question on exponentials (1 Viewer)

[hannahxxx]

Hey guys! I've done the following question about 6 times and it's driving me insane( the book says the answer is 7.4 but the only solutions I came up with was 40.7 and 303.8). The q is : Use simpson's rule with 3 function values to find an approximation for the primitive function of xe^x, between x=1 and x=2, correct to 1 dp.

Also, does anyone know why you get different answers if you use the multiple subinterval formula for simpson's rule, rather than the the one designed for just 3 subintervals(i.e. for one application). Cheers

 

[alexvincent]

Your 3 x values are 1.0 , 1.5 , 2.0
for x = 1.0 , y = e ---- y0
for x = 1.5 , y = 1.5e^1.5 ----- y1
for x = 2.0 , y = 2e^2 ----- yn

Simpson's rule: integral approx = h/3[ y0 + yn + 4(odd) + 2(even)]
= 0.5/3[ e + 2e^2 + 4(1.5e^1.5) + 0]
= 7.39775

Careful with your calculator work.

 

[alexvincent]

Also, does anyone know why you get different answers if you use the multiple subinterval formula for simpson's rule, rather than the the one designed for just 3 subintervals(i.e. for one application). Cheers

This function is a curve. When you use Simpson's rule you only take shapes/strips under the curve which are formed with straight lines as it sides. Meaning, you have boundaries for these strips (for this example 1.0, 1.5, 2.0). Vertical lines are drawn from these x values on the x-axis up to the function. Then these points where these vertical lines meet the function are joined together by constructing a straight line between them. Your are finding the area of all these shapes and not the area under the curve, using Simpson's rule you will get a value under the exact value.

Using more of these strips means you have more subintervals/boundaries. This minimises the area that you miss when you only find the area of the strips.

I think I might have been a bit confusing. But there should be stuff on this in any 2u text book.

 

[hannahxxx]

Your 3 x values are 1.0 , 1.5 , 2.0
for x = 1.0 , y = e ---- y0
for x = 1.5 , y = 1.5e^1.5 ----- y1
for x = 2.0 , y = 2e^2 ----- yn

Simpson's rule: integral approx = h/3[ y0 + yn + 4(odd) + 2(even)]
= 0.5/3[ e + 2e^2 + 4(1.5e^1.5) + 0]
= 7.39775

Careful with your calculator work.

Yea I can see how it works with that one-thanks for the help. But why use the multiple subinterval formula when there is a simpler one? why does it give different values(for me it does, anyway)

 

[alexvincent]

I don't know exactly what you mean... what simpler one? Different values for what?

 

[ThuanSUX]

There is no difference between the multiple subinterval and the simple formula. The complete Simpson's rule is given by:

f'(x) = h/3[(y0+yn)+4(odd)+2(even)]

Whereas the simplified version is designed for a function with ONLY 3 subintervals. Taking this into account, you only have y0, y1 and yn. Thus substituting it into the above formula and you get:

f'(x) = h/3[(y0+yn)+4(y1)] or h/3[y0+4y1+yn]

(note that this formula is a standard formula used in General Maths - they do not use the multiple subinterval formula). In 2u maths you should be using the first formula beacuse it's straight forward and can be done in one go. You can only use the simplified formula if the number of subintervals are in multiples of 3 (hence if you have 6, you can use the simplified formula twice...etc).

 

[hannahxxx]

K thanks 4 that