On this educational webpage, we're tasked with applying the Midpoint Rule to estimate the integral:\[\int_{-0.5}^{5.5} x^3 \, dx\]The instruction specifies using \( n = 6 \) subintervals for this approximation.Below the problem statement, there is an option for video assistance titled "Question Help," providing support for those who might need it. Users can submit their answers using the "Submit Question" button.The Midpoint Rule involves dividing the interval into equal parts and evaluating the function at the midpoint of each subinterval. These estimates are then used to approximate the area under the curve. This exercise helps in understanding numerical integration, a key concept in calculus.

Given integral is in the form of ∫abfxdx. Here, a=-0.5, b=5.5 & f(x)=x3. and n=6 (given). Now, we have to evaluate this integral by using mid-point rule. Formula for midpoint approximation is, ∫abf(x)dx≈∆x[fx0+x12+fx1+x22+fx2+x32+. . . . . . . +fxn-2+xn-12+fxn-1+xn2] where ∆x=b-anPlug the values in ∆x=b-an ∆x=5.5--0.56=5.5+0.56=66=1 Now let's divide the interval [-0.5, 5.5] into 6 subintervals of length ∆x=1. a=-0.5, 0.5, 1.5, 2.5, 3.5, 4.5, 5.5=b We will find the function at these endpoints, as we know f(x)=x3 Let's calculate one by one, fx0+x12=f-0.5+0.52=f0=0fx1+x22=f0.5+1.52=f1=13=1fx2+x32=f1.5+2.52=f2=23=8fx3+x42=f2.5+3.52=f3=33=27fx4+x52=f3.5+4.52=f4=43=64fx5+x62=f4.5+5.52=f5=53=125 Just plug the values in the formula ∫abfxdx≈∆xfx0+x12+fx1+x22+fx2+x32+fx3+x42+fx4+x52+fx5+x62 Then, it becomes ∫-0.55.5x3dx≈10+1+8+27+64+125=225Therefore, ∫-0.55.5x3dx=225 by midpoint rule with n=6