Find the approximate area of the shaded figure shown using Simpson's rule. Each of the equidistant parallel chords is measured from the base to a point on the curve. All units are expressed in km. enter image description here

So I tried this problems and they said the answer is $77km^2$ I got $137.76km^2$ I don't understand where I was wrong


$S.Rule=\frac{1}{3}2.89$$[(10+9)+4(22)+2(18)]$ I started from the right and got my interval by finding the base using soh cah toa by using $30^\circ$ and $10$ and by dividing by $6$ to get the interval $\frac{10\sqrt{3}}{6}$


There are 7 points (usually we need odd number of points for the Simpson's rule). The sum should be

$$\frac{10\cot30^{\circ}}{3} \times \frac{1}{6} (y_0 + 4 y_1 + 2y_2 + 4y_3 + 2y_4 + 4y_5 + y_6)$$

And the question actually asks for the shaded area, while the Simpson's rule give the total area of the shaded region and the right-angled triangle. So the formula is

$$\frac{10\cot30^{\circ}}{3} \times \frac{1}{6} (y_0 + 4 y_1 + 2y_2 + 4y_3 + 2y_4 + 4y_5 + y_6) - \frac{1}{2} \times 10 \times 10\cot30^{\circ}$$

The numerical answer is 76.980035892.

  • $\begingroup$ I have a question how did you get $\frac{10tan30^\circ}{3}$ this is your base right?? $\endgroup$ – Mickey Jan 1 '15 at 10:23
  • 1
    $\begingroup$ Sorry I made a mistake in simply copying and pasting. It is the base length and there is no need to be divided by 3. $\endgroup$ – Empiricist Jan 1 '15 at 10:33
  • $\begingroup$ $Tan30^\circ=\frac{10}{adjacent}$ so its $\frac{10}{tan30^\circ}$ right? $\endgroup$ – Mickey Jan 1 '15 at 10:36
  • $\begingroup$ Also where did $\frac{1}{6}$ come from? $\endgroup$ – Mickey Jan 1 '15 at 10:43
  • $\begingroup$ Just a different way to write the Simpson's rule. I usually regard it as $\frac{b-a}{6} (f(a) + 4f(\frac{a+b}{2}) + f(b)$. Say $a = x_0$ and $b = x_2$, with $\frac{a+b}{2} = x_1$ and $b-a = \frac{10 \tan 30^{\circ}}{3}$. $\endgroup$ – Empiricist Jan 1 '15 at 13:36

After solving multiple times and with the help of the comments

$S.Rule=\frac{1}{3}2.89$$[(10)+4(31)+2(18)]$ Starting from the left I divided my base to 6 and got $2.89$ as my interval. Finally I got $S.Rule=\frac{1}{3}2.89$$(170)$ - $86.62$ the area of the right triangle and finally got $77.14km^2$

  • $\begingroup$ I think it should be a good practice for not using approximation for intermediate values, i.e. the interval length and the area of the triangle. $\endgroup$ – Empiricist Jan 1 '15 at 13:42

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