step 2\nwe know that ( overline{x_{1}} ) represents the midpoint of the first subinterval ( left0…

step 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}=\frac{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}=\frac{3}{8} ).\nand\n( fleft(overline{x_{2}}\right)=2 cos left(\frac{9}{64}\right) ).\nstep 5\nthis gives us\n( 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 )\n( =quad ) (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=quad ).

step 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}=\frac{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}=\frac{3}{8} ).\nand\n( fleft(overline{x_{2}}\right)=2 cos left(\frac{9}{64}\right) ).\nstep 5\nthis gives us\n( 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 )\n( =quad ) (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=quad ).

Answer

Explanation:

Step1: Calculate the width of each sub - interval

The interval is ([a,b]=[0,1]) and (n = 4). The width of each sub - interval (\Delta x=\frac{b - a}{n}=\frac{1-0}{4}=\frac{1}{4})

Step2: Find the mid - points of sub - intervals

For (n = 4), the sub - intervals are ([0,\frac{1}{4}],[\frac{1}{4},\frac{1}{2}],[\frac{1}{2},\frac{3}{4}],[\frac{3}{4},1]) The mid - points (x_i^) are: (x_1^=\frac{0+\frac{1}{4}}{2}=\frac{1}{8}), (x_2^=\frac{\frac{1}{4}+\frac{1}{2}}{2}=\frac{3}{8}), (x_3^=\frac{\frac{1}{2}+\frac{3}{4}}{2}=\frac{5}{8}), (x_4^*=\frac{\frac{3}{4}+1}{2}=\frac{7}{8})

Step3: Evaluate the function at mid - points

The function (f(x)=2\cos(x^{2})) (f(x_1^)=2\cos((\frac{1}{8})^{2})=2\cos(\frac{1}{64})) (f(x_2^)=2\cos((\frac{3}{8})^{2})=2\cos(\frac{9}{64})) (f(x_3^)=2\cos((\frac{5}{8})^{2})=2\cos(\frac{25}{64})) (f(x_4^)=2\cos((\frac{7}{8})^{2})=2\cos(\frac{49}{64}))

Step4: Apply the Midpoint Rule formula

The Midpoint Rule formula is (M_n=\Delta x\sum_{i = 1}^{n}f(x_i^*)) Here, (\Delta x=\frac{1}{4}) and (n = 4) (M_4=\frac{1}{4}\left[2\cos(\frac{1}{64})+2\cos(\frac{9}{64})+2\cos(\frac{25}{64})+2\cos(\frac{49}{64})\right]) (M_4=\frac{1}{2}\left[\cos(\frac{1}{64})+\cos(\frac{9}{64})+\cos(\frac{25}{64})+\cos(\frac{49}{64})\right]) Using a calculator: (\cos(\frac{1}{64})\approx0.999922), (\cos(\frac{9}{64})\approx0.994008), (\cos(\frac{25}{64})\approx0.952413), (\cos(\frac{49}{64})\approx0.787714) (M_4=\frac{1}{2}(0.999922 + 0.994008+0.952413 + 0.787714)) (M_4=\frac{1}{2}(3.734057)) (M_4 = 1.867029)

Answer:

(\int_{0}^{1}2\cos(x^{2})dx\approx1.867029)