the midpoint rule says that\n int_{a}^{b} f(x) d x=m_{n}=delta xleftfleft(overline{x_{1}}\right)+fleft(overli…

the midpoint rule says that\n int_{a}^{b} f(x) d x=m_{n}=delta xleftfleft(overline{x_{1}}\right)+fleft(overline{x_{2}}\right)+ldots+fleft(overline{x_{n}}\right)\right \text { with } delta x=\frac{b-a}{n} \nwe again need to estimate ( int_{0}^{1} 2 cos left(x^{2}\right) d x ) with ( n=4 ) subintervals.\nfor this,\n delta x=\frac{1-0}{4}=0.25 \nstep 2\nwe know that ( overline{x_{1}} ) represents the midpoint of the first subinterval ( left0, \frac{1}{4}\right ).\nthe midpoint of ( left0, \frac{1}{4}\right ) is\n overline{x_{1}}=\frac{\frac{1}{4}+0}{2}=1 / 8 \nstep 3\nour function is ( f(x)=2 cos left(x^{2}\right) ). so we have\n fleft(overline{x_{1}}\right)=2 cos left(\frac{1}{64}\right) \nstep 4\nsimilarly, the second subinterval is ( left\frac{1}{4}, \frac{1}{2}\right ).\nthe midpoint of this subinterval is\n overline{x_{2}}=\frac{\frac{1}{4}+\frac{1}{2}}{2}=3 / 8 \nand\n fleft(overline{x_{2}}\right)=2 cos left(\frac{9}{64}\right) \nstep 5\nthis gives us\n \begin{aligned} m_{4} &=\frac{1}{4}left2 cos left(\frac{1}{64}\right)+2 cos left(\frac{9}{64}\right)+2 cos left(\frac{25}{64}\right)+2 cos left(\frac{49}{64}\right)\right \\ &= end{aligned} \n(rounded to six decimal places)\ntherefore, using the midpoint rule with ( n=4 ) and rounding to six decimal places we have\n int_{0}^{1} 2 cos left(x^{2}\right) d x=

the midpoint rule says that\n int_{a}^{b} f(x) d x=m_{n}=delta xleftfleft(overline{x_{1}}\right)+fleft(overline{x_{2}}\right)+ldots+fleft(overline{x_{n}}\right)\right \text { with } delta x=\frac{b-a}{n} \nwe again need to estimate ( int_{0}^{1} 2 cos left(x^{2}\right) d x ) with ( n=4 ) subintervals.\nfor this,\n delta x=\frac{1-0}{4}=0.25 \nstep 2\nwe know that ( overline{x_{1}} ) represents the midpoint of the first subinterval ( left0, \frac{1}{4}\right ).\nthe midpoint of ( left0, \frac{1}{4}\right ) is\n overline{x_{1}}=\frac{\frac{1}{4}+0}{2}=1 / 8 \nstep 3\nour function is ( f(x)=2 cos left(x^{2}\right) ). so we have\n fleft(overline{x_{1}}\right)=2 cos left(\frac{1}{64}\right) \nstep 4\nsimilarly, the second subinterval is ( left\frac{1}{4}, \frac{1}{2}\right ).\nthe midpoint of this subinterval is\n overline{x_{2}}=\frac{\frac{1}{4}+\frac{1}{2}}{2}=3 / 8 \nand\n fleft(overline{x_{2}}\right)=2 cos left(\frac{9}{64}\right) \nstep 5\nthis gives us\n \begin{aligned} m_{4} &=\frac{1}{4}left2 cos left(\frac{1}{64}\right)+2 cos left(\frac{9}{64}\right)+2 cos left(\frac{25}{64}\right)+2 cos left(\frac{49}{64}\right)\right \\ &= end{aligned} \n(rounded to six decimal places)\ntherefore, using the midpoint rule with ( n=4 ) and rounding to six decimal places we have\n int_{0}^{1} 2 cos left(x^{2}\right) d x=

Answer

Explanation:

Step1: Calculate the mid - points of the remaining sub - intervals

The third sub - interval is ([\frac{1}{2},\frac{3}{4}]). The mid - point (\overline{x}{3}=\frac{\frac{1}{2}+\frac{3}{4}}{2}=\frac{\frac{2 + 3}{4}}{2}=\frac{5}{8}). Then (f(\overline{x}{3})=2\cos((\frac{5}{8})^{2})=2\cos(\frac{25}{64})). The fourth sub - interval is ([\frac{3}{4},1]). The mid - point (\overline{x}{4}=\frac{\frac{3}{4}+1}{2}=\frac{\frac{3 + 4}{4}}{2}=\frac{7}{8}). Then (f(\overline{x}{4})=2\cos((\frac{7}{8})^{2})=2\cos(\frac{49}{64})).

Step2: Apply the Mid - point Rule formula

We know that (M_{n}=\Delta x\sum_{i = 1}^{n}f(\overline{x}{i})), where (\Delta x=\frac{1-0}{4}=\frac{1}{4}), (n = 4), (f(x)=2\cos(x^{2})), (\overline{x}{1}=\frac{1}{8}), (\overline{x}{2}=\frac{3}{8}), (\overline{x}{3}=\frac{5}{8}), (\overline{x}{4}=\frac{7}{8}). [ \begin{align*} M{4}&=\frac{1}{4}\left[2\cos\left(\frac{1}{64}\right)+2\cos\left(\frac{9}{64}\right)+2\cos\left(\frac{25}{64}\right)+2\cos\left(\frac{49}{64}\right)\right]\ &=\frac{1}{2}\left[\cos\left(\frac{1}{64}\right)+\cos\left(\frac{9}{64}\right)+\cos\left(\frac{25}{64}\right)+\cos\left(\frac{49}{64}\right)\right] \end{align*} ] Using a calculator: (\cos\left(\frac{1}{64}\right)\approx0.999805), (\cos\left(\frac{9}{64}\right)\approx0.994437), (\cos\left(\frac{25}{64}\right)\approx0.968912), (\cos\left(\frac{49}{64}\right)\approx0.891207) [ \begin{align*} M_{4}&=\frac{1}{2}(0.999805 + 0.994437+0.968912 + 0.891207)\ &=\frac{1}{2}(3.854361)\ &=1.927181 \end{align*} ]

Answer:

(1.927181)