QuestionCalculate the work done by the force $F(x)=x^{-1}+4x$ from $x=3$ to $x=5$ : $W=\int_{3}^{5}(x^{-1}+4x)dx$

Studdy Solution

STEP 1

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

STEP 2

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

STEP 3

We can split the integral into two separate integrals $W = \int_{3}^{5} x^{-1} dx + \int_{3}^{5}x dx$

STEP 4

Now, we need to find the antiderivative of each function.The antiderivative of $x^{-1}$ is $\ln|x|$ , and the antiderivative of $4x$ is $2x^2$ .
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 $[\ln|x|]_{3}^{5} = \ln|5| - \ln|3|$ For the second term, we have $[2x^2]_{3}^{5} =2(5)^2 -2(3)^2$

STEP 6

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

STEP 7

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

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