Using the midpoint rule

Question

QUestion:

needed help here. i had tried it out but i get an answer that is not in the question answer list. i get 0.846474251 and i just assumed 0.63689453.

see my workings below i used a tabular method of finding midpoint approx.

Answer 1

You have $\Delta x=\frac{1}{6}$ and taking the function value at interval midpoints you get $$\int_0^1\sin\pi xdx\approx\sum_{n=0}^5\Delta x\cdot\sin\left(\pi\left( \frac{\Delta x}{2}+\frac{n}{6}\right)\right)=\sum_{n=0}^5\frac{1}{6}\cdot\sin\left( \pi\left(\frac{1}{12}+\frac{n}{6}\right)\right)$$$$=\frac{1}{6}\cdot(0.258819+0.707107+0.965926+0.965926+0.707107+0.258819)$$$$\approx0.643951$$