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The derivative of the function 
f is defined by 
f^(')(x)=x^(2)cos(x-3). If 
f(-2)=-4, then use a calculator to find the value of 
f(6) to the nearest thousandth.
Answer:

The derivative of the function f f is defined by f(x)=x2cos(x3) f^{\prime}(x)=x^{2} \cos (x-3) . If f(2)=4 f(-2)=-4 , then use a calculator to find the value of f(6) f(6) to the nearest thousandth.\newlineAnswer:

Full solution

Q. The derivative of the function f f is defined by f(x)=x2cos(x3) f^{\prime}(x)=x^{2} \cos (x-3) . If f(2)=4 f(-2)=-4 , then use a calculator to find the value of f(6) f(6) to the nearest thousandth.\newlineAnswer:
  1. Integrate f(x)f'(x): To find the value of f(6)f(6), we need to integrate the derivative f(x)f'(x) to get the original function f(x)f(x). We will then use the initial condition f(2)=4f(-2) = -4 to find the constant of integration.
  2. Apply integration by parts: Integrate f(x)=x2cos(x3)f'(x) = x^2\cos(x-3) with respect to xx. This requires integration by parts or a special technique since it is a product of a polynomial and a trigonometric function.
  3. Simplify and integrate: Let u=x2u = x^2, which means du=2xdxdu = 2x\,dx. Let dv=cos(x3)dxdv = \cos(x-3)\,dx, which means v=cos(x3)dx=sin(x3)v = \int\cos(x-3)\,dx = \sin(x-3). Now we can apply integration by parts: udv=uvvdu\int u\,dv = uv - \int v\,du.
  4. Use initial condition: Using integration by parts, we get f(x)=x2sin(x3)2xsin(x3)dxf(x) = x^2\sin(x-3) - \int 2x\sin(x-3)\,dx. The second integral also requires integration by parts.
  5. Calculate trigonometric values: Let u=2xu = 2x, which means du=2dxdu = 2dx. Let dv=sin(x3)dxdv = \sin(x-3)dx, which means v=cos(x3)v = -\cos(x-3). Apply integration by parts again: udv=uvvdu\int u dv = uv - \int v du.
  6. Find constant of integration: We get f(x)=x2sin(x3)(2x(cos(x3))2cos(x3)dx)f(x) = x^2\sin(x-3) - (2x(-\cos(x-3)) - \int -2\cos(x-3)\,dx). Simplify and integrate the remaining term.
  7. Find f(6)f(6): Simplifying, we have f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx. The last integral is straightforward: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3).
  8. Find f(6)f(6): Simplifying, we have f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx. The last integral is straightforward: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3).Putting it all together, we have f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C, where CC is the constant of integration.
  9. Find f(6)f(6): Simplifying, we have f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx. The last integral is straightforward: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3).Putting it all together, we have f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C, where CC is the constant of integration.Now we use the initial condition f(2)=4f(-2) = -4 to solve for CC. Plug in x=2x = -2 into the equation: 4=(2)2sin(23)+2(2)cos(23)+2sin(23)+C-4 = (-2)^2\sin(-2-3) + 2(-2)\cos(-2-3) + 2\sin(-2-3) + C.
  10. Find f(6)f(6): Simplifying, we have f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx. The last integral is straightforward: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3). Putting it all together, we have f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C, where CC is the constant of integration. Now we use the initial condition f(2)=4f(-2) = -4 to solve for CC. Plug in x=2x = -2 into the equation: 4=(2)2sin(23)+2(2)cos(23)+2sin(23)+C-4 = (-2)^2\sin(-2-3) + 2(-2)\cos(-2-3) + 2\sin(-2-3) + C. Calculate the trigonometric values and simplify: 4=4sin(5)4cos(5)+2sin(5)+C-4 = 4\sin(-5) - 4\cos(-5) + 2\sin(-5) + C. Use a calculator to find the values of f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx00 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx11.
  11. Find f(6)f(6): Simplifying, we have f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx. The last integral is straightforward: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3). Putting it all together, we have f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C, where CC is the constant of integration. Now we use the initial condition f(2)=4f(-2) = -4 to solve for CC. Plug in x=2x = -2 into the equation: 4=(2)2sin(23)+2(2)cos(23)+2sin(23)+C-4 = (-2)^2\sin(-2-3) + 2(-2)\cos(-2-3) + 2\sin(-2-3) + C. Calculate the trigonometric values and simplify: 4=4sin(5)4cos(5)+2sin(5)+C-4 = 4\sin(-5) - 4\cos(-5) + 2\sin(-5) + C. Use a calculator to find the values of f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx00 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx11. After calculating, we find that f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx22 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx33. Now substitute these values into the equation: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx44.
  12. Find f(6)f(6): Simplifying, we have f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx. The last integral is straightforward: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3).Putting it all together, we have f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C, where CC is the constant of integration.Now we use the initial condition f(2)=4f(-2) = -4 to solve for CC. Plug in x=2x = -2 into the equation: 4=(2)2sin(23)+2(2)cos(23)+2sin(23)+C-4 = (-2)^2\sin(-2-3) + 2(-2)\cos(-2-3) + 2\sin(-2-3) + C.Calculate the trigonometric values and simplify: 4=4sin(5)4cos(5)+2sin(5)+C-4 = 4\sin(-5) - 4\cos(-5) + 2\sin(-5) + C. Use a calculator to find the values of f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx00 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx11.After calculating, we find that f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx22 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx33. Now substitute these values into the equation: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx44.Simplify the equation to find CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx66. Now solve for CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx88.
  13. Find f(6)f(6): Simplifying, we have f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx. The last integral is straightforward: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3). Putting it all together, we have f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C, where CC is the constant of integration. Now we use the initial condition f(2)=4f(-2) = -4 to solve for CC. Plug in x=2x = -2 into the equation: 4=(2)2sin(23)+2(2)cos(23)+2sin(23)+C-4 = (-2)^2\sin(-2-3) + 2(-2)\cos(-2-3) + 2\sin(-2-3) + C. Calculate the trigonometric values and simplify: 4=4sin(5)4cos(5)+2sin(5)+C-4 = 4\sin(-5) - 4\cos(-5) + 2\sin(-5) + C. Use a calculator to find the values of f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx00 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx11. After calculating, we find that f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx22 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx33. Now substitute these values into the equation: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx44. Simplify the equation to find CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx66. Now solve for CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx88. Calculate the value of CC: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3)00. This is the constant of integration.
  14. Find f(6)f(6): Simplifying, we have f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx. The last integral is straightforward: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3). Putting it all together, we have f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C, where CC is the constant of integration. Now we use the initial condition f(2)=4f(-2) = -4 to solve for CC. Plug in x=2x = -2 into the equation: 4=(2)2sin(23)+2(2)cos(23)+2sin(23)+C-4 = (-2)^2\sin(-2-3) + 2(-2)\cos(-2-3) + 2\sin(-2-3) + C. Calculate the trigonometric values and simplify: 4=4sin(5)4cos(5)+2sin(5)+C-4 = 4\sin(-5) - 4\cos(-5) + 2\sin(-5) + C. Use a calculator to find the values of f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx00 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx11. After calculating, we find that f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx22 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx33. Now substitute these values into the equation: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx44. Simplify the equation to find CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx66. Now solve for CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx88. Calculate the value of CC: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3)00. This is the constant of integration. Now that we have CC, we can find f(6)f(6) by plugging 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3)33 into the integrated function: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3)44.
  15. Find f(6)f(6): Simplifying, we have f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx. The last integral is straightforward: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3). Putting it all together, we have f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C, where CC is the constant of integration. Now we use the initial condition f(2)=4f(-2) = -4 to solve for CC. Plug in x=2x = -2 into the equation: 4=(2)2sin(23)+2(2)cos(23)+2sin(23)+C-4 = (-2)^2\sin(-2-3) + 2(-2)\cos(-2-3) + 2\sin(-2-3) + C. Calculate the trigonometric values and simplify: 4=4sin(5)4cos(5)+2sin(5)+C-4 = 4\sin(-5) - 4\cos(-5) + 2\sin(-5) + C. Use a calculator to find the values of f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx00 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx11. After calculating, we find that f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx22 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx33. Now substitute these values into the equation: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx44. Simplify the equation to find CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx66. Now solve for CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx88. Calculate the value of CC: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)00. This is the constant of integration. Now that we have CC, we can find f(6)f(6) by plugging 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)33 into the integrated function: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)44. Calculate the trigonometric values for 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)55 and 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)66 using a calculator, then substitute them into the equation to find f(6)f(6).
  16. Find f(6)f(6): Simplifying, we have f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx. The last integral is straightforward: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3). Putting it all together, we have f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C, where CC is the constant of integration. Now we use the initial condition f(2)=4f(-2) = -4 to solve for CC. Plug in x=2x = -2 into the equation: 4=(2)2sin(23)+2(2)cos(23)+2sin(23)+C-4 = (-2)^2\sin(-2-3) + 2(-2)\cos(-2-3) + 2\sin(-2-3) + C. Calculate the trigonometric values and simplify: 4=4sin(5)4cos(5)+2sin(5)+C-4 = 4\sin(-5) - 4\cos(-5) + 2\sin(-5) + C. Use a calculator to find the values of f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx00 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx11. After calculating, we find that f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx22 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx33. Now substitute these values into the equation: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx44. Simplify the equation to find CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx66. Now solve for CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx88. Calculate the value of CC: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)00. This is the constant of integration. Now that we have CC, we can find f(6)f(6) by plugging 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)33 into the integrated function: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)44. Calculate the trigonometric values for 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)55 and 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)66 using a calculator, then substitute them into the equation to find f(6)f(6). After calculating, we find that 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)88 and 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)99. Now substitute these values into the equation: f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C00.
  17. Find f(6)f(6): Simplifying, we have f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx. The last integral is straightforward: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3). Putting it all together, we have f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C, where CC is the constant of integration. Now we use the initial condition f(2)=4f(-2) = -4 to solve for CC. Plug in x=2x = -2 into the equation: 4=(2)2sin(23)+2(2)cos(23)+2sin(23)+C-4 = (-2)^2\sin(-2-3) + 2(-2)\cos(-2-3) + 2\sin(-2-3) + C. Calculate the trigonometric values and simplify: 4=4sin(5)4cos(5)+2sin(5)+C-4 = 4\sin(-5) - 4\cos(-5) + 2\sin(-5) + C. Use a calculator to find the values of f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx00 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx11. After calculating, we find that f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx22 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx33. Now substitute these values into the equation: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx44. Simplify the equation to find CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx66. Now solve for CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx88. Calculate the value of CC: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)00. This is the constant of integration. Now that we have CC, we can find f(6)f(6) by plugging 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)33 into the integrated function: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)44. Calculate the trigonometric values for 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)55 and 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)66 using a calculator, then substitute them into the equation to find f(6)f(6). After calculating, we find that 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)88 and 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)99. Now substitute these values into the equation: f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C00. Simplify the equation to find f(6)f(6): f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C22. Now calculate the sum to get the value of f(6)f(6).
  18. Find f(6)f(6): Simplifying, we have f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx. The last integral is straightforward: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3). Putting it all together, we have f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C, where CC is the constant of integration. Now we use the initial condition f(2)=4f(-2) = -4 to solve for CC. Plug in x=2x = -2 into the equation: 4=(2)2sin(23)+2(2)cos(23)+2sin(23)+C-4 = (-2)^2\sin(-2-3) + 2(-2)\cos(-2-3) + 2\sin(-2-3) + C. Calculate the trigonometric values and simplify: 4=4sin(5)4cos(5)+2sin(5)+C-4 = 4\sin(-5) - 4\cos(-5) + 2\sin(-5) + C. Use a calculator to find the values of f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx00 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx11. After calculating, we find that f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx22 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx33. Now substitute these values into the equation: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx44. Simplify the equation to find CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx66. Now solve for CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)\,dx88. Calculate the value of CC: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)00. This is the constant of integration. Now that we have CC, we can find f(6)f(6) by plugging 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)33 into the integrated function: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)44. Calculate the trigonometric values for 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)55 and 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)66 using a calculator, then substitute them into the equation to find f(6)f(6). After calculating, we find that 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)88 and 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)\,dx = -2\sin(x-3)99. Now substitute these values into the equation: f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C00. Simplify the equation to find f(6)f(6): f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C22. Now calculate the sum to get the value of f(6)f(6). Calculate the value of f(6)f(6): f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C55. Round to the nearest thousandth.
  19. Find f(6)f(6): Simplifying, we have f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx. The last integral is straightforward: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3).Putting it all together, we have f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C, where CC is the constant of integration.Now we use the initial condition f(2)=4f(-2) = -4 to solve for CC. Plug in x=2x = -2 into the equation: 4=(2)2sin(23)+2(2)cos(23)+2sin(23)+C-4 = (-2)^2\sin(-2-3) + 2(-2)\cos(-2-3) + 2\sin(-2-3) + C.Calculate the trigonometric values and simplify: 4=4sin(5)4cos(5)+2sin(5)+C-4 = 4\sin(-5) - 4\cos(-5) + 2\sin(-5) + C. Use a calculator to find the values of f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx00 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx11.After calculating, we find that f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx22 and f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx33. Now substitute these values into the equation: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx44.Simplify the equation to find CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx66. Now solve for CC: f(x)=x2sin(x3)+2xcos(x3)2cos(x3)dxf(x) = x^2\sin(x-3) + 2x\cos(x-3) - \int -2\cos(x-3)dx88.Calculate the value of CC: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3)00. This is the constant of integration.Now that we have CC, we can find f(6)f(6) by plugging 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3)33 into the integrated function: 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3)44.Calculate the trigonometric values for 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3)55 and 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3)66 using a calculator, then substitute them into the equation to find f(6)f(6).After calculating, we find that 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3)88 and 2cos(x3)dx=2sin(x3)\int -2\cos(x-3)dx = -2\sin(x-3)99. Now substitute these values into the equation: f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C00.Simplify the equation to find f(6)f(6): f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C22. Now calculate the sum to get the value of f(6)f(6).Calculate the value of f(6)f(6): f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C55. Round to the nearest thousandth.The value of f(6)f(6) to the nearest thousandth is approximately f(x)=x2sin(x3)+2xcos(x3)+2sin(x3)+Cf(x) = x^2\sin(x-3) + 2x\cos(x-3) + 2\sin(x-3) + C77.

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