Answer to Question #283851 in Quantitative Methods for Ar-Ar

Question

Evaluate the following integral: 𝑰 = ∫ 𝒙𝐬𝐢𝐧(𝒙)𝒅𝒙 𝝅/𝟒

𝟎

(1) analytically, (2) using single application of the trapezoidal rule, (3) using composite

trapezoidal rule with n = 2 and 4. For the numerical estimates (2) and (3), determine the true

percent relative error based on (1).

Answer 1

1)

$\int^{\pi/4}_0 xsinx dx=-xcosx|^{\pi/4}_0+\int^{\pi/4}_0 cosx dx=$

$=-xcosx|^{\pi/4}_0+sinx|^{\pi/4}_0=-\frac{\pi}{4\sqrt 2}+\frac{1}{\sqrt 2}=0.152$

2)

$f(x)=xsinx$

$I=(b-a)\frac{f(a)+f(b)}{2}=\frac{\pi}{4}\frac{\pi}{2\cdot4\sqrt 2}=0.436$

3)

for n = 2:

$I=\frac{\pi}{8}(\frac{f(0)+f(\pi/8)}{2}+\frac{f(\pi/8)+f(\pi/4)}{2})=\frac{\pi}{8}(\pi sin(\pi/8)/8+\pi sin(\pi/4)/8)=0.021$

for n = 4:

$I=\frac{\pi}{16}(\frac{f(0)+f(\pi/16)}{2}+\frac{f(\pi/16)+f(\pi/8)}{2}+\frac{f(\pi/8)+f(3\pi/16)}{2}+\frac{f(3\pi/16)+f(\pi/4)}{2})=$

$=\frac{\pi}{16}(\pi sin(\pi/16)/16+\pi sin(\pi/8)/8+3\pi sin(3\pi/16)/16+\pi sin(\pi/4)/8)=$

$=0.616(0.012+0.048+0.104+0.88)=0.155$

relative error:

for single application:

$\frac{0.436-0.152}{0.152}=1.87=187\%$

for n = 2:

$\frac{0.152-0.021}{0.152}=0.86=86\%$

for n = 4:

$\frac{0.155-0.152}{0.152}=0.02=2\%$