O edcc.instructure. XO Acing Google Co xO Kevawn «Redund xM Whole Course Re XM Invitation: CS&e PapagoC Get Homework E XGet HomeworkA variables2 - Solve XA WAMAP - Posts x5.2 Calc.pdf+A https://wamaps3.s3.amazonaws.com/ufiles/222249/5.2_Calc.pdfThese are equal by the distributive property. The-1other formulas are discussed in Appendix F.11(а, — b) —E b.a, -EXAMPLE 2 Evaluating an integral as a limit of Riemann sums(a) Evaluate the Riemann sum for f(x) = x' – 6.x, taking the sample points to be rightendpoints and a = 0, b = 3, and n = 6.(b) Evaluate (:x' - 6x) dx.SOLUTION(a) With n = 6 the interval width isb - a3 - 0Ar =%3D62and the right endpoints are x, = 0.5, r2 = 1.0, x; = 1.5, x, = 2.0, xs = 2.5, andX, = 3.0. So the Riemann sum isR, =i-= f(0.5) Ar + f(1.0) Ar + f(1.5) Ar + f(2.0) Ax + f(2.5) Ax + f(3.0) Ax5y =x'- 6x= }(-2,875 – 5 – 5.625 – 4 + 0.625 + 9)= -3.9375Notice that f is not a positive function and so the Riemann sum does not represent a sumof areas of rectangles. But it does represent the sum of the areas of the blue rectangles(above the x-axis) minus the sum of the areas of the gold rectangles (below the x-axis) inFIGURE 5Figure 5.347SECTION 5.2 THE DEFINITE INTEGRAL(b) With n subintervals we haveb - a3Thus Xp = 0, x, = 3/n, x, = 6/n, X= 9/n, and, in general, x, = 3i/n. Since we areusing right endpoints, we can use Theorem 4:3i 3[(r' - 6x) dx = lim E S(x,) Ax = limte sum, N is a constant (unlikenve 3n in front of the E signso we can= limEquation 9 withe오후 12:57P Type here to search^O G ) O2021-01-07 See Example 2 in Section 5.2. Suppose that in part (b) we change the upper limit of integration to 2,resulting in| (2 – 6z)dr. To find the value of this integral, work through the following step-by-stepprocess.a) We know that a =and b =b) Using n subintervals, Ar =c) Assume that the sample points in each interval are right endpoints. Find the following:21 = a + Az =22 = a + 2Az =23 = a + 3Az =In general, the ith sample point is r; = a + iAr =Note: simplify your answer for thisquestion!d) Now find the sum of the areas of n approximating rectangles. Note: your answer will be an expressionin terms of n only. You really need to use the information in example 2 to help you complete this question!f(r:)Az =i=1e) Finally, find the value of the integral by letting the number of rectangles approach infinity.[ (2" – 6z)dz = lim E f(#:)Az =1=1Question Help: O Message instructor D Post to forumSubmit Question

Question

O edcc.instructure. XO Acing Google Co xO Kevawn «Redund xM Whole Course Re XM Invitation: CS&e PapagoC Get Homework E XGet HomeworkA variables2 - Solve XA WAMAP - Posts x5.2 Calc.pdf+A https://wamaps3.s3.amazonaws.com/ufiles/222249/5.2_Calc.pdfThese are equal by the distributive property. The-1other formulas are discussed in Appendix F.11(а, — b) —E b.a, -EXAMPLE 2 Evaluating an integral as a limit of Riemann sums(a) Evaluate the Riemann sum for f(x) = x' – 6.x, taking the sample points to be rightendpoints and a = 0, b = 3, and n = 6.(b) Evaluate (:x' - 6x) dx.SOLUTION(a) With n = 6 the interval width isb - a3 - 0Ar =%3D62and the right endpoints are x, = 0.5, r2 = 1.0, x; = 1.5, x, = 2.0, xs = 2.5, andX, = 3.0. So the Riemann sum isR, =i-= f(0.5) Ar + f(1.0) Ar + f(1.5) Ar + f(2.0) Ax + f(2.5) Ax + f(3.0) Ax5y =x'- 6x= }(-2,875 – 5 – 5.625 – 4 + 0.625 + 9)= -3.9375Notice that f is not a positive function and so the Riemann sum does not represent a sumof areas of rectangles. But it does represent the sum of the areas of the blue rectangles(above the x-axis) minus the sum of the areas of the gold rectangles (below the x-axis) inFIGURE 5Figure 5.347SECTION 5.2 THE DEFINITE INTEGRAL(b) With n subintervals we haveb - a3Thus Xp = 0, x, = 3/n, x, = 6/n, X= 9/n, and, in general, x, = 3i/n. Since we areusing right endpoints, we can use Theorem 4:3i 3[(r' - 6x) dx = lim E S(x,) Ax = limte sum, N is a constant (unlikenve 3n in front of the E signso we can= limEquation 9 withe오후 12:57P Type here to search^O G ) O2021-01-07 See Example 2 in Section 5.2. Suppose that in part (b) we change the upper limit of integration to 2,resulting in| (2 – 6z)dr. To find the value of this integral, work through the following step-by-stepprocess.a) We know that a =and b =b) Using n subintervals, Ar =c) Assume that the sample points in each interval are right endpoints. Find the following:21 = a + Az =22 = a + 2Az =23 = a + 3Az =In general, the ith sample point is r; = a + iAr =Note: simplify your answer for thisquestion!d) Now find the sum of the areas of n approximating rectangles. Note: your answer will be an expressionin terms of n only. You really need to use the information in example 2 to help you complete this question!f(r:)Az =i=1e) Finally, find the value of the integral by letting the number of rectangles approach infinity.[ (2&#34; – 6z)dz = lim E f(#:)Az =1=1Question Help: O Message instructor D Post to forumSubmit Question

Accepted Answer

As per Bartleby guidelines, we are allowed to solve 3 parts of a question.
Given ∫02(x3-6x)dx