Pre-sessional Quantitative Methods
Extra integration worked examples
Pratibha Yadav†
† Department of Economics
Royal Holloway, University of London
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
1
Homework 7: Question 1
Integrate
R
x ln x dx
Let u = ln x and dv = x dx
Then du/dx = 1/x and v = x 2 /2, so:
Z
x ln x dx
=
Pre-sessional Quantitative Methods
x2 1
dx
2 x
Z
1 2
1
x ln x −
x dx
2
2
= ln x
x2
−
2
Z
=
1 2
1 x2
x ln x −
+c
2
2 2
=
1 2
x2
x ln x −
+c
2
4
Pratibha Yadav
September 2014
Extra integration worked examples
2
Homework 7: Question 2
Integrate
R
x 2 e3x dx
Let u = x 2 and dv = e3x dx
Then du/dx = 2x and v = e3x /3, so:
3x
e3x
2x dx
3
3
Z
1 2 3x 2
=
x e −
x e3x dx
3
3
R
Now we apply integration by parts to the term x e3x dx again
Z
2 3x
x e
Pre-sessional Quantitative Methods
dx
= x
Pratibha Yadav
2e
Z
−
September 2014
Extra integration worked examples
3
Homework 7: Question 2
Let u = x and dv = e3x dx
Then du/dx = 1 and v = e3x /3, so:
Z
Z 3x
e3x
e
3x
x e dx = x
−
dx
3
3
Z
1 3x 1
xe −
e3x dx
=
3
3
1 3x 1 e3x
=
xe −
3
3 3
So:
Z
2 3x
x e
dx
=
=
Pre-sessional Quantitative Methods
1 2 3x 2 1 3x e3x
x e −
xe −
+c
3
3 3
9
1 2 3x 2 3x
2
x e − x e + e3x + c
3
9
27
Pratibha Yadav
September 2014
Extra integration worked examples
4
Homework 7: Question 3
R
Integrate (ln x)3 dx
First, note that (ln x)3 6= ln x 3 = 3 ln x
Let u = (ln x)3 and dv = 1 dx
Then du/dx = 3(ln x)2 · 1/x and v = x, so:
Z
Z
3
(ln x)3 dx = (ln x)3 x − x (ln x)2 dx
x
Z
= x(ln x)3 − 3 (ln x)2 dx
Now we use integration by parts for the bold term
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
5
Homework 7: Question 3
R
We have (ln x)2 dx
Let u = (ln x)2 and dv = 1 dx
Then du/dx = 2(ln x) · 1/x and v = x, so:
Z
Z
2
2
2
(ln x) dx = (ln x) x − x (ln x) dx
x
Z
= x(ln x)2 − 2 ln x dx
We apply integration by parts to the bold term again
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
6
Homework 7: Question 3
R
From before, we know that ln x dx = x ln x − x, so we can now
solve for our original problem:
Z
Z
3
2
(ln x) dx = x(ln x) − 3 (ln x)2 dx
Z
2
= x(ln x) − 3 x(ln x) − 2 ln x dx
3
= x(ln x)3 − 3x(ln x)2 + 6 [x ln x − x] + c
= x(ln x)3 − 3x(ln x)2 + 6x ln x − 6x + c
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
7
Homework 7: Question 4
Integrate
R
2x
√
1 + x 2 dx
We use the substitution procedure to find this integral
Let u = 1 + x 2 , then du/dx = 2x and so dx = du/2x, so now the
integral becomes:
Z
√ du
=
2x u
2x
Z
√
u du =
u 3/2
+c
3/2
Now substitute the original expression for u back; therefore we get:
Z
p
2
2x 1 + x 2 dx = (1 + x 2 )3/2 + c
3
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
8
Homework 7: Question 5
Integrate
R
√ x
1−4x 2
dx
Let u = 1 − 4x 2 , then du/dx = −8x and so dx = −du/8x, so:
Z
Z
x du
1
du
√
−√
= −
8
u 8x
u
Pre-sessional Quantitative Methods
= −
1 u −1/2+1
+c
8 −1/2 + 1
= −
1 u 1/2
+c
8 1/2
= −
1p
1 − 4x 2 + c
4
Pratibha Yadav
September 2014
Extra integration worked examples
9
Homework 7: Question 6
Evaluate the area between the following curves in the domain [0, 1]:
f (x) = 2x − x 2 and g (x) = x 2 .
Both are quadratic functions. f (x) is opening downward at:
b
b
2
2
− ,f −
= −
,f −
= (1, 1)
2a
2a
2(−1)
2(−1)
and g (x) is opening upward at:
b
b
= (0, 0)
− ,g −
2a
2a
Now sketch these curves
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
10
Homework 7: Question 6
Notice that 2x − x 2 lies above x 2 on the domain [0, 1], so the area
contained in between the two curves on the domain [0, 1] is:
Z
1
2
2
1
Z
[(2x − x ) − x ] dx
(2x − 2x 2 ) dx
=
0
0
Z
= 2
1
(x − x 2 ) dx
0
x2 x3
−
= 2
2
3
1 1
= 2
−
2 3
1
0
= 1/3
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
11
Probability problem
Suppose f (x) = 6x − x 2 , for 0 ≤ x ≤ 1
The mean is the expected value, which is:
Z 1
Z 1
x f (x) dx =
x(6x − x 2 ) dx
0
0
=
6x 3 6x 4
−
3
4
1
0
= 1/2
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
12
Examples – integration
Determine the integrals
R√
√
xe
x
dx and
R
√1
2+ x−x
dx
We determine the first one using integration by substitution
√
√
Let u = x, hence du = 1/(2 x) dx, so dx = 2u du
So we now express the integral as:
Z
Z
Z
√ √x
u
xe dx = ue (2u du) = 2 u 2 eu du
Now we need to apply integration by parts, which recall is:
Z
Z
f dg = fg − g df
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
13
Examples – integration
So we obtain:
Z
Z
2 u
2 u
u
2 u e du = 2 u e − 2ue du
Z
2 u
u
u
= 2 u e − 2 ue − e du
= 2u 2 eu − 4ueu + 4eu + c
√
= 2xe
x
√
√ √
− 4 xe x + 4e x + c
√
√
= 2 x −2 x +2 e x +c
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
14
Examples – integration
To determine
Z
R
√1
dx
2+ x−x
p(x)
dx
q(x)
we use partial fractions:
Z =
B
A
+
(x − a1 ) (x − a2 )
dx
= A ln |x − a1 | + B ln |x − a2 | + c
√
√
First though, apply the substitution u = x, hence du = 1/(2 x) dx,
so dx = 2u du, resulting in:
Z
Z
Z
1
dx
2u du
√
√
√ =
dx =
(2 − u)(1 + u)
2+ x −x
(2 − x)(1 + x)
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
15
Examples – integration
So now determine the partial fractions:
A
B
2u
=
+
=⇒ 2u = A(1 + u) + B(2 − u)
(2 − u)(1 + u)
2−u 1+u
For u = −1, we get −2 = 3B, i.e. B = −2/3
For u = 2, we get 4 = 3A, i.e. A = 4/3
So:
Z
2u
du =
(2 − u)(1 + u)
Pre-sessional Quantitative Methods
Pratibha Yadav
Z 4/3
−2/3
+
2−u 1+u
September 2014
du
Extra integration worked examples
16
Examples – integration
Z −2/3
4/3
+
2−u 1+u
4
2
du = − ln |2 − u| − ln |1 + u| + c
3
3
√
√
2
4
= − ln |2 − x| − ln |1 + x| + c
3
3
√
√
2
2
= − ln(2 − x)2 − ln |1 + x| + c
3
3
√
√
2
= − ln |(2 − x)2 (1 + x)| + c
3
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
17
Examples – integration
Determine the following integral:
Z
π/4
√
0
Pre-sessional Quantitative Methods
Pratibha Yadav
1
dx
tan x cos2 x
September 2014
Extra integration worked examples
18
Examples – integration
Begin with the substitution u = tan x
Hence du = 1/(cos2 x) dx (this can be calculated using the quotient
rule), the fact that tan x = sin x/cos x and the fact that
sin2 x + cos2 x = 1
So the integral is:
Z
0
1
√ 1
1
√ du = 2 u 0 = 2
u
Note the limits changed, using the fact that tan π/4 = 1
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
19
Examples – integration
Determine the following integrals:
(a)
Z
2t + 4
dt
t 2 + 4t + 5
(b)
Z
t2
(c)
Z
dt
−4
√
t 1 − t dt
(d)
Z
Pre-sessional Quantitative Methods
Pratibha Yadav
t 3 et dt
September 2014
Extra integration worked examples
20
Examples – integration
For (a), use the substitution u = t 2 + 4t + 5, so that:
du = (2t + 4) dt
so:
Z
2t + 4
dt =
2
t + 4t + 5
Z
1
du = ln |u| + c = ln |t 2 + 4t + 5| + c
u
Note that since t 2 + 4t + 5 = (t + 2)2 + 1 > 0, we can drop the
absolute value ‘bars’ in the answer to give just ln(t 2 + 4t + 5) + c
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
21
Examples – integration
For (b), note that we can write the integrand as:
t2
1
A
B
1
=
=
+
−4
(t − 2)(t + 2)
t −2 t +2
for some constants A and B using partial fractions
To find these constants note that cross-multiplying means the
numerators must satisfy:
1 = A(t + 2) + B(t − 2)
for all t and so using t = 2 gives us A = 1/4 and t = −2 gives
B = −1/4
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
22
Examples – integration
Therefore we have:
Z
1
dt =
2
t −4
=
=
1
4
Z 1
1
−
t −2 t +2
dt
1
(ln |t − 2| − ln |t + 2|) + c
4
1 t − 2 ln
+c
4 t + 2
Note here we cannot drop the absolute values
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
23
Examples – integration
For (c), we will see that different substitutions can be used to solve
the same integral
R √
We require t 1 − t dt, so let’s first use the substitution u = 1 − t
so that du = −dt:
Z
Z
t(1 − t)1/2 dt = − (1 − u)u 1/2 du
Z = −
u 1/2 − u 3/2 du
!
u 3/2 u 5/2
= −
−
+c
3/2
5/2
2
2
= − (1 − t)3/2 + (1 − t)5/2 + c
3
5
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
24
Examples – integration
Now let’s instead use the substitution u 2 = 1 − t or u =
that 2u du = −dt:
Z
Z
t(1 − t)1/2 dt =
(1 − u 2 )u(−2u) du
Z
(u 2 − u 4 ) du
u3 u5
−
3
5
= −2
= −2
√
1 − t so
+c
2
2
= − (1 − t)3/2 + (1 − t)5/2 + c
3
5
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
25
Examples – integration
For (d),
R
t 3 et dt, we use integration by parts, i.e.:
Z
Z
0
u(t)v (t) dt = u(t) v (t) − u 0 (t) v (t) dt
So set u(t) = t 3 and v 0 (t) = et to get:
Z
Z
3 t
3 t
t e dt = t e − 3t 2 et dt
Now apply parts again with u(t) = 3t 2 and v 0 (t) = et to get:
Z
Z
t 3 et dt = t 3 et − 3t 2 et − 6tet dt
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
26
Examples – integration
Now apply parts (again!) with u(t) = 6t and v 0 (t) = et to get:
Z
Z
3 t
3 t
2 t
t
t
t e dt = t e − 3t e + 6te − 6e dt
So the answer is:
Z
t 3 et dt = t 3 et − 3t 2 et + 6tet − 6et + c
= (t 3 − 3t 2 + 6t − 6)et + c
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
27
And finally...
ALL THE BEST WITH YOUR EXAM!
Pre-sessional Quantitative Methods
Pratibha Yadav
September 2014
Extra integration worked examples
28