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Calculus with Algebra and Trigonometry IILecture 10
The definite integral
Feb 26, 2015
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 1 / 17
The definite integral
Given a function f (x) defined on an interval [a, b], the definite integral off (x) from a to b written as ∫ b
af (x) dx
It represents the signed area of the region bounded byx = a, x = b, y = 0, y = f (x). The function f (x) is called the integrand.
In the diagram above∫ b
af (x) dx > 0
∫ c
bf (x) dx < 0
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 2 / 17
Some examples
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 3 / 17
The graph consists of a triangle and a quarter circle.
Area of triangle =1
2(4)(2) = 4 Area of quarter circle =
1
4(π(2)2) = π
so ∫ 6
0f (x) = 4 + π
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 4 / 17
Properties of the definite integral
∫ b
a(f (x) + g(x)) dx =
∫ b
af (x) dx +
∫ b
ag(x) dx∫ b
ak f (x) dx = k
∫ b
af (x) dx
For example given the f (x) that consists of the triangle and quarter circlethen∫ 6
0(f (x) + 2) dx =
∫ 6
0f (x) , dx +
∫ 6
02 dx = 4 + π + 2(6) = 16 + π
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 5 / 17
One consequence of this is obtained by putting a = b∫ c
af (x) dx +
∫ a
cf (x) dx =
∫ a
af (x) dx = 0
this implies ∫ a
cf (x) dx = −
∫ c
af (x) dx
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 6 / 17
If m ≤ f (x) ≤ M for a ≤ x ≤ b then
m(b − a) ≤∫ b
af (x) dx ≤ M(b − a)
The average value of f (x) on [a, b] is defined to be
Average =
∫ ba f (x) dx
b − a
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 7 / 17
Area under a parabola
To calculate ∫ a
−ax2 dx
we will use Archimedes’ result relating the area between a parabola and achord and an inscribed triangle
The area between the segment from (−a, a2) to (a, a2) and the parabola is4/3 the area of the triangle shown, so
Area of parabolic sector =4
3a3
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 8 / 17
Then the area we want (in blue)∫ a
−ax2 dx = Area of the rectangle− Area of parabolic sector
= 2a3 − 4
3a3 =
2
3a3
By symmetry we can deduce ∫ a
0x2 dx =
1
3a3
and ∫ b
ax2 dx =
∫ b
0x2 dx −
∫ a
0x2 dx =
1
3b3 − 1
3a3
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 9 / 17
Another example
Given ∫ π
0sin x dx = 2
Find
(a)
∫ 2π
0sin x dx (b)
∫ π2
0sin x dx
(c)
∫ π2
0cos x dx (d)
∫ π2
0sin 2x dx
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 10 / 17
(a)∫ 2π
0 sin x dx = 0, since by the properties of sin x∫ 2π
πsin x dx = −
∫ π
0sin x dx
(b) By symmetry ∫ π2
0sin x dx =
1
2
∫ π
0sin x dx = 1
(c) Since the graph of cos x is the same as the graph of sin x shifted π2 to
the left ∫ π2
0cos x dx =
∫ π
π2
sin x dx = 1
(d) Since the graph of sin 2x is the same as the graph of sin x withdistances in the x direction shrunk by a factor of 2∫ π
2
0sin 2x dx =
1
2
∫ π
0sin x dx = 1
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 11 / 17
Riemann Sums
In general to calculate ∫ b
af (x) dx
Partition the interval [a, b] into n subintervals, that is pick n − 1 points,a = x0 < x1 < x2 < · · · < xk < · · · < xn = b. The kth interval is[xk−1, xk ]. Pick any point ck ∈ [xk−1, xk ] then∫ xk
xk−1
f (x) dx ≈ f (ck)(xk − xk−1) = f (ck)∆xk
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 12 / 17
Let the maximum value of ∆xk be Mn then we can define the definiteintegral to be given by∫ b
af (x) dx = lim
n→∞lim
Mn→0
n∑k=1
f (ck)∆xk
Various choices of the ck give rise to special Riemann sums
Upper sum: Choose ck so f (ck) has the maximum value on [xk−1, xk ]
Lower sum: Choose ck so f (ck) has the minimum value on [xk−1, xk ]
Left sum: Choose ck = xk−1
Right sum: Choose ck = xk
Midpoint sum: Choose ck = (xk−1 + xk)/2.
A closely related sum is the trapezoidal sum∫ b
af (x) dx ≈
n∑k=1
1
2(f (xk−1) + f (xk)) ∆xk
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 13 / 17
An example
Using four subintervals find an approximate value for
I =
∫ 4
0
x + 2
x + 1dx
(a) Upper sum
I ≈ 1.f (0) + 1.f (1) + 1.f (2) + 1.f (3) = 2 +3
2+
4
3+
5
4= 6.083 · · ·
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 14 / 17
(b) Lower sum
I ≈ 1.f (1) + 1.f (2) + 1.f (3) + 1.f (4) =3
2+
4
3+
5
4+
6
5= 5.283 · · ·
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 15 / 17
(c) Midpoint sum
I ≈ 1.f (0.5) + 1.f (1.5) + 1.f (2.5) + 1.f (3.5) =5
3+
7
5+
9
7+
11
9= 5.575 · · ·
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 16 / 17
(d) Trapezoidal sum
I ≈ 1
2(f (0) + f (1)) +
1
2(f (1) + f (2)) +
1
2(f (2) + f (3)) +
1
2(f (3) + f (4))
= 1 +3
2+
4
3+
5
4+
3
5= 5.683 · · ·
Calculus with Algebra and Trigonometry II Lecture 10The definite integralFeb 26, 2015 17 / 17