MAT4MDS Assignment 4: Calculus Integration, PDF, Solutions, 2019

This document presents the complete solutions to a MAT4MDS assignment, focusing on calculus and probability. The assignment comprises three main questions. The first question involves evaluating definite integrals using various techniques, including algebraic manipulation and substitution. The second question delves into probability density functions (PDFs) and cumulative distribution functions (CDFs), requiring the calculation of the CDF and the mean of a given distribution through integration. The final question explores numerical integration using the trapezoidal rule, with graphical representations and applications to real-world scenarios like bio-fuel production. The solutions demonstrate step-by-step working and explanations, providing a valuable resource for students studying calculus and related topics. The document covers a range of integration techniques and applications, including definite integrals, probability density functions, cumulative distribution functions, and numerical integration methods like the trapezoidal rule. The document provides detailed solutions to each part of the questions, showing all the required steps and working to obtain the final answers. The solutions are well-organized and easy to follow, making them a valuable resource for students studying calculus.

Q 1.
Answer
(a)
I =
Z 3
−3
2x(x + 3) dx
Multiplying and expanding:
I =
Z 3
−3
2x2 + 6x dx
[ Standard Integral:
Z
xn dx = xn+1
n + 1]
∴ I = 2 ·x3
3 + 6 ·x2
2
3
−3
= 2
3(33 − (−3)3) + 3(32 − (−32)) = 36 + 0
= 36
(b)
I =
Z e−3
−2
2x
x + 3dx
Substitute:x + 3 = t=⇒ dx = dt
Evaluate the indefinite integral, that is integral without limits first.
Therefore,

I indef = 2
Z t − 3
t dt
= 2
Z
1 −3
t dt
[Standard Integral:
Z
1 dx = x &
Z 1
x dx = ln x]
= 2 (t − 3 ln t)
[re-substitute: t = x + 3 ]
= 2 x + 3 − 3 ln (x + 3)
Finally, re-introduce the limits to evaluate the definite integral I:
I =
h
2 x + 3 − 3 ln (x + 3)
i e−3
−2
= 2 · [(e − 3 + 3) − (−2 + 3)] − 6[ln (e − 3 + 3) − ln (−2 + 3)]
= 2(e − 1) − 6
= 2e − 8
(c)
I =
Z ∞
1
1
x3 e1
x dx
Substitute: 1
x = t =⇒ − 1
x2 dx = dt

∴ 1
x3 e1
x dx = 1
x · 1
x2 dx · e
1
x = t · (−dt) · et
Evaluate the indefinite integral first.
I indef = −
Z
tet dt
[Let, u = t & v0 = et. Integration by Parts:
Z
uv0 = uv −
Z
vdu]
= −(tet −
Z
et dt)
= et − tet
[re-substitute:t = 1
x]
= e1
x − e1
x
x
Finally, re-introduce the limits to evaluate the definite integral I:
I =
h
e1
x − e1
x
x
i ∞
1
= lim
x→∞ e1
x − e1
x
x
!
− (e1 − e1
1 )
[ lim
x→∞ e1
x = 1 & lim
x→∞
1
xe1
x = 0]
∴ I = (1 − 0) − (e − e) = 1

Q 2.
Answer.
(a)
Probability density function (pdf) is:
f (x) = e−x
(1 + e−x)2
Cumulative density function (cdf) F (x),of a continuous pdf is determined
using the formula:
F (x) =
Z x
−∞
f (t) dt
Therefore, cdf for the given function is:
F (x) =
Z x
−∞
e−t
(1 + e−t)2 dt
Evaluate the indefinite integral first and then apply limits:
I indef =
Z e−t
(1 + e−t)2 dt
[Let, 1 + e−t = u =⇒ e−t = du]
=
Z
− 1
u2 du
= − u−2+1
−2 + 1
= 1
u
[re-substitute: u = 1 + e−t]
∴ I indef = 1
1 + e−t

Applying limits:
F (x) = 1
1 + e−t
x
−∞
= 1
1 + e−x − lim
t→−∞
1
1 + e−t
= 1
1 + e−x − 0 = 1
1 + e−x
(b)
Mean of a distribution is given by:
μ =
Z ∞
−∞
xf (x) dx
For the given pdf:
μ =
Z ∞
−∞
xe−x
(1 + e−x)2 dx
Substitute:
u = 1
1 + e−x =⇒ du = e−x
(1 + e−x)2
Now,
x = ln u
1 − u
Limits:
x → −∞ =⇒ u → 0 & x → ∞ =⇒ u → 1

Therefore,
μ =
Z 1
0
ln u
1 − u du
=
Z 1
0
ln u du −
Z 1
0
ln (1 − u) du
[Property:
Z a
0
g(x) dx =
Z a
0
g(a − x) dx]
=
Z 1
0
ln u du −
Z 1
0
ln u du
∴ μ = 0
Q 3.
Answer
(a)
1. Graph for trapezoidal rule with one trapezium:

2. Graph for trapezoidal rule with two trapezia:
3. Graph for trapezoidal rule with four trapezia:

(b)
1. Total globalproduction of bio-fuelby trapezoidalrule using one trapez-
ium:
Refer the graph with one trapezium in part (a).
Total global production = Area under the curve for global production
By Trapezoidal rule:
Area under the curve = Area of trapezia
From the graph:
base (h) of the trapezium is the length along x- axis between points 1 and 2
= 20 years = 20 x 365 = 7300 days
Height at point 1 = y1 = 0.1429 million barrels/day
Height at point 2 = y2 = 1.1429 million barrels/day
Area of trapezium =
h
2(y1+y2) = 7300
2 (0.1429+1.1429)·106 = 4.6932 billion barrels
2. Total globalproduction of bio-fuelby trapezoidalrule using two trapez-
ium:
Refer the graph with two trapezium in part (a).

base (h) of the two trapeziums is equaland is the length along x- axis be-
tween points 1 and 2 = 10 years = 10 x 365 = 3650 days
Height at point 1 = y1 = 0.1429 million barrels/day
Height at point 2 = y2 = 0.185 million barrels/day
Height at point 2 = y2 = 0.1429 million barrels/day
Area of trapezium =
h
2(y1+y3+2y2) = 3650
2 (0.1429+1.1429+2×0.185)·106
= 3.0218 billion barrels
3. Total global production of bio-fuel by trapezoidal rule using four trapez-
ium:
Refer the graph with four trapezium in part (a).
base (h) of the four trapeziums is equaland is the length along x- axis be-
tween points 1 and 2 = 5 years = 5 x 365 = 1825 days
Height at point 1 = y1 = 0.1429 million barrels/day
Height at point 2 = y2 = 0.185 million barrels/day
Height at point 2 = y3 = 0.185 million barrels/day
Height at point 2 = y4 = 0.386 million barrels/day
Height at point 2 = y5 = 0.1429 million barrels/day
Area of trapezium =
h
2(y1 + y5 + 2y2 + 2y3 + 2y4)
= 1825
2 (0.1429+1.1429+2×0.185+2×0.185+2×0.386)·106 = 2.553 billion barrels