Trapezoidal Rule: h = _______ y 0 = ____ y 1 = _______ y2=____ y 3 = ______ y 4

Question

Trapezoidal Rule:

h = _______ y0 = ____ y1 = _______

y2=____ y3 = ______ y4 = _______

            Show the Trapezoidal formula that you used to find this value.

Simpsons Rule:      Use decimal values correct to six places for intermediate values and round

your final answer to four places.

h = _________ y0 = ______ y1 = ____________

y2 = _____ y3 = ___ y4 = ________________

4)Approximation using Simpsons rule

Show the formula used to find this value correct to 4 decimal places.

Which formula, Trapezoidal or Simpsons, gives the better approximation when using the same number of terms in the expansion? Why?

Explanation / Answer

2)

Trapezoidal :

a = 0 , b = 2 , n = 4

h = (b - a)/n = (2 - 0)/4 = 0.5

So, the endpoints are : 0 , 1/2 , 1 , 3/2 , 2

y0 = f(0) =(0 - 3)^2 + 1 = 10

y1 = f(1/2) = (1/2 - 3)^2 + 1 = 29/4

y2 = f(1) = (1 - 3)^2 + 1 = 5

y3 = f(3/2) = (3/2 - 3)^2 + 1 = 13/4

y4 =(2 - 3)^2 + 1 = 2

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3) Now, the formula goes :

delta(x)/2 * (y0 + 2y1 + 2y2 +2y3 + y4) --> formula used

(0.5)/2 * (10 + 2(29/4) + 2(5) + 2(13/4) + 2)

10.75

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The exact same values of h,y0,y1,y2,y3 and y4 hold even for the simpson rule

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4)

But for simpson's rule, we use the formula as

delta(x)/3 * (y0 + 4y1 + 2y2 +4y3 + y4) ---> formula used

(0.5)/3 * (10 + 4(29/4) + 2(5) + 4(13/4) + 2)

10.6667 ----> ANSWER

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5) Exact value

The function (x - 3)^2 + 1 can be simplified as x^2 - 6x + 10

Now, we need to integrate :

(int from 0 to 2) x^2 - 6x + 10

(1/3)x^3 - 3x^2 + 10x (over 0 to 2)

(1/3)*(2^3 - 0^3) - 3(2^2 - 0^2) + 10(2 - 0)

10.6667 ---> EXACT VALUE

The simpson rule gave us this exact same answer. So, the simpson rule gave us a better approximation

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6)

f(x) = x^3 - 8

a = -1 , b = 2 , n = 6

delta(x) = (b - a)/n --> (2 - (-1)) / 6 = 0.5

So, endpoints are : -1 , -0.5 , 0 , 0.5 , 1 , 1.5 and 2

f(-1) = (-1)^3 - 8 = -9
f(-0.5) = (-0.5)^3 - 8 = -8.125
f(0) = 0^3 - 8 = -8
f(0.5) = (0.5)^3 - 8 = -7.875
f(1) = 1^3 - 8 = -7
f(1.5) = (1.5)^3 - 8 = -4.625
f(2) = (2)^3 - 8 = 0

Now, using trapezoidal, add them likt this :

-9 + 2(-8.125) + 2(-8) + 2(-7.875) + 2(-7) + 2(-4.625) + 0

-80.25

Now, to this, multiply delta(x)/2 :

-80.25 * 0.5/2

-20.0625 ---> ANSWER

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7) f(x) = sqrt(x) - 5

a = 0 , b = 3 , n = 6

delta(x) = (3 - 0)/6 = 0.5

f(0) = sqrt0 - 5 =-5
f(0.5) = sqrt(0.5) - 5 = -4.2928932188134525
f(1) = sqrt(1) - 5 = -4
f(1.5) = sqrt(1.5) - 5 = -3.775255128608411
f(2) = sqrt(2) - 5 = -3.585786437626905
f(2.5) = sqrt(2.5) - 5 = -3.4188611699158103
f(3) = sqrt(3) - 5 = -3.2679491924311227

Now, we add them likt this :

-5 + 4(-4.2928932188134525) + 2(-4) + 4(-3.775255128608411) + 2(-3.585786437626905) + 4(-3.4188611699158103) + -3.2679491924311227

-69.3875601370356279

And to this, multiply delta(x)/3 :

-69.3875601370356279 * (0.5) / 3

-11.56459335617260465 ------> ANSWER

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