# Find the approximate area using Simpson's Rule

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. 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

$S.Rule=\frac{1}{3}d[(y^1+y^6)+4(y^3+y^5)+2(y^2+y^4)]$

$S.Rule=\frac{1}{3}2.89$$[(10+9)+4(22)+2(18)] I started from the right and got my interval by finding the base using soh cah toa by using 30^\circ and 10 and by dividing by 6 to get the interval \frac{10\sqrt{3}}{6} ## 2 Answers There are 7 points (usually we need odd number of points for the Simpson's rule). The sum should be$$\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}$$The numerical answer is 76.980035892. • I have a question how did you get \frac{10tan30^\circ}{3} this is your base right?? – Mickey Jan 1 '15 at 10:23 • Sorry I made a mistake in simply copying and pasting. It is the base length and there is no need to be divided by 3. – Empiricist Jan 1 '15 at 10:33 • Tan30^\circ=\frac{10}{adjacent} so its \frac{10}{tan30^\circ} right? – Mickey Jan 1 '15 at 10:36 • Also where did \frac{1}{6} come from? – Mickey Jan 1 '15 at 10:43 • Just a different way to write the Simpson's rule. I usually regard it as \frac{b-a}{6} (f(a) + 4f(\frac{a+b}{2}) + f(b). Say a = x_0 and b = x_2, with \frac{a+b}{2} = x_1 and b-a = \frac{10 \tan 30^{\circ}}{3}. – Empiricist Jan 1 '15 at 13:36 After solving multiple times and with the help of the comments S.Rule=\frac{1}{3}2.89$$[(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 $77.14km^2$

• 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. – Empiricist Jan 1 '15 at 13:42