c z c z c z c - · pdf filedx = 2x2 1 3 x3 4 0 = 32 64 3 (0) = 32 3 (b) z x=6 x=0 2x x2 4x...
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MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 1
5.1 Calculate the following indefinite integrals:
(a)
∫x8 dx (b)
∫ (3x8 − 5
x6
)dx (c)
∫ (5x6 − 2x5 − 8x+ 5
)dx
(d)
∫1
x8dx (e)
∫(ex + sinx+ cosx) dx (f)
∫ (1 +
1
x+
1
x2
)dx
Outline Solution
(a)
∫x8 dx =
x9
9+ C
(d)
∫ (3x8 − 5
x6
)dx =
1
3x9 +
1
x5+ C
(c)
∫ (5x6 − 2x5 − 8x+ 5
)dx =
5
7x7 − 1
3x6 − 4x2 + 5x+ C
(d)
∫1
x8dx = − 1
7x7+ C
(e)
∫(ex + sinx+ cosx) dx = ex − cos(x) + sinx+ C
(f)
∫ (1 +
1
x+
1
x2
)dx = x+ lnx− 1
x+ C
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 2
5.2 Calculate the following definite integrals:
(a)
∫ 1
0t7 dt (b)
∫ 2
1
1
x7dx (c)
∫ 1
−1t(1− t2
)dt
(d)
∫ −1
−2
(24w7 +
12
w4
)dw (c)
∫ 2
1
(x+
1
x
)2
dx (f)
∫ 4
1
3x− 2√x
xdx
Outline Solution
(a)
∫ 1
0t7 dt =
[t8
8
]10
=1
8
(b)
∫ 2
1
1
x7dx =
[− 1
6x6
]21
= − 1
384+
1
6=
63
384
(c)
∫ 1
−1t(1− t2
)dt =
∫ 1
−1
(t− 2t2 + t3
)dt =
[t2
2− 2t3
3+t4
4
]1−1
=
(1
2− 2
3+
1
4
)−(
1
2+
2
3+
1
4
)=
−4
3
(d)
∫ −1
−2
(24w7 +
12
w4
)dw =
[3w8 − 4
w3
]−1
−2
= 3 + 4− (768 +1
2) = −761.5
(e)
∫ 2
1
(x+
1
x
)2
dx =
∫ 2
1
(x2 + 2 +
1
x2
)dx =
[x3
3+ 2x− 1
x
]21
=
(8
3+ 4− 1
2
)−(
1
3+ 2− 1
)=
10
3
(f)
∫ 4
1
3x− 2√x
xdx =
∫ 4
1
(3− 2x−
12
)dx =
[3x− 2 · 2x 1
2
]41
=[3x− 4
√x]41
= 4− (−1) = 5
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 3
5.3 Use integration by substitution to calculate each of the following integrals:
(a)
∫1
x lnxdx (b)
∫x6e5x
7−2 dx (c)
∫ √√x+ 4√x
dx
(d)
∫ 4
1
x
3x2 + 1dx (e)
∫ 2
1x√x− 1 dx (f)
∫ 4
0
x dx√1 + 2x
Outline Solution
(a) Substitution:
u = lnx⇒ du
dx=
1
x⇒ du =
1
xdx
so that ∫1
x lnxdx =
∫1
lnx· dxx
=
∫1
udu
= lnu+ C
= ln(lnx) + C
(b) Substitution:
u = 5x7 − 2 ⇒ du
dx= 35x6 ⇒ du = 35x6dx ⇒ 1
35du = x6dx
so that ∫x6e5x
7−2 dx =
∫e5x
7−2(x6 dx
)=
∫eu(
1
35du
)=
1
35
∫eu du
=1
35eu + C
=1
35e5x
7−2 + C
(c) Substitution:
u =√x+ 4⇒ du
dx=
1
2x−
12 ⇒ 2du =
1√xdx
so that ∫ √√x+ 4√x
dx
=
∫ √u · 2 du
= 2
∫u
12 du
= 2
(2
3u
32
)=
4
3
(√x+ 4
) 32
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 4
(d) Substitution:
u = 3x2 + 1 ⇒ du
dx= 6x ⇒ du = 6xdx ⇒ 1
6du = x dx
Change the limits of integration:
x = 1 ⇒ u = 3x2 + 1 = 4 ⇒ x = 4 ⇒ u = 3x2 + 1 = 49
so that ∫ 4
1
x
3x2 + 1dx =
∫ 4
1
1
3x2 + 1(x dx)
=
∫ 49
4
1
u
(1
6du
)=
1
6
∫ 49
4
1
udu
=1
6[lnu]494
=1
6(ln 49− ln 4)
(e)
∫ 2
1x√x− 1 dx:
Let u = x− 1 ⇒ 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 ∫ 2
1x√x− 1 dx =
∫ 1
0(u+ 1)
√u du
=
∫ 1
0
(u
32 + u
12
)du
=
[2
5u
52 +
2
3u
32
]10
=2
5+
2
3
=16
15
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 5
(f)
∫ 4
0
x dx√1 + 2x
:
Let u = 1 + 2x ⇒ du = 2 dx ⇒ dx =1
2du and x =
1
2(u− 1)
Change the limits of integration:
x = 0⇒ u = 1 + 2x = 1 x = 4⇒ u = 1 + 2x = 9
so that ∫ 4
0
x dx√1 + 2x
=
∫ 9
1
12(u− 1)√
u· 1
2du
=1
4
∫ 9
1
(u
12 − u− 1
2
)du
=1
4
[2
3u
32 − 2u
12
]91
=1
6
[u
32 − 3u
12
]91
=1
6[(27− 9)− (1− 3)]
=10
3
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 6
5.4 Use integration by parts to calculate each of the following integrals:
(a)
∫x lnx dx (b)
∫x e2x dx (c)
∫x cos 5x dx
(d)
∫ 3
1r3 ln r dr (e)
∫ 1
0
(x2 + 1
)e−x dx (f)
∫ 1
0tan−1 x dx
Outline Solution
(a) Integration by parts:
u = ln(x) ⇒ du =1
xdx
dv = x dx ⇒ v =1
2x2
and the formula yields∫u dv =
∫ln(x) · x dx = uv −
∫v du
= ln(x) · 1
2x2 −
∫1
2x2 · 1
xdx
=1
2x2 ln(x)− 1
2
∫x dx
=1
2x2 ln(x)− 1
4x2 + C
=1
2x2[ln(x)− 1
2
]+ C
(b) Integration by parts:
u = x ⇒ du = dx
dv = e2x dx ⇒ v =1
2e2x
and the formula yields∫u dv =
∫x · e2x dx = uv −
∫v du
= x
(1
2e2x)−∫
1
2e2xdx
=1
2x e2x − 1
4e2x + C
(c)
∫x cos 5x dx: Integration by parts:
u = x ⇒ du = dx
dv = cos 5x dx ⇒ v =1
5sin 5x
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 7
and the formula yields∫u dv =
∫x cos 5x dx = uv −
∫v du
=1
5x sin 5x−
∫1
5sin 5x dx
=1
5x sin 5x+
1
25cos 5x+ C
(d)
∫ 3
1r3 ln r dr: Integration by parts:
u = ln r ⇒ du =1
rdr
dv = r3 dr ⇒ v =1
4r4
and the formula yields∫u dv =
∫ 3
1r3 ln r dr = uv −
∫v du
=
[1
4r4 ln r
]31
−∫ 3
1
1
4r3 dr
=81
4ln 3− 0− 1
4
[1
4r4]31
=81
4ln 3− 1
16(81− 1)
=81
4ln 3− 5
(e)
∫ 1
0
(x2 + 1
)e−x dx Integration by parts:
u = x2 + 1 ⇒ du = 2x
dv = e−x ⇒ v = −e−x
and the formula yields∫u dv =
∫ 1
0
(x2 + 1
)e−x dx = uv −
∫v du
=[−(x2 + 1
)e−x]10
+
∫ 1
02x e−x dx
= −2e−1 + 1 + 2
∫ 1
0x e−x dx
Separately, we apply integration by parts to
∫ 1
0x e−x dx to obtain∫ 1
0x e−x dx =
[−x e−x
]10−∫ 1
0e−x dx = −e−1 +
[−e−x
]10
= −e−1 −−e−1 + 1 = −2e−1 + 1
Combining both results, we obtain:∫ 1
0
(x2 + 1
)e−x dx = −2e−1 + 1 + 2
∫ 1
0x e−x dx = −2e−1 + 1 + 2
[−2e−1 + 1
]= −6e−1 + 3
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 8
(f)
∫ 1
0tan−1 x dx: Integration by parts:
u = tan−1 x ⇒ du =dx
1 + x2
dv = dx ⇒ v = x
and the formula yields∫u dv =
∫ 1
0tan−1 x dx = uv −
∫v du
=[1 · tan−1 x
]10−∫ 1
0
x
1 + x2dx
= 1 · tan−1 1− 0 · tan−1 0−∫ 1
0
x
1 + x2dx
=π
4−∫ 1
0
x
1 + x2dx
To evaluate this integral, we use substitution:
v = 1 + x2 ⇒ dv = 2x dx ⇒ x dx =1
2du
and change the limits of integration:
x = 0⇒ v = 1 + x2 = 1 and x = 1⇒ v = 1 + x2 = 2
to obtain ∫ 1
0
x
1 + x2dx =
1
2
∫ 2
1
1
vdv =
[1
2ln v
]21
=1
2ln 2− 1
2ln1 =
1
2ln 2
Combining both results, we obtain∫ 1
0tan−1 x dx =
π
4−∫ 1
0
x
1 + x2dx =
π
4− 1
2ln 2
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 9
5.5 Find the area of the shaded regions:
(a) (b)
(c) (d)
Outline Solution
(a) ∫ x=4
x=0(yT − yB) dx =
∫ 4
0
[(5x− x2
)− x]dx
=
∫ 4
0
(4x− x2
)dx
=
[2x2 − 1
3x3]40
=
(32− 64
3
)− (0) =
32
3(b) ∫ x=6
x=0
[2x−
(x2 − 4x
)]dx =
∫ 6
0
(6x− x2
)dx
=
[3x2 − 1
3x3]60
= (108− 72)− (0) = 36(c) ∫ y=1
y=0(xR − xL) dy =
∫ 1
0
[√y −
(y2 − 1
)]dy
=
∫ 1
0
(√y − y2 + 1
)dy
=
[2
3y
32 − 1
3y3 + y
]10
=
(2
3− 1
3+ 1
)− (0) =
4
3
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 10
(d) ∫ y=3
y=0
[(2y − y2
)−(y2 − 4y
)]dy =
∫ 3
0
(−2y2 + 6y
)dy
=
[−2
3y3 + 3y2
]20
= (−18 + 27)− (0) = 9
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 11
5.6 In each of the following, sketch the region enclosed by the given curves, decide whether to integratewith respect to x or y and then find the area of the region:
(a) y = x, y = x2 (b) 4x+ y2 = 12, x = y
(c) y = x2 − x− 2, y = 4− x2, x ∈ [−2, 3] (d) y = cos(x), y = 0, x ∈[0,
3π
2
](e) y = x(x− 2)(x− 4), y = 0, x ∈ [0, 4] (f) 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:
A =
∫ 1
0
(x− x2
)dx =
[x2
2− x3
3
]10
=1
2− 1
3=
1
6
(b) 4x+ y2 = 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
A =
∫ 2
−6
[(−1
4y2 + 3
)− y]dy =
∫ 2
−6
(−1
4y2 − y + 3
)dy
=
[− 1
12y3 − 1
2y2 + 3y
]2−6
=
(−2
3− 2 + 6
)− (18− 18− 18) =
64
3
(c) The solutions of the equation f(x) = g(x), i.e. x2 − x− 2 = 4− x2 are x = −32 and x = 2,
so these are the two points of intersection on the graph.
−3 −2 −1 1 2 3
−4
−2
2
4
6
x
yf(x) = x2 − x− 2
g(x) = 4− x2
Notice that f is above g on[−2,−3
2
], g is above f on
[−3
2 , 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
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 12
area between the graphs (without requiring the absolute value formulation):∫ 3
−2|f(x)− g(x)| dx
=
∫ − 32
−2[f(x)− g(x)] dx+
∫ 2
− 32
[g(x)− f(x)] dx+
∫ 3
2[f(x)− g(x)] dx
=
∫ − 32
−2
(2x2 − x− 6
)dx+
∫ 2
− 32
−(2x2 − x− 6
)dx+
∫ 3
2
(2x2 − x− 6
)dx
=
[2
3x3 − 1
2x2 − 6x
]− 32
−2
+
[−(
2
3x3 − 1
2x2 − 6x
)]2− 3
2
+
[2
3x3 − 1
2x2 − 6x
]32
=23
24+
343
24+
100
24=
466
24
(d) y = cos(x), y = 0, x ∈[0,
3π
2
]:
A =
∫ 3π2
0|cos(x)| dx
=
∫ π2
0cos(x) dx+
∫ 3π2
π2
[− cos(x)] dx
= sin(x)|π20 − sin(x)|
3π2π2
= (1− 0)− (−1− 1) = 3
(e) The curve intersects the x-axis at x = 0, x = 2 and x = 4.
1 2 3 4
−4
−2
2
4f(x) = x3 − 6x2 + 8x
The total area is the sum of the two areas A1 and A2 where
A1 =
∫ 2
0
(x3 − 6x2 + 8x
)dx
A2 =
∣∣∣∣∫ 4
2
(x3 − 6x2 + 8x
)dx
∣∣∣∣Area A2 is on the negative side of the x-axis and, without taking absolute values, the definiteintegral will be a negative number and that is why, in such cases, we regard the area as the
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 13
absolute value of the definite integral. We therefore obtain
A1 =
∫ 2
0
(x3 − 6x2 + 8x
)dx
=
[x4
4− 2x3 + 4x2
]20
= (4− 16 + 16)− (0)
= 4
Also
A2 =
∣∣∣∣∫ 4
2
(x3 − 6x2 + 8x
)dx
∣∣∣∣=
∣∣∣∣∣[x4
4− 2x3 + 4x2
]42
∣∣∣∣∣= |(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 andcosine cross at x = π
4 and x = 5π4 Therefore,
A =
∫ π4
0[cos(x)− sin(x)] dx+
∫ 5π4
π4
[sin(x)− cos(x)] dx+
∫ 2π
5π4
[cos(x)− sin(x)] dx
= [sin(x) + cos(x)]π40 + [− (cos(x) + sin(x))]
5π4π4
+ [(sin(x) + cos(x))]2π5π4
=(√
2− 1)
+(√
2 +√
2)
+(
1 +√
2)
= 4√
2
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 14
5.7 Find the area of the plane region lying to the right of the parabola x = y2− 12 and to the left ofthe straight line y = x.
Outline Solution
Find the intersections of the curves x = y2 − 12 and y = x:
y2 − 12 = x = y ⇒ y2 − y − 12 = 0⇒ (y − 4)(y + 3) = 0⇒ y = 4 or y = −3
Noting that y2 − 12 ≤ y for −3 ≤ y ≤ 4, the area is
A =
∫ 4
−3
[y −
(y2 − 12
)]dy
=
∫ 4
−3
(y − y2 + 12
)dy
=
[y2
2− y3
3+ 12y
]43
=343
6
Note that the same result could have been obtained by integrating in the x-direction but theintegral would have been more complicated:
A =
∫ −3
−12
[√12 + x−
(−√
12 + x)]dx+
∫ 4
−3
[√12 + x− x
]dx
= 2
∫ −3
−12
[√12 + x
]dx+
∫ 4
−3
[√12 + x− x
]dx
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 5 15
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
A =
∫ 3
0
(3x− x2
)dx
=
(3x2
2− x3
3
)3
0
=
(27
2− 27
3
)− (0− 0) =
9
2
(b) The points of intersection of y = 1 and y = 5x2+1
are obtained as follows:
1 =5
x2 + 1⇒ x = ±2
Hence, the area is given by
R =
∫ 2
−2
5
x2 + 1dx− 4
= 2
∫ 2
0
5
x2 + 1dx− 4 =
(10 tan−1 x
)20− 4 = 10 tan−1 2− 4
JC MS121: IT Mathematics Semester 2 Tutorial Sheet 5 28/01/15