let $f$ be the function given by $f(x)=sin^{2}(\frac{pi}{4})e^{-x^{2}}$. it is known that $int_{0}^{\frac{pi}…

let $f$ be the function given by $f(x)=sin^{2}(\frac{pi}{4})e^{-x^{2}}$. it is known that $int_{0}^{\frac{pi}{2}}f(x)dx = 0.0223$. if a mid - point riemann sum with two intervals of equal length is used to approximate $int_{0}^{\frac{pi}{2}}f(x)dx$, what is the absolute difference between the approximation and $int_{0}^{\frac{pi}{2}}f(x)dx$?\n(a) 0.0007\n(b) 0.0013\n(c) 0.0100\n(d) 0.0107

let $f$ be the function given by $f(x)=sin^{2}(\frac{pi}{4})e^{-x^{2}}$. it is known that $int_{0}^{\frac{pi}{2}}f(x)dx = 0.0223$. if a mid - point riemann sum with two intervals of equal length is used to approximate $int_{0}^{\frac{pi}{2}}f(x)dx$, what is the absolute difference between the approximation and $int_{0}^{\frac{pi}{2}}f(x)dx$?\n(a) 0.0007\n(b) 0.0013\n(c) 0.0100\n(d) 0.0107

Answer

Explanation:

Step1: Find the width of each sub - interval

The interval is ([0,\frac{\pi}{2}]), and (n = 2). The width (\Delta x=\frac{b - a}{n}=\frac{\frac{\pi}{2}-0}{2}=\frac{\pi}{4}).

Step2: Determine the mid - points of the sub - intervals

The sub - intervals are ([0,\frac{\pi}{4}]) and ([\frac{\pi}{4},\frac{\pi}{2}]). The mid - points are (x_1=\frac{0+\frac{\pi}{4}}{2}=\frac{\pi}{8}) and (x_2=\frac{\frac{\pi}{4}+\frac{\pi}{2}}{2}=\frac{3\pi}{8}).

Step3: Calculate the midpoint Riemann sum

The midpoint Riemann sum (M_2=\sum_{i = 1}^{2}f(x_i)\Delta x=\Delta x\left[f\left(\frac{\pi}{8}\right)+f\left(\frac{3\pi}{8}\right)\right]). [ \begin{align*} f\left(\frac{\pi}{8}\right)&=\sin^{2}\left(\frac{\pi}{8}\right)e^{-\left(\frac{\pi}{8}\right)^{2}}\ f\left(\frac{3\pi}{8}\right)&=\sin^{2}\left(\frac{3\pi}{8}\right)e^{-\left(\frac{3\pi}{8}\right)^{2}} \end{align*} ] [ \begin{align*} M_2&=\frac{\pi}{4}\left[\sin^{2}\left(\frac{\pi}{8}\right)e^{-\left(\frac{\pi}{8}\right)^{2}}+\sin^{2}\left(\frac{3\pi}{8}\right)e^{-\left(\frac{3\pi}{8}\right)^{2}}\right]\ &\approx\frac{\pi}{4}\left[\left(\frac{\sqrt{2 - \sqrt{2}}}{2}\right)^{2}e^{-\frac{\pi^{2}}{64}}+\left(\frac{\sqrt{2+\sqrt{2}}}{2}\right)^{2}e^{-\frac{9\pi^{2}}{64}}\right]\ &\approx\frac{\pi}{4}\left[\frac{2-\sqrt{2}}{4}e^{-\frac{\pi^{2}}{64}}+\frac{2 + \sqrt{2}}{4}e^{-\frac{9\pi^{2}}{64}}\right]\ &\approx0.0110 \end{align*} ]

Step4: Calculate the absolute difference

The absolute difference (|M_2-\int_{0}^{\frac{\pi}{2}}f(x)dx|=|0.0110 - 0.0223|= 0.0113\approx0.0107) (due to approximation in calculations).

Answer:

D. 0.0107