Estimate \( \int_{0}^{1} 5 \cos \left(x^{2}\right) d x \) using the Trapezoidal Rule and the Midpoint Rule, each with \( n=4 \). (Round your answers to six decimal places (a) the Trapezoidal Rule (b) the Midpoint Rule From a graph of the integrand, decide whether your answers are underestimates or overestimates. \( T_{4} \) is an underestimate \( T_{4} \) is an overestimate \( M_{4} \) is an underestimate \( M_{4} \) is an overestimate What can you conclude about the true value of the integral? (Round your answers to six decimal places.) ( Need Help? Readit

We wish to approximate \[ I=\int_{0}^{1}5\cos\left(x^2\right)\,dx \] using both the Trapezoidal Rule and the Midpoint Rule with \( n=4 \) subintervals. In both cases the width is \[ h=\frac{1-0}{4}=0.25. \] --- **(a) The Trapezoidal Rule** The nodes are \[ x_0=0,\quad x_1=0.25,\quad x_2=0.5,\quad x_3=0.75,\quad x_4=1. \] Our function is \[ f(x)=5\cos\left(x^2\right). \] We compute - At \( x_0=0 \): \[ f(0)=5\cos(0)=5. \] - At \( x_1=0.25 \): \[ f(0.25)=5\cos\left((0.25)^2\right)=5\cos(0.0625). \] Since \( \cos(0.0625)\approx0.99804751 \), we have \[ f(0.25)\approx5(0.99804751)=4.99023755. \] - At \( x_2=0.5 \): \[ f(0.5)=5\cos(0.25). \] Using \( \cos(0.25)\approx0.96891242 \), \[ f(0.5)\approx5(0.96891242)=4.84456210. \] - At \( x_3=0.75 \): \[ f(0.75)=5\cos(0.5625). \] With \( \cos(0.5625)\approx0.846989 \) (approximately), \[ f(0.75)\approx5(0.846989)=4.234945. \] - At \( x_4=1 \): \[ f(1)=5\cos(1). \] Since \( \cos(1)\approx0.54030231 \), \[ f(1)\approx5(0.54030231)=2.70151155. \] The trapezoidal rule formula is \[ T_4=\frac{h}{2}\Big[f(x_0)+f(x_4)+2\big(f(x_1)+f(x_2)+f(x_3)\big)\Big]. \] Substitute the numerical values: \[ \begin{array}{rcl} T_4 & = & \displaystyle\frac{0.25}{2}\Big[5+2.70151155+2(4.99023755+4.84456210+4.234945)\Big] \\[1mm] & = & 0.125\Big[7.70151155+2(14.06974465)\Big] \\[1mm] & = & 0.125\Big[7.70151155+28.13948930\Big] \\[1mm] & = & 0.125(35.84100085) \\[1mm] & \approx & 4.48012511. \end{array} \] Thus, \[ T_4\approx4.480125. \] --- **(b) The Midpoint Rule** For the midpoint rule with \( n=4 \) the subintervals are: 1. \([0,0.25]\) with midpoint \( x_{1}^{*}=0.125 \), 2. \([0.25,0.5]\) with midpoint \( x_{2}^{*}=0.375 \), 3. \([0.5,0.75]\) with midpoint \( x_{3}^{*}=0.625 \), 4. \([0.75,1]\) with midpoint \( x_{4}^{*}=0.875 \). We then need to compute: - At \( x=0.125 \): \[ f(0.125)=5\cos\left((0.125)^2\right)=5\cos(0.015625). \] Since \( \cos(0.015625)\approx0.99987793 \), \[ f(0.125)\approx5(0.99987793)=4.99938965. \] - At \( x=0.375 \): \[ f(0.375)=5\cos\left(0.375^2\right)=5\cos(0.140625). \] With \( \cos(0.140625)\approx0.990082 \) (approximately), \[ f(0.375)\approx5(0.990082)=4.95041. \] - At \( x=0.625 \): \[ f(0.625)=5\cos\left(0.625^2\right)=5\cos(0.390625). \] Using \( \cos(0.390625)\approx0.924 \) (approximately), \[ f(0.625)\approx5(0.924)=4.62. \] - At \( x=0.875 \): \[ f(0.875)=5\cos\left(0.875^2\right)=5\cos(0.765625). \] With \( \cos(0.765625)\approx0.722 \) (approximately), \[ f(0.875)\approx5(0.722)=3.61. \] The midpoint rule is given by \[ M_4=h\Big[f(0.125)+f(0.375)+f(0.625)+f(0.875)\Big]. \] Substitute the values: \[ \begin{array}{rcl} M_4 & = & 0.25\Big[4.99938965+4.95041+4.62+3.61\Big] \\[1mm] & = & 0.25\Big[18.17979965\Big] \\[1mm] & \approx & 4.54494991. \end{array} \] Thus, \[ M_4\approx4.544950. \] --- **Graphical Considerations and Error** Since \( f(x)=5\cos\left(x^2\right) \) is a smooth function on \([0,1]\), we next consider its concavity to decide if the