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Calculus: Simpson’s Rule The figure shows the graph of y=ax2+bx+c. Suppose that the points localid="1647353664782" -h,y0,0,yandh,y2are on the graph. It can be shown that the area enclosed by the parabola, the localid="1647353732361" x-axis, and the lineslocalid="1647353669382" x=-h,x=his Area localid="1647353674576" =h32ah2+6c.Show that this area may also be given by Area localid="1647353678398" =h3y0+4y1+y2.

Short Answer

Expert verified

y=ax2+bx+cIt is proven that arealocalid="1647353590732" =h3y0+4y1+y2.

Step by step solution

01

Step 1. Given information

The given figure is,

The figure shows the graph y=ax2+bx+c.Suppose that the point-h,y0,0,yandh,y2is on the graph and the area is an area=h32ah2+6c.We need to show that the area may be given by Area=h3y0+4y1+y2.

02

Step 2. Simplify

The figure is y=ax2+bx+c.

For the point -h,y0,the function is,

y0=a·-h2+b-h+c.

role="math" localid="1647352990982" y0=h2-bh+c⋯(1).

For the point 0,y1,the function is,

y1=c⋯(2).

For the point h,y2the function is,

y2=ah2+bh+c⋯(3).

Add equations (1),(3) and include equation (2).

y0+y2=2·a·h2+2·y1.

2·a·h2=y0+y2-2y1⋯(4).

Now include equation (4) in the equation of the area.

Area role="math" localid="1647353241247" =h32ah2+6c.

=h3y0+4y1-2y1+6y1.

=h3y0+4y1+y2.

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