Calculus Area between curve and x-axis: curve y=a+bx+cx^2 drawn through (-h,y_1),(0,y_2),(h,y_3); find a,b,c in terms of y_1,y_2,y_3,h, and...

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reggiwilliams

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I am revising Calculus at present and area between curve and axis.
Came across this question which involves integration but not sure how to prove
Can anyone assist pleaseIMG_5918.jpeg
 
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reggiwilliams said:
I am revising Calculus at present and area between curve and axis.
Came across this question which involves integration but not sure how to prove
Can anyone assist please
Click to expand...
First task is to calculate the values of a, b & c .

You are given co-ordinates of three points. Using those co-ordinates calculate the equation of the curve passing through those three points.

Please show your work.
 
reggiwilliams said:
So I get b=y1-y3/2h and c =3y^3-2y^2-y1/h^2
Where do I go from there?
Click to expand...
You're missing some parentheses.
b = (y1-y3)/(2h)

c isn't correct. You shouldn't have any square or cube terms in the numerator.

Use the underscore to denote subscripts e.g. y_2

After that, you can set up the integral and then integrate.
 
reggiwilliams said:
No matter I do can’t get to answer Hate this question
Click to expand...
First, y1−y3=−2bh  ⟹  b=y1−y3−2hy_1 - y _3 = -2bh \implies b = \dfrac{y_1-y_3}{-2h}

It follows c=y1−2y2+y32h2c = \dfrac{y_1-2y_2+y_3}{2h^2}

I didn't catch the negative sign from bb. The rest should flow through and things will cancel.
 
I’m sorry but even following your calculations for b and c still can’t arrive at answer given for area
Could I please ask you for your full workings so that I can understand where I have gone wrong
 
reggiwilliams said:
I’m sorry but even following your calculations for b and c still can’t arrive at answer given for area
Could I please ask you for your full workings so that I can understand where I have gone wrong
Click to expand...
I suggest that you first integrate the original function (using a, b, and c) from -h to h and write an expression for the area in terms of a, b, c; then replace those with the expressions you were given for them.

It looks like you swapped y2 and y3 in the last term in the first line of your work (and in your previous work). Use the expressions in #14.

Then show us all of your work (starting after we got expressions for a, b, c, so we can check for errors (and you can, too).
 
∫−hhy2+(y3−y12h)x+(y1−2y2+y32h2)x2 dx\int_{-h}^h y_2 +\left(\dfrac{y_3-y_1}{2h}\right)x + \left(\dfrac{y_1-2y_2+y_3}{2h^2}\right)x^2 \, dx
=y2x+(y3−y14h)x2+(y1−2y2+y36h2)x3∣−hh=[y2h+(y3−y14h)h2+(y1−2y2+y36h2)h3]−[−y2h+(y3−y14h)(−h)2−(y1−2y2+y36h2)h3]=y_2x +\left(\dfrac{y_3-y_1}{4h}\right)x^2+\left(\dfrac{y_1-2y_2+y_3}{6h^2}\right)x^3 \biggr\rvert_{-h}^{h} \\ =\left[ y_2h +\cancel{\left(\dfrac{y_3-y_1}{4h}\right)h^2}+\left(\dfrac{y_1-2y_2+y_3}{6h^2}\right)h^3\right] - \left[ -y_2h + \cancel{\left(\dfrac{y_3-y_1}{4h}\right)(-h)^2}-\left(\dfrac{y_1-2y_2+y_3}{6h^2}\right)h^3\right]
=[h×6y2+y1−2y2+y36]−[h×−6y2−y1+2y2−y36]=\left[h\times \dfrac{6y_2+y_1-2y_2+y_3}{6}\right] - \left[h \times \dfrac{-6y_2-y_1+2y_2-y_3}{6}\right]
h6[(4y2+y1+y3)−(−4y2−y1−y3)]=h6(8y2+2y1+2y3)=h3(4y2+y1+y3)\dfrac{h}{6} \left[\left(4y_2+y_1+y_3\right)-(-4y_2-y_1-y_3)\right] = \dfrac{h}{6}(8y_2+2y_1+2y_3)= \dfrac{h}{3}(4y_2+y_1+y_3)
 
Dr.Peterson said:
Note that the result differs from the claim in the problem, which lacked a coefficient of 4.

The correct result amounts to Simpson's Rule for n=2.

I confirmed it in Desmos:

Desmos | Graphing Calculator

www.desmos.com www.desmos.com
Click to expand...
My workings are shown and clearly answer should be h/3(4y2+y1+y3) No wonder I could never prove answer given in question
Thanks so much for your help
Hopefully I won’t get such a tricky question in my exam
 
BigBeachBanana said:
∫−hhy2+(y3−y12h)x+(y1−2y2+y32h2)x2 dx\int_{-h}^h y_2 +\left(\dfrac{y_3-y_1}{2h}\right)x + \left(\dfrac{y_1-2y_2+y_3}{2h^2}\right)x^2 \, dx
=y2x+(y3−y14h)x2+(y1−2y2+y36h2)x3∣−hh=[y2h+(y3−y14h)h2+(y1−2y2+y36h2)h3]−[−y2h+(y3−y14h)(−h)2−(y1−2y2+y36h2)h3]=y_2x +\left(\dfrac{y_3-y_1}{4h}\right)x^2+\left(\dfrac{y_1-2y_2+y_3}{6h^2}\right)x^3 \biggr\rvert_{-h}^{h} \\ =\left[ y_2h +\cancel{\left(\dfrac{y_3-y_1}{4h}\right)h^2}+\left(\dfrac{y_1-2y_2+y_3}{6h^2}\right)h^3\right] - \left[ -y_2h + \cancel{\left(\dfrac{y_3-y_1}{4h}\right)(-h)^2}-\left(\dfrac{y_1-2y_2+y_3}{6h^2}\right)h^3\right]
=[h×6y2+y1−2y2+y36]−[h×−6y2−y1+2y2−y36]=\left[h\times \dfrac{6y_2+y_1-2y_2+y_3}{6}\right] - \left[h \times \dfrac{-6y_2-y_1+2y_2-y_3}{6}\right]
h6[(4y2+y1+y3)−(−4y2−y1−y3)]=h6(8y2+2y1+2y3)=h3(4y2+y1+y3)\dfrac{h}{6} \left[\left(4y_2+y_1+y_3\right)-(-4y_2-y_1-y_3)\right] = \dfrac{h}{6}(8y_2+2y_1+2y_3)= \dfrac{h}{3}(4y_2+y_1+y_3)
Click to expand...
Thanks so much for your help
 
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