Question: The cost function for a product is C'(x) = 0.6x + 150x + 200. Find average cost over [0, 700]. Answer: The average value

The cost function for a product is C'(x) = 0.6x + 150x+ 200. Find average cost over [0, 700]. Answer: The average valueof the function v(x) = 5/x on the interval [1, c] isequal to 1. Find the value of c. C= Previously, we developedthe Product Rule and studied how it is employed to differentiate aproduct of two functions. In particular, recall that if and g aredifferentiable functions of x, then d dx [f(x) g(x)] = f(x) g'(x)+ g(x) f'(x). (a) For each of the following functions, use theProduct Rule to find the function's derivative. Notice the label of thederivative (e.g., the derivative of g(x) should be labeled g'(x)). (i) Ifg(x): = x sin(x), then g'(x) (ii) If h(x) = xex, then(iii) If p(x) = xln(x), then = (iv.) If q(x) = xcos(x), then (v.) If r(x) = e* sin(x), then = Previously, wedeveloped the Product Rule and studied how it is employed to differentiatea product of two functions. In particular, recall that if and gare differentiable functions of x, then d dx -[f(x) g(x)] = f(x)g(x) + g(x) f'(x). (a) For each of the following functions, usethe Product Rule to find the function's derivative. Notice the label ofthe derivative (e.g., the derivative of g(x) should be labeled g'(x)). (i)If g(x) = x sin(x), then g'(x) (ii) If h(x) = xex,then (iii) If p(x) = x In(x), then (iv.) If q(x) =

The cost function for a product is C'(x) = 0.6x + 150x + 200. Find average cost over [0, 700]. Answer: The average value of the function v(x) = 5/x on the interval [1, c] is equal to 1. Find the value of c. C= Previously, we developed the Product Rule and studied how it is employed to differentiate a product of two functions. In particular, recall that if and g are differentiable functions of x, then d dx [f(x) g(x)] = f(x) g'(x) + g(x) f'(x). (a) For each of the following functions, use the Product Rule to find the function's derivative. Notice the label of the derivative (e.g., the derivative of g(x) should be labeled g'(x)). (i) If g(x): = x sin(x), then g'(x) (ii) If h(x) = xex, then (iii) If p(x) = xln(x), then = (iv.) If q(x) = x cos(x), then (v.) If r(x) = e* sin(x), then = Previously, we developed the Product Rule and studied how it is employed to differentiate a product of two functions. In particular, recall that if and g are differentiable functions of x, then d dx -[f(x) g(x)] = f(x) g(x) + g(x) f'(x). (a) For each of the following functions, use the Product Rule to find the function's derivative. Notice the label of the derivative (e.g., the derivative of g(x) should be labeled g'(x)). (i) If g(x) = x sin(x), then g'(x) (ii) If h(x) = xex, then (iii) If p(x) = x In(x), then (iv.) If q(x) = x cos(x), then (v.) If r(x) = e* sin(x), then = = (b) Use your work in (a) to help you evaluate the following indefinite integrals. Use differentiation to check your work. (Don't forget the "+C".) (i) ze"+"de (ii) = i) / e (sin(x) + cos(x)) d = (iii) 1/2 2x cos(x) x sin(x) dx (iv) x cos(x) + sin(x) dx =| = (iii) 12 (iv) (v) 2x cos(x) = x sin(x) dx |_ x cos(x) + sin(x) dr 11 1+In(x) dx = = (c) Observe that the examples in (b) work nicely because of the derivatives you were asked to calculate in (a). Each integrand in (b) is precisely the result of differentiating one of the products of basic functions found in (a). To see what happens when an integrand is still a product but not necessarily the result of differentiating an elementary product, we consider how to evaluate (i) First, observe that x cos(x) dx. d dx [x sin(x)] = = x cos(x) + sin(x). Integrating both sides indefinitely and using the fact that the integral of a sum is the sum of the integrals, we find that (sin(x))) dz = cos(x) dz + sin(x) dr. [ x / d dx In this last equation, evaluate the indefinite integral on the left side: (sin(x)]) dx = Now evaluate the indefinite integral on the right side: dx+ sin(x) dx: Integrating both sides indefinitely and using the fact that the integral of a sum is the sum of the integrals, we find that d x dx (sin()]) dr = 2008 (2) dr +[sin(z) dir. [ cos(x) In this last equation, evaluate the indefinite integral on the left side: ([x sin(x)]) dx = Now evaluate the indefinite integral on the right side: sin(x) dx == (ii) Given the information calculated in (i), we can now determine that x cos(x) dx = = (iii) For which product of basic functions have you now found the antiderivative? Answer: Evaluate the indefinite integral. x sin (8x) dx = +C. Hint: Integrate by parts with u = x. Use integration by parts to evaluate the integral. +C 64 64x cos(4x)dx Evaluate the following integral: 2 7 ln(x) dx x2 Evaluate the integral 0 1 -2x5* dx 4 Suppose that f(1) = 6, (4) = 2, '(1) = 5, '(4) = 6, and f" is continuous. Find the value of xf"(x) dx. Evaluate the indefinite integral. Answer: : +C cos x tan x dx Evaluate Is Answer: sin 10 x cos x dx. 3 Evaluate the indefinite integral. tan (x) sec (x) dx = |+C. Evaluate the integral /4 -1 cos (0) tan n (0) do Evaluate the integral 3(cos(x) + sin(x)) sin(2x) dx Note: Use an upper-case "C" for the constant of integration. Use the formula and the reduction formula sin (x) dx = 1 X sin(2x) 2 4 n- 1 [sin" (e) dr = cos(r) sin" (a) + "= / sin^-^(s) dr sin(x) dx n n to evaluate the integral sin (x) dx. Note: Use an upper-case "C" for the constant of integration. Evaluate the indefinite integral. S sin(6x) cos(15x) dx = -0- +C Evaluate the indefinite integral. 3c 3 cos (5x) dx +C Evaluate the indefinite integral. Star 7 tan (x) sec (x) dx = +0 +C. Find the average value of the function f(x) = 3e* on the interval [-2, In 4]. fave = =

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