1S11: Calculus for students in Science
Dr. Vladimir Dotsenko
TCD
Michaelmas Term 2013
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
1 / 13
Substitutions in definite integrals
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
2 / 13
Substitutions in definite integrals
Recall the u-substitution method for computing antiderivatives: given an
integral of the form
Z
f (g (x))g ′ (x) dx,
we denote u = g (x) so that du = g ′ (x) dx, so that the integral becomes
Z
f (u) du.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
2 / 13
Substitutions in definite integrals
Recall the u-substitution method for computing antiderivatives: given an
integral of the form
Z
f (g (x))g ′ (x) dx,
we denote u = g (x) so that du = g ′ (x) dx, so that the integral becomes
Z
f (u) du.
In order to use this method to evaluate definite integrals of the same form
Z
b
f (g (x))g ′ (x) dx,
a
we need to take appropriate care of the effect of that on the limits of
integration.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
2 / 13
Substitutions in definite integrals
Recall the u-substitution method for computing antiderivatives: given an
integral of the form
Z
f (g (x))g ′ (x) dx,
we denote u = g (x) so that du = g ′ (x) dx, so that the integral becomes
Z
f (u) du.
In order to use this method to evaluate definite integrals of the same form
Z
b
f (g (x))g ′ (x) dx,
a
we need to take appropriate care of the effect of that on the limits of
integration. There are two ways to deal with it, which we shall now outline.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
2 / 13
Substitutions in definite integrals
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
3 / 13
Substitutions in definite integrals
Method 1. Use u-substitutions only on the level of indefinite integrals:
evaluate
Z
f (g (x))g ′ (x) dx,
and then use the formula
b
Z
Z b
f (g (x))g ′ (x) dx .
f (g (x))g ′ (x) dx =
a
Dr. Vladimir Dotsenko (TCD)
a
1S11: Calculus for students in Science
Michaelmas Term 2013
3 / 13
Substitutions in definite integrals
Method 1. Use u-substitutions only on the level of indefinite integrals:
evaluate
Z
f (g (x))g ′ (x) dx,
and then use the formula
b
Z
Z b
f (g (x))g ′ (x) dx .
f (g (x))g ′ (x) dx =
a
a
Method 2. Use the relationship u = g (x) to modify the limits:
Z
b
a
Dr. Vladimir Dotsenko (TCD)
′
f (g (x))g (x) dx =
Z
g (b)
f (u) du.
g (a)
1S11: Calculus for students in Science
Michaelmas Term 2013
3 / 13
Substitutions in definite integrals
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
4 / 13
Substitutions in definite integrals
Example 1. Let us use the first method to evaluate
Z 2
x(x 2 + 1)3 dx.
0
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
4 / 13
Substitutions in definite integrals
Example 1. Let us use the first method to evaluate
Z 2
x(x 2 + 1)3 dx.
0
We denote u = x 2 + 1, so that du = 2x dx, and
Z
Z
1
(x 2 + 1)4
u4
x(x 2 + 1)3 dx =
+C =
+ C.
u 3 du =
2
8
8
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
4 / 13
Substitutions in definite integrals
Example 1. Let us use the first method to evaluate
Z 2
x(x 2 + 1)3 dx.
0
We denote u = x 2 + 1, so that du = 2x dx, and
Z
Z
1
(x 2 + 1)4
u4
x(x 2 + 1)3 dx =
+C =
+ C.
u 3 du =
2
8
8
Therefore,
Z
0
2
x(x 2 + 1)3 dx =
Z
Dr. Vladimir Dotsenko (TCD)
x(x 2 + 1)3 dx
2
0
=
(x 2 + 1)4
8
1S11: Calculus for students in Science
2
0
=
625 1
− = 78.
8
8
Michaelmas Term 2013
4 / 13
Substitutions in definite integrals
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
5 / 13
Substitutions in definite integrals
Example 2. Let us use the second method to evaluate the same integral
Z 2
x(x 2 + 1)3 dx.
0
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
5 / 13
Substitutions in definite integrals
Example 2. Let us use the second method to evaluate the same integral
Z 2
x(x 2 + 1)3 dx.
0
We denote u = x 2 + 1, so that du = 2x dx, and
for x = 0, u = 1,
for x = 2, u = 5.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
5 / 13
Substitutions in definite integrals
Example 2. Let us use the second method to evaluate the same integral
Z 2
x(x 2 + 1)3 dx.
0
We denote u = x 2 + 1, so that du = 2x dx, and
for x = 0, u = 1,
for x = 2, u = 5.
Therefore,
Z
0
2
1
x(x + 1) dx =
2
2
Dr. Vladimir Dotsenko (TCD)
3
Z
1
5
1
u du =
2
3
54 14
−
4
4
1S11: Calculus for students in Science
= 78.
Michaelmas Term 2013
5 / 13
Substitutions in definite integrals
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
6 / 13
Substitutions in definite integrals
Example 3. Let us evaluate the integral
Z 3
cos(π/x)
dx.
x2
1
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
6 / 13
Substitutions in definite integrals
Example 3. Let us evaluate the integral
Z 3
cos(π/x)
dx.
x2
1
We put u = πx , so that du = − xπ2 dx, in other words,
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
1
x2
dx = − π1 du.
Michaelmas Term 2013
6 / 13
Substitutions in definite integrals
Example 3. Let us evaluate the integral
Z 3
cos(π/x)
dx.
x2
1
We put u = πx , so that du = − xπ2 dx, in other words,
We note that
1
x2
dx = − π1 du.
for x = 1, u = π,
for x = 3, u = π/3.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
6 / 13
Substitutions in definite integrals
Example 3. Let us evaluate the integral
Z 3
cos(π/x)
dx.
x2
1
We put u = πx , so that du = − xπ2 dx, in other words,
We note that
1
x2
dx = − π1 du.
for x = 1, u = π,
for x = 3, u = π/3.
Therefore,
Z
3
1
Z
1 π/3
cos(π/x)
dx = −
cos u du =
x2
π π
√
π/3
1
3
1
= − (sin(π/3) − sin π) = −
≈ −0.276.
= − sin u
π
π
2π
π
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
6 / 13
Substitutions in definite integrals
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
7 / 13
Substitutions in definite integrals
Example 4. Let us evaluate the integral
Z
Dr. Vladimir Dotsenko (TCD)
π/4 √
0
tan x
1
dx.
cos2 x
1S11: Calculus for students in Science
Michaelmas Term 2013
7 / 13
Substitutions in definite integrals
Example 4. Let us evaluate the integral
Z
π/4 √
1
dx.
cos2 x
1
cos2 x
dx.
0
We put u = tan x, so that du =
Dr. Vladimir Dotsenko (TCD)
tan x
1S11: Calculus for students in Science
Michaelmas Term 2013
7 / 13
Substitutions in definite integrals
Example 4. Let us evaluate the integral
Z
π/4 √
tan x
1
dx.
cos2 x
1
cos2 x
dx. We note that
0
We put u = tan x, so that du =
for x = 0, u = 0,
for x = π/4, u = 1.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
7 / 13
Substitutions in definite integrals
Example 4. Let us evaluate the integral
Z
π/4 √
tan x
1
dx.
cos2 x
1
cos2 x
dx. We note that
0
We put u = tan x, so that du =
for x = 0, u = 0,
for x = π/4, u = 1.
Therefore,
Z
0
π/4 √
1
tan x
dx =
cos2 x
Dr. Vladimir Dotsenko (TCD)
Z
0
1√
u 3/2
u du =
3/2
1S11: Calculus for students in Science
#1
0
=
2
.
3
Michaelmas Term 2013
7 / 13
Substitutions in definite integrals
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
8 / 13
Substitutions in definite integrals
Example 5. Let us prove, without evaluating integrals, that
Z
Dr. Vladimir Dotsenko (TCD)
π/2
0
sinn x dx =
Z
π/2
cosn x dx.
0
1S11: Calculus for students in Science
Michaelmas Term 2013
8 / 13
Substitutions in definite integrals
Example 5. Let us prove, without evaluating integrals, that
Z
π/2
sinn x dx =
0
In the second integral, we put u =
Dr. Vladimir Dotsenko (TCD)
π
2
Z
π/2
cosn x dx.
0
− x, so that du = −dx.
1S11: Calculus for students in Science
Michaelmas Term 2013
8 / 13
Substitutions in definite integrals
Example 5. Let us prove, without evaluating integrals, that
Z
π/2
sinn x dx =
0
In the second integral, we put u =
that
π
2
Z
π/2
cosn x dx.
0
− x, so that du = −dx. We note
for x = 0, u = π/2,
for x = π/2, u = 0,
and that cos x = cos(π/2 − u) = sin u.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
8 / 13
Substitutions in definite integrals
Example 5. Let us prove, without evaluating integrals, that
Z
π/2
sinn x dx =
0
In the second integral, we put u =
that
π
2
Z
π/2
cosn x dx.
0
− x, so that du = −dx. We note
for x = 0, u = π/2,
for x = π/2, u = 0,
and that cos x = cos(π/2 − u) = sin u. Therefore,
Z
π/2
0
n
cos x dx = −
Z
0
n
sin u du =
π/2
Z
π/2
sinn x dx,
0
as required.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
8 / 13
Substitutions in definite integrals
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
9 / 13
Substitutions in definite integrals
Example 6. Let us examine the integral
Z 1
1
dx.
1
+
x2
−1
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
9 / 13
Substitutions in definite integrals
Example 6. Let us examine the integral
Z 1
1
dx.
1
+
x2
−1
Let us perform the substitution u = x1 , so that du = − x12 dx, in other
words, du = −u 2 dx and dx = − u12 du.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
9 / 13
Substitutions in definite integrals
Example 6. Let us examine the integral
Z 1
1
dx.
1
+
x2
−1
Let us perform the substitution u = x1 , so that du = − x12 dx, in other
words, du = −u 2 dx and dx = − u12 du. We note that
for x = −1, u = −1,
and that
1
1+x 2
=
1
1+(1/u)2
Dr. Vladimir Dotsenko (TCD)
=
and for x = 1, u = 1,
u2
.
1+u 2
1S11: Calculus for students in Science
Michaelmas Term 2013
9 / 13
Substitutions in definite integrals
Example 6. Let us examine the integral
Z 1
1
dx.
1
+
x2
−1
Let us perform the substitution u = x1 , so that du = − x12 dx, in other
words, du = −u 2 dx and dx = − u12 du. We note that
for x = −1, u = −1,
and that
1
1+x 2
Z
1
−1
=
1
1+(1/u)2
=
1
dx = −
1 + x2
u2
.
1+u 2
Z
1
−1
and for x = 1, u = 1,
Therefore,
u2 1
du = −
1 + u2 u2
Z
1
−1
1
du,
1 + u2
so the integral is equal to its negative and hence equal to zero.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
9 / 13
Substitutions in definite integrals
Example 6. Let us examine the integral
Z 1
1
dx.
1
+
x2
−1
Let us perform the substitution u = x1 , so that du = − x12 dx, in other
words, du = −u 2 dx and dx = − u12 du. We note that
for x = −1, u = −1,
and that
1
1+x 2
Z
1
−1
=
1
1+(1/u)2
=
1
dx = −
1 + x2
u2
.
1+u 2
Z
1
−1
and for x = 1, u = 1,
Therefore,
u2 1
du = −
1 + u2 u2
Z
1
−1
1
du,
1 + u2
so the integral is equal to its negative and hence equal to zero. How is it
possible?
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
9 / 13
Substitutions in definite integrals
Example 6. Let us examine the integral
Z 1
1
dx.
1
+
x2
−1
Let us perform the substitution u = x1 , so that du = − x12 dx, in other
words, du = −u 2 dx and dx = − u12 du. We note that
for x = −1, u = −1,
and that
1
1+x 2
Z
1
−1
=
1
1+(1/u)2
=
1
dx = −
1 + x2
u2
.
1+u 2
Z
1
−1
and for x = 1, u = 1,
Therefore,
u2 1
du = −
1 + u2 u2
Z
1
−1
1
du,
1 + u2
so the integral is equal to its negative and hence equal to zero. How is it
possible? Of course, it happened because u = g (x) was not defined on all
the interval [−1, 1], having a singularity at x = 0.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
9 / 13
Substitutions in definite integrals
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
10 / 13
Substitutions in definite integrals
Example 7. (High school maths intervarsity competitions in Russia)
Let us evaluate the integral
Z π/2
(sin2 (sin x) + cos2 (cos x)) dx.
0
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
10 / 13
Substitutions in definite integrals
Example 7. (High school maths intervarsity competitions in Russia)
Let us evaluate the integral
Z π/2
(sin2 (sin x) + cos2 (cos x)) dx.
0
We shall split this integral as a sum
Z π/2
Z
sin2 (sin x) dx +
0
Dr. Vladimir Dotsenko (TCD)
π/2
cos2 (cos x) dx,
0
1S11: Calculus for students in Science
Michaelmas Term 2013
10 / 13
Substitutions in definite integrals
Example 7. (High school maths intervarsity competitions in Russia)
Let us evaluate the integral
Z π/2
(sin2 (sin x) + cos2 (cos x)) dx.
0
We shall split this integral as a sum
Z π/2
Z
sin2 (sin x) dx +
0
π/2
cos2 (cos x) dx,
0
and transform the second integral using the substitution u =
that du = −dx,
for x = 0, u = π/2,
π
2
− x, so
for x = π/2, u = 0,
cos(x) = cos(π/2 − u) = sin u
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
10 / 13
Substitutions in definite integrals
Example 7. (High school maths intervarsity competitions in Russia)
Let us evaluate the integral
Z π/2
(sin2 (sin x) + cos2 (cos x)) dx.
0
We shall split this integral as a sum
Z π/2
Z
sin2 (sin x) dx +
0
π/2
cos2 (cos x) dx,
0
and transform the second integral using the substitution u =
that du = −dx,
for x = 0, u = π/2,
Dr. Vladimir Dotsenko (TCD)
− x, so
for x = π/2, u = 0,
cos(x) = cos(π/2 − u) = sin u, leading to
Z
Z 0
Z π/2
cos2 (sin u) du =
cos2 (cos x) dx = −
0
π
2
π/2
1S11: Calculus for students in Science
π/2
cos2 (sin x) dx.
0
Michaelmas Term 2013
10 / 13
Substitutions in definite integrals
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
11 / 13
Substitutions in definite integrals
Therefore,
Z
π/2
(sin2 (sin x) + cos2 (cos x)) dx =
0
=
=
Z
π/2
2
sin (sin x) dx +
π/2
cos2 (cos x) dx =
0
0
Z
Z
π/2
2
sin (sin x) dx +
0
=
Z
π/2
Z
π/2
cos2 (sin x) dx =
0
(sin2 (sin x) + cos2 (sin x)) dx =
0
=
Z
π/2
1 dx =
0
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
π
.
2
11 / 13
An application of definite integrals
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
12 / 13
An application of definite integrals
Sums of the form 1 +
formulas.
Dr. Vladimir Dotsenko (TCD)
1
2
+
1
3
+ ··· +
1
n
often appear in mathematical
1S11: Calculus for students in Science
Michaelmas Term 2013
12 / 13
An application of definite integrals
Sums of the form 1 + 12 + 13 + · · · + n1 often appear in mathematical
formulas. It is beneficial to have a good estimate of such a sum, avoiding
adding up the terms directly.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
12 / 13
An application of definite integrals
Sums of the form 1 + 12 + 13 + · · · + n1 often appear in mathematical
formulas. It is beneficial to have a good estimate of such a sum, avoiding
adding up the terms directly.
y
y
1 2 3 4 5 6
Dr. Vladimir Dotsenko (TCD)
x
1 2 3 4 5 6
1S11: Calculus for students in Science
x
Michaelmas Term 2013
12 / 13
An application of definite integrals
Sums of the form 1 + 12 + 13 + · · · + n1 often appear in mathematical
formulas. It is beneficial to have a good estimate of such a sum, avoiding
adding up the terms directly.
y
y
x
x
1 2 3 4 5 6
1 2 3 4 5 6
These two figures prove that
Z n+1
1 1
1
1
1 + + + ··· + ≥
dx = ln(n + 1) − ln(1) = ln(n + 1),
2 3
n
x
1
Z n+1
1
1
1
1 1
+ + ··· + +
≤
dx = ln(n + 1) − ln(1) = ln(n + 1),
2 3
n n+1
x
1
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
12 / 13
An application of definite integrals
Sums of the form 1 + 12 + 13 + · · · + n1 often appear in mathematical
formulas. It is beneficial to have a good estimate of such a sum, avoiding
adding up the terms directly.
y
y
x
x
1 2 3 4 5 6
1 2 3 4 5 6
These two figures prove that
Z n+1
1 1
1
1
1 + + + ··· + ≥
dx = ln(n + 1) − ln(1) = ln(n + 1),
2 3
n
x
1
Z n+1
1
1
1
1 1
+ + ··· + +
≤
dx = ln(n + 1) − ln(1) = ln(n + 1),
2 3
n n+1
x
1
so we have
ln(n + 1) ≤ 1 +
Dr. Vladimir Dotsenko (TCD)
1 1
1
1
+ + · · · + ≤ ln(n + 1) + 1 −
.
2 3
n
n+1
1S11: Calculus for students in Science
Michaelmas Term 2013
12 / 13
Summary of integral calculus
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
13 / 13
Summary of integral calculus
Indefinite integral / antiderivative. Methods: look up in the table of
derivatives, simplify by u-substitution, simplify using integration by
parts, combine the above methods.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
13 / 13
Summary of integral calculus
Indefinite integral / antiderivative. Methods: look up in the table of
derivatives, simplify by u-substitution, simplify using integration by
parts, combine the above methods.
Definite integral: originally motivated by computing areas. Every
continuous function, and even every piecewise continuous (bounded)
function can be integrated.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
13 / 13
Summary of integral calculus
Indefinite integral / antiderivative. Methods: look up in the table of
derivatives, simplify by u-substitution, simplify using integration by
parts, combine the above methods.
Definite integral: originally motivated by computing areas. Every
continuous function, and even every piecewise continuous (bounded)
function can be integrated.
Main method for integration: use fundamental theorem of calculus
(compute an antiderivative and subtract its values). Sometimes
antiderivatives are not available, but a u-substitution applied to a part
of the integral would help.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
13 / 13
Summary of integral calculus
Indefinite integral / antiderivative. Methods: look up in the table of
derivatives, simplify by u-substitution, simplify using integration by
parts, combine the above methods.
Definite integral: originally motivated by computing areas. Every
continuous function, and even every piecewise continuous (bounded)
function can be integrated.
Main method for integration: use fundamental theorem of calculus
(compute an antiderivative and subtract its values). Sometimes
antiderivatives are not available, but a u-substitution applied to a part
of the integral would help.
Definite integral is defined via Riemann sums. Sometimes, handling a
sum is easier if you interpret it as a Riemann sum, and examine the
respective integral instead.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
13 / 13
Summary of integral calculus
Indefinite integral / antiderivative. Methods: look up in the table of
derivatives, simplify by u-substitution, simplify using integration by
parts, combine the above methods.
Definite integral: originally motivated by computing areas. Every
continuous function, and even every piecewise continuous (bounded)
function can be integrated.
Main method for integration: use fundamental theorem of calculus
(compute an antiderivative and subtract its values). Sometimes
antiderivatives are not available, but a u-substitution applied to a part
of the integral would help.
Definite integral is defined via Riemann sums. Sometimes, handling a
sum is easier if you interpret it as a Riemann sum, and examine the
respective integral instead.
Next time: applications of the definite integral in geometry, science, and
engineering.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Michaelmas Term 2013
13 / 13