# How do you use the trapezoidal rule with n=4 to approximate the area between the curve sqrt(x) sinx from pi/2 to pi?

Nov 9, 2016

${\int}_{\frac{\pi}{2}}^{\pi} \sqrt{x} \sin x \mathrm{dx} \approx 1.430$ (3dp)

#### Explanation:

The values of y are tabulated as follows (using Excel)

Using the trapezoidal rule:
${\int}_{a}^{b} y \mathrm{dx} \approx \frac{h}{2} \left\{\left({y}_{0} + {y}_{n}\right) + 2 \left({y}_{1} + {y}_{2} + \ldots + {y}_{n - 1}\right)\right\}$

we have:
${\int}_{\frac{\pi}{2}}^{\pi} \sqrt{x} \sin x \mathrm{dx} \approx \frac{\frac{\pi}{8}}{2} \left\{\left(1.25331 + 0.00000\right) + 2 \left(1.29458 + 1.08540 + 0.63448\right)\right\}$
$\therefore {\int}_{\frac{\pi}{2}}^{\pi} \sqrt{x} \sin x \mathrm{dx} \approx \frac{\pi}{16} \left\{1.25331 + 2 \left(3.01446\right)\right\}$
$\therefore {\int}_{\frac{\pi}{2}}^{\pi} \sqrt{x} \sin x \mathrm{dx} \approx \frac{\pi}{16} \left\{1.25331 + 6.02892\right\}$
$\therefore {\int}_{\frac{\pi}{2}}^{\pi} \sqrt{x} \sin x \mathrm{dx} \approx \frac{\pi}{16} \left\{7.28223\right\}$
$\therefore {\int}_{\frac{\pi}{2}}^{\pi} \sqrt{x} \sin x \mathrm{dx} \approx 1.42986$
$\therefore {\int}_{\frac{\pi}{2}}^{\pi} \sqrt{x} \sin x \mathrm{dx} \approx 1.430$ (3dp)