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