Lesson 7-7

Numerical Approximations of Integrals-- What we do when we can’t integrate a

function

Riemann SumsTrapezoidal Rule

Strategies for Integrals that we can’t do

• Use Riemann Sums – left-hand or right-hand, to approximate the area under the curve (i.e., the value of the definite integral)

• Use the Trapezoidal Rule to approximate the area under the curve– The trapezoidal rule gives us an average of the

right and left hand Riemann Sum approximations– The graph of f The trapezoidal rule will:

concave down underestimate the area concave up overestimate the area.

Trapezoidal Rule

From Geometry: Area of a Trapezoid = ½h•(b1 + b2)Bases will the “height” or the functional value and h is the width, (b-a)/n, of each trapezoidSince only the first base and the last base will be used only once Area = ½h•(b1 + 2b2 + … + 2bn-1 + bn)

Example 1

Use the trapezoidal rule with n = 4 to approximate ∫ sin x dxx=0

x=π

h = (b-a)/n = (π-0)/4 = π/4

n f(x) 0 sin(0) =01 sin(π/4) = 0.70712 sin(π/2) = 13 sin(3π/4) = 0.70714 sin(π) = 0

12 3

4

Area = ½ h (b1 + b2)

Area1 = ½ h (b1 + b2) = π/8(0 + 0.7071)Area2 = ½ h (b1 + b2) = π/8(0.7071 + 1)Area3 = ½ h (b1 + b2) = π/8(1 + 0.7071)Area4 = ½ h (b1 + b2) = π/8(0.7071 + 0)AreaT = Ai

= (π/8)[2(0.7071) + 2(1.7071)] = 1.89612

∫ sin x dx = 2x=0

x=π

Example 2

Use the trapezoidal rule with n = 5 to approximate ∫ ex² dxx=0

x=1

h = (b-a)/n = (1-0)/5 = 1/5

n f(x) 0 f(0) = 11 f(1/5) = 1.22142 f(2/5) = 1.49183 f(3/5) = 1.82214 f(4/5) = 2.22555 f(1) = 2.7183

Area = ½ h (b1 + b2)

Area1 = ½ h (b1 + b2) = 1/10(1 + 1.2214)Area2 = ½ h (b1 + b2) = 1/101.2214 + 1.4918)Area3 = ½ h (b1 + b2) = 1/10(1.4918 + 1.8221)Area4 = ½ h (b1 + b2) = 1/10(1.8221 + 2.2255)Area5 = ½ h (b1 + b2) = 1/10(2.2255 + 2.7183) AreaT = Ai

= (1/10)[1 + 2(1.2214) + 2(1.4918) + 2(1.8221) + 2(2.2255) + 2.7183] = 1.72399

∫ ex² dx = 1.46265x=0

x=1

123 4 5

note: calculator did a numeric apx

Pond Example

40

5045

3045

40

Area = ½ h (b1 + b2)

Area1 = ½ h (b1 + b2) = 10(0 + 40) = 400Area2 = ½ h (b1 + b2) = 10(40 + 50) = 900Area3 = ½ h (b1 + b2) = 10(50 + 45) = 950Area4 = ½ h (b1 + b2) = 10(45 + 30) = 750Area5 = ½ h (b1 + b2) = 10(30 + 45) = 750Area6 = ½ h (b1 + b2) = 10(45 + 40) = 850Area7 = ½ h (b1 + b2) = 10(40 + 0) = 400 AreaT = Ai = 5000

h = 20 =(b-a)/n n = 7

Approximating the Error of the Estimate

If f has a continuous second derivative on [a,b], then

the error E in approximating by the

trapezoidal rule is:

∫ f(x) dxx=a

x=b

xfnabE

max12 2

3

(b-a)³E ≤ -------- max |f ’’(x)| where a ≤ x ≤ b 12n²

(b – a)³n² ≤ ---------- max |f ’’(x)| (always round up to 12 E next whole # when calculating n)

Example 3

1 ∫ --------- dx x + 1x=0

x=1

Use the error formula to find the maximum possible error in approximating the integral, with n = 4.

(1-0)³E ≤ -------- |2| (at x = 0 ) 12•4²

E ≤ 0.01042

f(x) = (x + 1)-1

f’(x) = -(x+1)-2

f’’(x) = 2/(x+1)-3

(b-a)³E ≤ -------- max |f ’’(x)| where a ≤ x ≤ b 12n²

Example 4Use the error formula to find n so that the error in the approximation of the definite integral is less than 0.00001.n

(b – a)³n² ≥ ---------- max |f ’’(x)| (always round up to 12 E next whole # when calculating n)

1 ∫ --------- dx x + 1x=0

x=1 f(x) = (x + 1)-1

f’(x) = -(x+1)-2

f’’(x) = 2/(x+1)-3

(1 – 0)³ [2]n² ≥ ------------------- (always round up to 12 (0.00001) next whole # when calculating n)n² ≥ 16666.67

n ≥ 129.1 n = 130

Summary & Homework• Summary:

– Riemann Sums can be used to get approximations to definite integrals that we don’t know how to integrate

– The error used to approximate the integral can be calculated

– The number of sub-intervals required to keep the error bounded can be calculated

• Homework: – pg 527 – 529: 7, 8, 9