be organized and please slove the problem in paper step by step my exam in coming i understand more if its in paper wrriten please thank you

Course: ECO10002 Course: ECO10002 Instructor: Marian Melnyk Instructor: Marian Melnyk Problem Set 8 Problem 6 Due date: 11/14/2023 (11:59 PM) Use integration by parts to evaluate the following: Problem 1 (a) Suppose that the demand and supply curves for a commodity are P = f(Q) = 300-0.30 and P = g(Q) = 200+0.20, respectively. Find the equilibrium quantity and compute the consumer and producer surplus. Problem 7 (This problem is for practice and illustration only; it will not be graded and you can (b) Suppose that the demand and supply curves for a commodity are P = f(Q) = 10- VO and P = 8(0) = skip it) 20, respectively. Find the equilibrium quantity and compute the consumer and producer surplus. Use integration by parts to evaluate the following: Problem 2 (a ) Vilnidi (b ) [ (x - 2)e - 3 /2 dx Let K(t) denote the capital stock of an economy at time . Then net investment at time t, denoted (9 / 3-x3'dx by I(1), is given by the rate of increase K(1) of K(t). (a) If I(1) = 312 + 2t + 5 fort 2 0, what is the total increase in the capital stock during the interval from t = 0 to t = 5? (b) If K(to) = Ko, find an expression for the total increase in the capital stock from time t = to to 1 = T when the investment function /(t) is as in part (a). Problem 3 Suppose that the inverse demand and supply curves are, respectively, P = f(Q) = 200 - 0.20 and P = g(Q) = 20 + 0.1Q. Find the equilibrium price and quantity, then compute the consumer and producer surplus. Problem 4 Suppose the inverse demand and supply curves for a particular commodity are, respectively, P = f(Q) = 6000/(Q + 50) and P = 8(Q) = Q + 10. Find the equilibrium price and quantity, then compute the consumer and producer surplus. Problem 5 Use integration by parts to evaluate the following: (a) xedx ( b ) 3 xett dxIf '11. (hr = no I u (in Here, n. and dig are chosen parts ofthe integrand, and then you differentiate it. to get (in. and integrate do to get 1:.Let's applythis techniqueto the given integrals: (a) f :1: - 6\"\" do: Let's choose it. = :1: and do = e (in. = dz: [differentiate it] { integ tt-e do; Now apply the integration by parts formula: {whet _. C" is the constant of integraticm) (b)f 3:1: - e43 do: Let's CHOOSE "LE. 2 321? and d?) = (in. = ' ' '2 ( differentiate it] ate do } Apply the integration by parts formula: :1: = no l \"t! (in (c)f(1 + or") -e_"' do: Let's choose it. = l l 3:2 and Eh! = t Atthis point, you can usethe result from part (a) to evaluate the integral involving :1: . e:"'"'. Substituting that result, you can complete the solution for part (c). Sudv = uv - fv du Let's use this formula to solve the given integrals: (a) fx In(x + 2) da Let's choose u = In(x + 2) and du = x dx: du = #2 dx (differentiate u) v = -x2 (integrate du) Now apply the integration by parts formula: S' In(2 +2) da = wv|- S', vdu Evaluate uv at the limits: = x2 In(2+2) - S13x2.212 Now integrate the remaining term: = 1/am(2 +2) da Solve for the integral on the right side: Sal(x +2) da = 3 (ja In(2 + 2)|', - S', ga2 . 4, da) Now integrate the remaining term again: = x3 In(2+2) - 6532 1, Repeat this process until the integrals on the right side become manageable. The limits of integration can be substituted at the end. (b) So x . 2" da Let's choose u = x and du = 2 dx: du = dx (differentiate u) v = In(2) ' . 2" (integrate du) Apply the integration by parts formula: So a . 2" da = wv - fo v du Evaluate uv at the limits: In( 2) . X - 2" - So m 2) . 2" dx Now integrate the remaining term: = In(2) In(2) So 23 da Evaluate moy . 2" at the limits and solve the remaining integral.(@) fox2 . er da Let's choose u = x2 and du = e" do: du = 2x dx (differentiate u) U = e" (integrate du) Apply the integration by parts formula: fox . ex da = uv So v du Evaluate uv at the limits: Jo er . 2a da Now integrate the remaining term: = - 2 fox . e da Now, you can apply integration by parts again to evaluate the remaining integral. Substitute the limits at the end.(a) fi Vt . In(t) dt Let's choose u = In(t) and du = vt dt: du = : dt (differentiate u) v = 2+3/2 (integrate du) Apply the integration by parts formula: Si vt . In(t) at = uv - fi v du Evaluate uv at the limits: - 313/2 In(t) "- S1 313/2 . } at Now integrate the remaining term: = 3 Si t1/2 at Evaluate the integral on the right side and substitute the limits. (b) So (x - 2) . e-2/2 da Let's choose u = x - 2 and du = e-x/2 da: du = dx (differentiate u) v = -2e-#/2 (integrate du) Apply the integration by parts formula: So(x - 2) . e-x/2 da = uv - fo v du Evaluate uv at the limits: =-2e-x/2(x - 2) - 53(-2e-2/2) da Now integrate the remaining term: =-2e-x/2(x - 2) +2 50 e x/2 da Evaluate the integral on the right side and substitute the limits. (c) So (3 -z) . 3"' da Let's choose u = 3 - x and du = 3 dx: du = -dx (differentiate u) v = In(3) . 3 (integrate du) Apply the integration by parts formula: So (3 - x) . 3' da = uv - So v du Evaluate uv at the limits: In(3 . (3* - x . 3) + So In(3) mey . 3 da Now integrate the remaining term: = In(3) . (33 - 3 . 33) + may So 3* da Evaluate the integral on the right side and substitute the limits