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Use the Trapezium rule to estimate the area between the curve $y = x^2 -8x + 18$ and the $x$ axis from $x = 2$ to $x = 6$. Use $4$ strips of equal width.

What I did: height $= \frac{(b - a)}{n}$ $= \frac{(6 - 2)}{4} = 1$

$y_0 = 6 , y_1 = 3 , y_2 = 0 , y_3 = 3 , y_4 = 6$

$I = \frac{1}{2} [6 + 2(3+0+3)+6]$ $= 15 $ square units

But it says the answer is $14$.

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  • $\begingroup$ $1/2 [6 + 2(3+0+3)+6] = 12$, not $15$, but neither is it $14$, so I would double-check your values of $y_0,\dots,y_4$. $\endgroup$ – Matthew Leingang May 5 at 16:12
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You made a mistake in evaluating the function at $x=4$. You should have $y_2 = 2$ rather than $y_2 = 0$, so that you have

$$\frac{1}{2} [6 + 2(3+2+3)+6] = 14$$

as expected.

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  • $\begingroup$ Can you explain me how you get the value of y1 y2 ..... $\endgroup$ – Abhishek Kumar May 5 at 16:26
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    $\begingroup$ You substitute the values of x from 2 to 6 to the equation of $x^2−8x+18$ and your y values give you y0, y1, y2, y3 and y4. $\endgroup$ – xx_Gcsemathstudent_xx May 5 at 16:28
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First of all

height $= \frac{b-a}{n} = \frac{6-2}{4} = 1 $

Then your $x_0 = 2, x_1 = 3, x_2 = 4, x_3 = 5, x_4 = 6$

So the result is $I = \frac{1}{2} [y_0 + y_4 + 2(y_1 +y_2 +y_3)] =\frac{1}{2} [(6+6)+2(3+2+3)]= 14$

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