Evaluate the Riemann sum for f(x) = 3x − 1, −6 ≤ x ≤ 4, with five subintervals, taking the sample points to be right endpoints.

Answer:The Riemann sum equals -10.Step-by-step explanation:The right Riemann Sum uses the right endpoints of a sub-interval:[tex]\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_1)+f(x_2)+f(x_3)+...+f(x_{n-1})+f(x_{n})\right)[/tex]where[tex]\Delta{x}=\frac{b-a}{n}[/tex]To find the Riemann sum for [tex]\int\limits^{4}_{-6} {3x-1} \, dx[/tex] with n = 5 rectangles, using right endpoints you must:We know that a = -6, b = 4 and n = 5, so [tex]\Delta{x}=\frac{4-\left(-6\right)}{5}=2[/tex]We need to divide the interval −6 ≤ x ≤ 4 into n = 5 sub-intervals of length [tex]\Delta{x}=2[/tex][tex]a=\left[-6, -4\right], \left[-4, -2\right], \left[-2, 0\right], \left[0, 2\right], \left[2, 4\right]=b[/tex]Now, we just evaluate the function at the right endpoints:[tex]f\left(x_{1}\right)=f\left(-4\right)=-13=-13[/tex][tex]f\left(x_{2}\right)=f\left(-2\right)=-7=-7[/tex][tex]f\left(x_{3}\right)=f\left(0\right)=-1=-1[/tex][tex]f\left(x_{4}\right)=f\left(2\right)=5=5[/tex][tex]f\left(x_{5}\right)=f(b)=f\left(4\right)=11=11[/tex]Finally, just sum up the above values and multiply by 2[tex]2(-13-7-1+5+11)=-10[/tex]The Riemann sum equals -10