1. Lesson 7-7 Numerical Approximations of Integrals -- What we do when we can’t integrate a function Riemann Sums Trapezoidal Rule

2. 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.

3. Trapezoidal Rule From Geometry: Area of a Trapezoid = ½h(b 1 + b 2 ) Bases will the “height” or the functional value and h is the width, (b-a)/n, of each trapezoid Since only the first base and the last base will be used only once  Area = ½h(b 1 + 2b 2 + … + 2b n-1 + b n )

4. Example 1 Use the trapezoidal rule with n = 4 to approximate ∫ sin x dx x=0 x=π h = (b-a)/n = (π-0)/4 = π/4 n f(x) 0 sin(0) =0 1 sin(π/4) = 0.7071 2 sin(π/2) = 1 3 sin(3π/4) = 0.7071 4 sin(π) = 0 1 23 4 Area = ½ h (b 1 + b 2 ) Area 1 = ½ h (b 1 + b 2 ) = π/8(0 + 0.7071) Area 2 = ½ h (b 1 + b 2 ) = π/8(0.7071 + 1) Area 3 = ½ h (b 1 + b 2 ) = π/8(1 + 0.7071) Area 4 = ½ h (b 1 + b 2 ) = π/8(0.7071 + 0) Area T =  A i = (π/8)[2(0.7071) + 2(1.7071)] = 1.89612 ∫ sin x dx = 2 x=0 x=π

5. Example 2 Use the trapezoidal rule with n = 5 to approximate ∫ e x² dx x=0 x=1 h = (b-a)/n = (1-0)/5 = 1/5 n f(x) 0 f(0) = 1 1 f(1/5) = 1.2214 2 f(2/5) = 1.4918 3 f(3/5) = 1.8221 4 f(4/5) = 2.2255 5 f(1) = 2.7183 Area = ½ h (b 1 + b 2 ) Area 1 = ½ h (b 1 + b 2 ) = 1/10(1 + 1.2214) Area 2 = ½ h (b 1 + b 2 ) = 1/101.2214 + 1.4918) Area 3 = ½ h (b 1 + b 2 ) = 1/10(1.4918 + 1.8221) Area 4 = ½ h (b 1 + b 2 ) = 1/10(1.8221 + 2.2255) Area 5 = ½ h (b 1 + b 2 ) = 1/10(2.2255 + 2.7183) Area T =  A i = (1/10)[1 + 2(1.2214) + 2(1.4918) + 2(1.8221) + 2(2.2255) + 2.7183] = 1.72399 ∫ e x² dx = 1.46265 x=0 x=1 1 2 3 4 5 note: calculator did a numeric apx

6. Pond Example 40 50 45 30 45 40 Area = ½ h (b 1 + b 2 ) Area 1 = ½ h (b 1 + b 2 ) = 10(0 + 40)= 400 Area 2 = ½ h (b 1 + b 2 ) = 10(40 + 50)= 900 Area 3 = ½ h (b 1 + b 2 ) = 10(50 + 45)= 950 Area 4 = ½ h (b 1 + b 2 ) = 10(45 + 30)= 750 Area 5 = ½ h (b 1 + b 2 ) = 10(30 + 45)= 750 Area 6 = ½ h (b 1 + b 2 ) = 10(45 + 40)= 850 Area 7 = ½ h (b 1 + b 2 ) = 10(40 + 0)= 400 Area T =  A i = 5000 h = 20 =(b-a)/n n = 7

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

8. Example 3 1 ∫ --------- dx x + 1 x=0 x=1 Use the error formula to find the maximum possible error in approximating the integral, with n = 4. (1-0)³ E ≤ -------- |2| (at x = 0 ) 124² E ≤ 0.01042 f(x) = (x + 1) -1 f’(x) = -(x+1) -2 f’’(x) = 2/(x+1) -3 (b-a)³ E ≤ -------- max |f ’’(x)| where a ≤ x ≤ b 12n²

9. Example 4 Use the error formula to find n so that the error in the approximation of the definite integral is less than 0.00001. (b – a)³ n² ≥ ---------- max |f ’’(x)| (always round up to 12 E next whole # when calculating n) 1 ∫ --------- dx x + 1 x=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

10. 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