# Trapezium Rule Exam Question

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 $$14$$.

• $1/2 [6 + 2(3+0+3)+6] = 12$, not $15$, but neither is it $14$, so I would double-check your values of $y_0,\dots,y_4$. – Matthew Leingang May 5 at 16:12

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.

• Can you explain me how you get the value of y1 y2 ..... – Abhishek Kumar May 5 at 16:26
• 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. – xx_Gcsemathstudent_xx May 5 at 16:28

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