1
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
5.1 Calculate the following indefinite integrals:
Z
(a)
x dx
Z
(d)
8
1
dx
x8
Z (b)
Z
(e)
5
3x − 6
x
8
Z
dx
(ex + sin x + cos x) dx
(c)
(f)
Outline Solution
x9
x8 dx =
+C
9
Z 5
1
1
8
3x − 6 dx = x9 + 5 + C
(d)
x
3
x
Z
1
5
(c)
5x6 − 2x5 − 8x + 5 dx = x7 − x6 − 4x2 + 5x + C
7
3
Z
1
1
(d)
dx = − 7 + C
8
x
7x
Z
(e)
(ex + sin x + cos x) dx = ex − cos(x) + sin x + C
Z 1
1
1
(f)
1 + + 2 dx = x + ln x − + C
x x
x
Z
(a)
5x6 − 2x5 − 8x + 5 dx
Z 1
1
1 + + 2 dx
x x
2
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
5.2 Calculate the following definite integrals:
Z
1
t dt
Z0 −1 12
7
24w + 4 dw
w
−2
(a)
(d)
(b)
(c)
2
1
dx
7
x
Z1 2 1 2
x+
dx
x
1
Z
7
Z
(c)
(f)
1
t 1 − t2 dt
√
Z−1
4
3x − 2 x
dx
x
1
Outline Solution
1
t8
1
t dt =
=
8 0 8
0
Z 2
1
1 2
1
1
63
dx = − 6 = −
+ =
7
6x 1
384 6
384
1 x
2
1
Z 1
Z 1
t
2t3 t4
1 2 1
1 2 1
2
2
3
t 1 − t dt =
t − 2t + t dt =
−
+
=
− +
−
+ +
=
2
3
4 −1
2 3 4
2 3 4
−1
−1
4
−
3
Z −1 1
12
4 −1
7
8
= 3 + 4 − (768 + ) = −761.5
24w + 4 dw = 3w − 3
w
w −2
2
−2
2
3
2 Z 2
Z 2
1
1
x
8
1
1
1
2
x+
x + 2 + 2 dx =
=
−
dx =
+ 2x −
+4−
+2−1 =
x
x
3
x 1
3
2
3
1
1
10
3
√
Z 4
Z 4
h
i
√ 4
1
1 4
3x − 2 x
3 − 2x− 2 dx = 3x − 2 · 2x 2 = 3x − 4 x 1 = 4 − (−1) = 5
dx =
x
1
1
1
Z
(a)
(b)
(c)
(d)
(e)
(f)
1
7
3
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
5.3 Use integration by substitution to calculate each of the following integrals:
Z
Z
Z p√
1
x+4
6 5x7 −2
√
(a)
dx
(b)
x e
dx
(c)
dx
x
Z 4x ln x
Z 2
Z 4
√
x
x dx
√
x x − 1 dx
(d)
dx
(e)
(f)
2
1 + 2x
1 3x + 1
1
0
Outline Solution
(a) Substitution:
u = ln x ⇒
du
1
1
= ⇒ du = dx
dx
x
x
so that
Z
1
dx =
x ln x
1 dx
·
ln x x
Z
1
du
=
u
= ln u + C
Z
= ln(ln x) + C
(b) Substitution:
u = 5x7 − 2
du
= 35x6
dx
⇒
⇒
du = 35x6 dx
⇒
so that
Z
6 5x7 −2
x e
(c) Substitution:
u=
√
Z
x+4⇒
so that
=
=
=
=
7
e5x −2 x6 dx
Z
1
u
=
e
du
35
Z
1
=
eu du
35
1 u
e +C
=
35
1 5x7 −2
=
e
+C
35
dx =
du
1 1
1
= x− 2 ⇒ 2du = √ dx
dx
2
x
Z p√
x+4
√
dx
x
Z
√
u · 2 du
Z
1
2 u 2 du
2 3
2
u
2
3
3
4 √
x+4 2
3
1
du = x6 dx
35
4
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
(d) Substitution:
u = 3x2 + 1
du
= 6x
dx
⇒
⇒
⇒
du = 6xdx
1
du = x dx
6
Change the limits of integration:
x=1
⇒
u = 3x2 + 1 = 4
⇒
x=4
u = 3x2 + 1 = 49
⇒
so that
4
Z
1
x
dx =
2
3x + 1
=
=
=
=
Z
(e)
2
x
√
1
x − 1 dx:
Let u = x − 1
⇒
4
1
(x dx)
+1
1
Z 49 1 1
du
u 6
4
Z
1 49 1
du
6 4 u
1
[ln u]49
4
6
1
(ln 49 − ln 4)
6
Z
3x2
x = u + 1 and du = dx
Change the limits of integration:
x=1⇒u=x−1=0
x=2⇒u=x−1=1
so that
Z
2
x
1
√
x − 1 dx =
Z
1
√
(u + 1) u du
0
Z
=
1
3
1
u2 + u2
du
0
=
=
=
2 5 2 3
u2 + u2
5
3
2 2
+
5 3
16
15
1
0
5
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
Z
(f)
0
4
√
x dx
:
1 + 2x
⇒
Let u = 1 + 2x
du = 2 dx
⇒
dx =
1
1
du and x = (u − 1)
2
2
Change the limits of integration:
x = 0 ⇒ u = 1 + 2x = 1
x = 4 ⇒ u = 1 + 2x = 9
so that
Z
0
4
√
x dx
1 + 2x
=
=
=
=
=
9 1 (u −
2 √
1) 1
· du
2
u
1
Z 9
1
1
1
u 2 − u− 2 du
4 1
9
1
1 2 3
u2 − 2 u2
4 3
1
i
1 9
1h 3
u2 − 3 u2
6
1
1
[(27 − 9) − (1 − 3)]
6
10
3
Z
=
6
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
5.4 Use integration by parts to calculate each of the following integrals:
Z
(a)
Z
x ln x dx
Z
(d)
3
1
Z
r3 ln r dr
Z
x e2x dx
(b)
(e)
(c)
x2 + 1 e−x dx
1
(f)
tan−1 x dx
0
0
1
x cos 5x dx
Z
Outline Solution
(a) Integration by parts:
u = ln(x)
⇒
dv = x dx
⇒
du =
1
dx
x
1
v = x2
2
and the formula yields
Z
Z
Z
u dv = ln(x) · x dx = uv − v du
=
=
=
=
Z
1 2
1 2 1
ln(x) · x −
x · dx
2
2
x
Z
1
1 2
x ln(x) −
x dx
2
2
1 2
1
x ln(x) − x2 + C
2 4 1 2
1
x ln(x) −
+C
2
2
(b) Integration by parts:
⇒
u=x
dv = e2x dx
and the formula yields
Z
Z
u dv =
du = dx
1
v = e2x
2
⇒
2x
x·e
Z
dx = uv − v du
Z
1 2x
1 2x
= x
e
−
e dx
2
2
1 2x 1 2x
=
xe − e + C
2
4
Z
(c)
x cos 5x dx: Integration by parts:
u=x
dv = cos 5x dx
⇒
⇒
du = dx
1
v = sin 5x
5
7
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
and the formula yields
Z
Z
Z
u dv = x cos 5x dx = uv − v du
Z
1
1
x sin 5x −
sin 5x dx
5
5
1
1
x sin 5x +
cos 5x + C
5
25
=
=
3
Z
r3 ln r dr: Integration by parts:
(d)
1
u = ln r
⇒
dv = r3 dr
⇒
and the formula yields
Z
Z
u dv =
3
1
1
Z
(e)
du =
1
dr
r
1
v = r4
4
Z
r3 ln r dr = uv − v du
3 Z 3
1 3
1 4
=
r ln r −
r dr
4
1 4
1
1 1 4 3
81
ln 3 − 0 −
r
=
4
4 4
1
81
1
=
ln 3 − (81 − 1)
4
16
81
ln 3 − 5
=
4
x2 + 1 e−x dx Integration by parts:
0
u = x2 + 1
dv = e
and the formula yields
Z
Z
u dv =
1
2
⇒
−x
du = 2x
v = −e−x
⇒
−x
dx = uv −
x +1 e
0
Z
v du
Z 1
−x 1
2
= − x +1 e
+
2x e−x dx
0
0
Z 1
= −2e−1 + 1 + 2
x e−x dx
0
Z
Separately, we apply integration by parts to
1
x e−x dx to obtain
0
Z
1
−x
xe
0
dx =
1
−x e−x 0
−
Z
0
1
1
e−x dx = −e−1 + −e−x 0 = −e−1 − −e−1 + 1 = −2e−1 + 1
Combining both results, we obtain:
Z 1
Z
−x
2
−1
x + 1 e dx = −2e + 1 + 2
0
0
1
x e−x dx = −2e−1 + 1 + 2 −2e−1 + 1 = −6e−1 + 3
8
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
Z
(f)
1
tan−1 x dx: Integration by parts:
0
u = tan−1 x
⇒
du =
dv = dx
⇒
v=x
and the formula yields
Z
Z
u dv =
1
−1
tan
0
x dx = uv −
dx
1 + x2
Z
v du
1
x
dx
2
0 1+x
Z 1
x
= 1 · tan−1 1 − 0 · tan−1 0 −
dx
2
0 1+x
Z 1
π
x
=
−
dx
2
4
0 1+x
=
−1
1 · tan
x
1
−
0
Z
To evaluate this integral, we use substitution:
v = 1 + x2
⇒
dv = 2x dx
⇒
x dx =
1
du
2
and change the limits of integration:
x = 0 ⇒ v = 1 + x2 = 1
to obtain
Z
0
1
x
1
dx =
2
1+x
2
Z
1
2
and
x = 1 ⇒ v = 1 + x2 = 2
2
1
1
1
1
1
dv =
ln v = ln 2 − ln1 = ln 2
v
2
2
2
2
1
Combining both results, we obtain
Z 1
Z 1
π
x
π 1
tan−1 x dx = −
dx = − ln 2
2
4
1
+
x
4 2
0
0
9
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
5.5 Find the area of the shaded regions:
(a)
(b)
(c)
(d)
Outline Solution
(a)
Z
x=4
x=0
(yT − yB ) dx =
4
Z
0
5x − x2 − x dx
4
Z
4x − x2 dx
0
1 3 4
2
= 2x − x
3
0
32
64
− (0) =
=
32 −
3
3
=
(b)
Z
x=6 x=0
2
6x − x2 dx
0
1 3 6
2
= 3x − x
3
0
= (108 − 72) − (0) = 36
(c)
Z
y=1
y=0
6
Z
2x − x − 4x dx =
(xR − xL ) dy =
Z
1
√
0
Z
1
y − y2 − 1
√
dy
y − y 2 + 1 dy
0
1
2 3 1 3
2
=
y − y +y
3
3
0
2 1
4
=
− + 1 − (0) =
3 3
3
=
10
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
(d)
Z
y=3 y=0
2y − y
2
2
− y − 4y
Z
dy =
0
3
−2y 2 + 6y dy
2
2 3
2
= − y + 3y
3
0
= (−18 + 27) − (0) = 9
11
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
5.6 In each of the following, sketch the region enclosed by the given curves, decide whether to integrate
with respect to x or y and then find the area of the region:
(a)
y = x, y = x2
2
(b)
4x + y 2 = 12, x = y
2
(c)
y = x − x − 2, y = 4 − x , x ∈ [−2, 3]
(d)
(e)
y = x(x − 2)(x − 4), y = 0, x ∈ [0, 4]
(f)
3π
y = cos(x), y = 0, x ∈ 0,
2
y = sin(x), y = cos(x), x ∈ [0, 2π]
Outline Solution
(a) y = x, y = x2 : The curves intersect when x = x2 , i.e. when x(x − 1) = 0 ⇒ x = 0 or x = 1:
Z
1
x2 x3
−
2
3 0
1 1
1
− =
2 3
6
1
x−x
A=
0
2
dx =
=
(b) 4x + y 2 = 12, x = y: The curves intersect when 4x + x2 = 12, i.e.
x2 − 4x − 12 = 0 ⇒ (x + 6)(x − 2) = 0 ⇒ x = −6 or x = 2
With x = −6 or x = 2, we have correspondingly y = −6 or y = 2 and
Z 2
1
1
− y 2 + 3 − y dy =
− y 2 − y + 3 dy
4
4
−6
−6
2
1 3 1 2
= − y − y + 3y
12
2
−6
2
64
=
− − 2 + 6 − (18 − 18 − 18) =
3
3
Z
A=
2
(c) The solutions of the equation f (x) = g(x), i.e. x2 − x − 2 = 4 − x2 are x = − 23 and x = 2,
so these are the two points of intersection on the graph.
6
4
y
f (x) = x2 − x − 2
g(x) = 4 − x2
2
x
−3
−2
−1
1
2
3
−2
−4
Notice that f is above g on −2, − 32 , g is above f on − 23 , 2 and f is again above g on
[2, 3]. On each of these smaller intervals, we can use the definite integral to calculate the
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MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
area between the graphs (without requiring the absolute value formulation):
Z
=
3
|f (x) − g(x)| dx
−2
Z −3
2
−2
Z
=
− 32
−2
=
[f (x) − g(x)] dx +
− 32
2
2x − x − 6 dx +
2 3 1 2
x − x − 6x
3
2
− 3
2
−2
2
Z
Z
[g(x) − f (x)] dx +
3
Z
2
2
2
− 23
− 2x − x − 6 dx +
[f (x) − g(x)] dx
Z
3
2
2x2 − x − 6 dx
2
3
2 3 1 2
2 3 1 2
+ −
x − x − 6x
+ x − x − 6x
3
2
3
2
−3
2
2
466
23 343 100
+
+
=
=
24
24
24
24
3π
(d) y = cos(x), y = 0, x ∈ 0,
:
2
Z
A =
3π
2
0
Z
=
π
2
|cos(x)| dx
0
π
=
3π
2
Z
cos(x) dx +
π
2
3π
2
π
2
[− cos(x)] dx
sin(x)|02 − sin(x)|
= (1 − 0) − (−1 − 1) = 3
(e) The curve intersects the x-axis at x = 0, x = 2 and x = 4.
4
f (x) = x3 − 6x2 + 8x
2
1
2
3
4
−2
−4
The total area is the sum of the two areas A1 and A2 where
Z 2
A1 =
x3 − 6x2 + 8x dx
Z0 4
3
2
A2 = x − 6x + 8x dx
2
Area A2 is on the negative side of the x-axis and, without taking absolute values, the definite
integral will be a negative number and that is why, in such cases, we regard the area as the
13
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
absolute value of the definite integral. We therefore obtain
Z 2
A1 =
x3 − 6x2 + 8x dx
0
2
x4
3
2
=
− 2x + 4x
4
0
= (4 − 16 + 16) − (0)
= 4
Also
Z 4
A2 = x3 − 6x2 + 8x dx
2
4 x4
− 2x3 + 4x2 = 4
2
= |(64 − 128 + 64) − (4 − 16 + 16)|
= |−4| = 4
Therefore, the required area is
A = A1 + A2 = 4 + 4 = 8
(f) y = sin(x), y = cos(x), x ∈ [0, 2π]:
The region A is shaded in the diagram shown. Between 0 and 2π, the graphs of sine and
cosine cross at x = π4 and x = 5π
4 Therefore,
Z
A =
0
π
4
[cos(x) − sin(x)] dx +
π
5π
4
Z
π
4
[sin(x) − cos(x)] dx +
5π
Z
2π
5π
4
[cos(x) − sin(x)] dx
= [sin(x) + cos(x)]04 + [− (cos(x) + sin(x))] π4 + [(sin(x) + cos(x))]2π
5π
4
4
√
√
√ √ √
=
2−1 +
2+ 2 + 1+ 2 =4 2
14
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
5.7 Find the area of the plane region lying to the right of the parabola x = y 2 − 12 and to the left of
the straight line y = x.
Outline Solution
Find the intersections of the curves x = y 2 − 12 and y = x:
y 2 − 12 = x = y ⇒ y 2 − y − 12 = 0 ⇒ (y − 4)(y + 3) = 0 ⇒ y = 4 or y = −3
Noting that y 2 − 12 ≤ y for −3 ≤ y ≤ 4, the area is
Z 4
A =
y − y 2 − 12 dy
−3
4
Z
=
−3
y2
y − y 2 + 12 dy
y3
−
+ 12y
2
3
343
6
=
=
4
3
Note that the same result could have been obtained by integrating in the x-direction but the
integral would have been more complicated:
Z −3
Z 4
√
√
√
A =
12 + x − − 12 + x dx +
12 + x − x dx
= 2
−12
Z −3
√
−12
−3
Z
4
12 + x dx +
−3
√
12 + x − x dx
15
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5
5.8 Find the area of the plane region lying:
(a) above the x-axis and under the curve y = 3x − x2 ;
(b) above the line y = 1 and under the curve y =
5
.
x2 +1
Outline Solution
(a) We need to find the points where the curve y = 3x − x2 meets the x-axis:
0 = 3x − x2
⇒
x = 0, x = 3
Hence, the area is given by
3
Z
A =
0
=
3x − x2 dx
3x2 x3
−
2
3
3
=
0
(b) The points of intersection of y = 1 and y =
1=
5
x2 + 1
27 27
−
2
3
5
x2 +1
⇒
− (0 − 0) =
9
2
are obtained as follows:
x = ±2
Hence, the area is given by
Z 2
5
R =
dx − 4
2
−2 x + 1
Z 2
2
5
−1
= 2
dx
−
4
=
10
tan
x
− 4 = 10 tan−1 2 − 4
0
2+1
x
0
JC
MS121: IT Mathematics
Semester 2 Tutorial Sheet 5
28/01/15