(1) We first notice that limn→∞ e1n = 0 so the divergence test implies nothing. Consider now the function
f (x) = e1x which is decreasing, continuous on [1, ∞], represents the sequence at integer points and
is always positive. Looking at the improper integral
Z ∞
−x ∞
1
1
1
1
dx
=
−e
=
lim
−
=
x
t
1
t→∞
e
e
e
e
1
we know that the sequence converges!
Notice that
2 3 4
∞
X
1
1
1
1
1
= +
+
+
+ ...
n
e
e
e
e
e
n=1
which is a geometric series with a = 1e , r =
∞
1
X
1
e
=
n
e
1
−
n=1
(c)
(d)
(e)
(f)
< 1. So,
1
e
1
= ·
=
e
e−1
e−1
3n2
n2 +1
= 3 6= 0 so series diverges by divergence test.
n−1
P∞
This series can be re-expressed as 31 n=1 23
∞ so is geometric with a = 13 and r = 32 < 1.
Hence, converges
1/x
1/x
Consider f (x) = ex2 and we know 1 ≥ e1/x so x12 ≥ ex2 . Now, the p-series x12 converges so by
comparison test,
our seriesconverges!
n−1
P P
which is geometric with r = 12 < 1, so converges.
Check the n (−1)n+1 23n = 32 n 21
Now, since the series is absolutely convergent, it is thus, convergent.
This is a sum of zeros since sin(nπ) = 0 for all n. Converges.
limn→∞ ln n1 = −∞ so series diverges.
(2) (a) Here limn→∞
(b)
1
e
1
e
20
(g) Consider f (x) = x(ln1x)3 which is continuous, decreasing, positive and matches the sequence. So
with the substitution u = ln x, du = x1 dx we find
−2 ∞
Z ∞
Z ∞
−u
1
−3
dx =
u du =
3
x(ln
x)
2
0
1
0
this limit does not exist because of the lower limit of the integral which tends to infinity.
21
Tutorial 6
Determine absolute, conditional convergence or divergence of the following series.
(1)
(5)
∞
X
∞
X
(−1)n n2
3n
n=1
1
n + sin n
2
n=1
(2)
(6)
∞
X
(−1)n n!
10n
n=1
∞
X
(−1)n
(2n + 5)3/2
n=1
(3)
(7)
∞
X
(−1)n+1 6n
5n+1
n=1
(4)
∞
X
(−1)n 23n
nn
n=1
(8)
√
∞
X
(−1)n 2n − 1
n
n=1
∞ X
sec n 3/4 n
n=1
Solutions:
(1) Since
1
2n
1
1
·
= lim
=
=1
n→∞ 2n + sin n
n→∞ 1 + sinnn
1
1
+
0
2
P∞ 1
the series converges because n=1 2n converges.
lim
(2)
(n + 1)! 10n = lim 10n = ∞
lim
·
n→∞ 10(n + 1)
n! n→∞ 10
so, by the ratio test, this series diverges.
P∞ 6 −6 n−1
(3) This series can be rewritten as n=1 25
which geometric with |r| = 65 > 1, so diverges.
5
P∞ √2n−1
to check for absolute convergence. Comparing the sequence
(4) Consider first, the series n=1 n
√
1
√
an = 2n−1
as
with
b
=
n
n
n
r
p
√
an
2n − 1
lim
= lim
= lim
2 − 1/n = 2
n→∞ bn
n→∞
n→∞
n
P
By the limit comparison test, then, both series diverge since
bn diverges as a p-series with p =
1/2 < 1. So the series is not absolutely convergent.
√
1−x
Now, to check for conditional convergence, let f (x) = 2x−1
so that f 0 (x) = x2 √
≤ 0 for all
x
2x−1
√
= 0. By the alternating series
x ≥ 1. This implies our sequence is decreasing and limn→∞ 2n−1
n
test, this series converges.
Therefore, this series is conditionally convergent.
(5) This is absolutely convergent by the ratio test:
2
(n + 1)2 3n an+1 n + 2n + 1 1
= <1
lim
= lim n+1 · 2 = lim 3
n→∞
n→∞ an n→∞
3
n
3n2
(6) This series will be absolutely convergent by the limit comparison test with the sequence bn =
(7) Notice that
r
n
p
8
n 8
lim n |an | = lim
= lim
=0<1
n→∞
n→∞
n→∞ n
nn
so the sequence is convergent by the root
test.
1
n
(8) Note that | sec n| ≥ 1 so, n3/4
≤ sec
and
the sequence diverges by the comparison test.
3/4
n
22
1
n3/2
Tutorial 7
(1) Show that
3
2 x
e − 1 + x + x ≤ x
2! 3!
on the interval [0, 1/2].
R 1/2 2
(2) Estimate 0 ex dx and give an error bound on your approximation.
R 1/3 1
−4
(3) Estimate 0 1+x
.
4 dx correct to 10
(4) Find the power series representation of f (t) = arcsin(t) using the Binomial Theorem.
Solutions:
(1) Recall that f (x) = ex =
xn
n≥0 n!
P
and T2 (x) = 1 + x +
x2
2! .
So we should show that |R2 (x)| =
3
|f (x)−T2 (x)| ≤
We know that |R2 (x)| ≤ k|x|
where k = max[0,1/2]
√ 3 3! 3
ex
x
x
since e is increasing. Hence, R2 (x) = 3! < 3! as required.
x3
3!
|f (3) (x)| = max[0,1/2] ex = e1/2
(2) We know that
1/2
Z
2
ex dx =
Z
0
1/2
0
X x2n
n!
n≥0
and
Z
Z
f (x)dx ≤ |f (x)| dx
so
Z
Z
Z 1/2 X
Z 1/2 2(N +1)
N
N
1/2 2
1/2 2n 2n X
2
x
x
x
x
x
e dx −
dx.
≤
e −
dx ≤
0
n!
n!
(N + 1)!
0
0
0
n=0
n=0
Now, lets approximate with T2 (x).
1/2
Z 1/2 1
x3
x5
1
1
523
x4
= +
dx = x +
+
+
=
1 + x2 +
2!
3
5
·
2
2
24
320
960
0
0
with an error bound of
Z
1/2
R2 (x) ≤
0
7 1/2
x
1
x6
dx =
= 7
3!
7 · 3! 0
2 · 7 · 3!
(3) First, write
X
X
1
1
=
=
(−x4 )n =
(−1)n x4n
4
4
1+x
1 − (−x )
n≥0
n≥0
with R = 1. Now,
4n+1 1/3 X
Z 1/3
Z 1/3 X
X
(−1)n
1
n 4n
n x
dx
=
(−1)
x
=
(−1)
=
1 + x4
4n + 1 0
34n+1 (4n + 1)
0
0
n≥0
n≥0
n≥0
1
34n+1 (4n+1)
This is an alternating series with cn =
which is non-increasing and limits to 0. Hence,
the alternating series theorem applies. So, |Rn | ≤ cn+1 < 10−4 if an only if
1
< 10−4 .
34(n+1)+1 (4(n + 1) + 1)
Suppose, that n + 1 = 2, then c2 =
1
X
n=0
1
39 ·9
< 10−4 and hence, a valid approximation here is
(−1)n
1
1
= −
+ 1)
3 1215
34n+1 (4n
23
(4)
Z
arcsin t =
√
1
dt
1 − t2
Z
(1 + (−t2 ))−1/2 dt
Z
∞ X
−1/2
=
=
n=0
n
!
2 n
(−t )
dt

Z
=

X 1 · 3 · 5 · · · (2n − 1)t2n
1 +
 dt
2n · n!
n≥0
X 1 · 3 · 5 · · · (2n − 1)
=t+
t2n+1
2n (2n + 1)n!
n≥0
24
Tutorial 8
(1) Determine if the improper integral
Z
∞
0
xex
dx
2x + 1
converges.
(2) Calculate the length of the curve y = sin x on [0, π/2] using the midpoint rule within 0.01 units of
accuracy.
Solutions:
xex
x
(1) Observe that 2x+1
≥ 2x+1
on [0, ∞] and with the substitution u = 2x + 1 so that x = u−1
and
2
du = 2dx. Note, when x = 0, t that u = 1, 2t + 1
Z x=t u−1 1 Z ∞
x
2
2 du
dx = lim
t→∞ x=0
2x + 1
u
0
Z 2t+1
u−1
= lim
du
t→∞ 1
4u
Z s
1
u−1
= lim
du
s→∞ 4 1
u
s
1
1
u+ 2
= lim
s→∞ 4
u 1
1
1
= lim
s−1+ 2 −1
s→∞ 4
s
=∞
So this integral diverges and, by comparison test, also the original integral of interest must diverge.
(2) The length formula says that we must compute
Z π/2 p
L=
1 + cos2 xdx.
0
The error for the midpoint formula is given by
|E(Mn )| ≤
where
k(b − a)3
24n2
2 p
d
2
k = max 2 1 + cos x = 2.
[0,π/2] dx
|
{z
}
2 x−cos4 x
(1+cos2 x)3/2
= − cos
So,
2( π2 )3
24n2
1 π 3 1
=
12 3
n2
|E(Mn )| ≤
and this is < 0.01 whenever, n2 > 32.2 which is, whenever n > 5.6. So, when n = 6 we should have
π
our desired approximation. Now, n = 6 ∆x = π2 · n1 = 12
and
Z π/2 p
1 + cos2 xdx ≈ M6 = ∆x (f (x1 ) + f (x2 ) + · · · + f (x6 )) = 1.804142
0
25