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Math

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PROBLEM

Calculate the work done by the force F(x)=x1+4xF(x)=x^{-1}+4x from x=3x=3 to x=5x=5: W=35(x1+4x)dxW=\int_{3}^{5}(x^{-1}+4x)dx

STEP 1

Assumptions1. The force acting on the object is given by the function (x)=x1+4x(x) = x^{-1} +4x.
. The object is moved along the x-axis from x=3x=3 to x=5x=5.
3. The work done by the force is given by the definite integral of the force function from x=ax=a to x=bx=b.

STEP 2

We need to evaluate the definite integral of the force function from x=x= to x=5x=5. The integral is given byW=5F(x)dx=5(x1+4x)dxW=\int_{}^{5} F(x) d x = \int_{}^{5} (x^{-1}+4x) dx

STEP 3

We can split the integral into two separate integralsW=35x1dx+35xdxW = \int_{3}^{5} x^{-1} dx + \int_{3}^{5}x dx

STEP 4

Now, we need to find the antiderivative of each function.The antiderivative of x1x^{-1} is lnx\ln|x|, and the antiderivative of 4x4x is 2x22x^2.
So, we have$$W = [\ln|x|]_{3}^{} + [2x^2]_{3}^{}
$$

STEP 5

Next, we need to evaluate each antiderivative at the upper and lower limits of the integral.
For the first term, we have$$[\ln|x|]_{3}^{5} = \ln|5| - \ln|3|
Forthesecondterm,wehaveFor the second term, we have[2x^2]_{3}^{5} =2(5)^2 -2(3)^2$$

STEP 6

Now, we can simplify each term$$\ln|5| - \ln|3| = \ln\left(\frac{5}{3}\right)
andand2(5)^2 -2(3)^2 =2(25 -9) =2(16) =32$$

SOLUTION

Finally, we add the two terms together to find the total work doneW=ln(53)+32W = \ln\left(\frac{5}{3}\right) +32So, the work done by the force (x)=x1+4x(x)=x^{-1}+4 x moving an object along the xx-axis from x=3x=3 to x=5x=5 is ln(53)+32\ln\left(\frac{5}{3}\right) +32.

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