Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places. 64 0 sin( x ) dx, n = 4

Answer:[tex]\int\limits^{64}_0 {sin(x)} \, dx\approx 4.9168[/tex]Step-by-step explanation:We want to approximate the integral[tex]\int\limits^{64}_0 {sin(x)} \, dx[/tex]using n = 4.The Midpoint Rule is given by[tex]\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right)[/tex]where [tex]\Delta{x}=\frac{b-a}{n}[/tex]We know that a = 0, b = 64 and n = 4. Therefore, [tex]\Delta{x}=\frac{64-0}{4}=16[/tex].We need to divide the interval [0, 64] into 4 sub-intervals of length [tex]\Delta x =16[/tex]; 0, 16, 32, 48, 64. Now, we just evaluate the function at these endpoints:[tex]f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{\left(0\right)+\left(16\right)}{2}\right)=f\left(8\right)=\sin{\left(8 \right)}=0.9894[/tex][tex]f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{\left(16\right)+\left(32\right)}{2}\right)=f\left(24\right)=\sin{\left(24 \right)}=-0.9056[/tex][tex]f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{\left(32\right)+\left(48\right)}{2}\right)=f\left(40\right)=\sin{\left(40 \right)}=0.7451[/tex][tex]f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\left(48\right)+\left(64\right)}{2}\right)=f\left(56\right)=\sin{\left(56 \right)}=-0.5216[/tex]Finally, just sum up the above values and multiply by [tex]\Delta x =16[/tex][tex]\int\limits^{64}_0 {sin(x)} \, dx \approx 16(0.9894-0.9056+0.7451-0.5216)=4.9168[/tex]