Solving Integrals using Standard Integral Formulas

This document explains how to solve integrals using standard integral formulas. It covers the concept of indefinite and definite integrals and provides examples of solving different types of integrals.

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