I need help using simpsons rule

[kggirl]

evaluate the integral from 0 to 1 xsin[x] dx where n = 4 steps.

I don't know where to begin with this one.

 

[soroban]

Hello, kggirl!

Simpson's Rule
Evaluate: \(\displaystyle \L\int^{\;\;\;\;1}_0 x\cdot\sin(x)\,dx\;\) where $\displaystyle \,n = 4\,$ steps.

I don't know where to begin with this one. . . . . really?

Do you know the formula for Simpson's Rule?

$\displaystyle A\;\approx\;\frac{h}{3}\left(y_0\,+\,4y_1\,+\,2y_2\,+\,4y_3\,+\,\cdots\,+\,2y_{n-2}\,+\,4y_{n-1}\,+\,y_n\right)$

where $\displaystyle h$ is the width of the intervals, and $\displaystyle y_i$ are the "heights" of the function at each $\displaystyle x_i$.


With 4 steps, we have: $\displaystyle \;\begin{array}{ccccc}x_0\,=\,0.00\\x_1\,=\,0.25 \\x_2\,=\,0.50\\x_3\,=\,0.75\\x_4\,=\,1.00\end{array}\;\;\Rightarrow\;\;\begin{array}{cccc}y_0\,=\,0.00\cdot\sin(0.00)\,= 0.0000\\ y_1\,=\,0.25\cdot\sin(0.25)\,\approx\,0.0619 \\ y_2\,=\,0.50\cdot\sin(0.50)\,\approx\,0.2397\\ y_3\,=\,0/75\cdot\sin(0.75)\,\approx\,0.5112 \\ y_4\,=\,1.00\cdot\sin(1.00)\,\approx\,0.8415\end{array}$

. . and we have: $\displaystyle \;h\,=\,0.25$


Hence: $\displaystyle \;A\;\approx\;\frac{0.25}{3}[0\,+\,4(0.0619)\,+\,2(0.2397)\,+\,4(0.5122)\,+\,0.8415]\;=\;0.301108333$

Therefore: \(\displaystyle \L\;A\;\approx\;0.3011\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The actual area is: \(\displaystyle \L\int^{\;\;\;\;1}_0 x\cdot\sin(x)\,dx\;=\;0.301168679\)

. . Our approximation is very good!

 

[kggirl]

Thank you, i was trying to solve using h = 1/4 so my answers are different than yours. Thanks again.