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 (1 / 64) \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 (9 / 64) \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 (1 / 64) \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 (9 / 64) \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 (M_4)

First, we know that (M_{n}=\Delta x\left[f(\overline{x}{1})+f(\overline{x}{2})+\cdots +f(\overline{x}_{n})\right]) with (\Delta x=\frac{1 - 0}{4}=\frac{1}{4}), and (f(x)=2\cos(x^{2})).

We have (f(\overline{x}{1}) = 2\cos\left(\frac{1}{64}\right)), (f(\overline{x}{2})=2\cos\left(\frac{9}{64}\right)), (f(\overline{x}{3}) = 2\cos\left(\frac{25}{64}\right)), (f(\overline{x}{4})=2\cos\left(\frac{49}{64}\right))

[ \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)\approx0.994427), (\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 + 0.994427+0.968912 + 0.891207)\ &=\frac{1}{2}(3.854351)\ &=1.927176 \end{align*} ]

Answer:

(1.927176)