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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


Some examples


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


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


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


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


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


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


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


(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


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


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


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 · · ·


(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 · · ·


(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 · · ·


(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 10 the ...frooney/m217_11_defint.pdfcalculus with...

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