Question

A variable force of 9x−29x−2 pounds moves an object along a straight line when it is xx feet from the origin. Calculate the work done in moving the object from x=1ft, to, x=19ft. (Round your answer to two decimal places.)

Integrals
ANSWERED
asked 2021-03-11

A variable force of \(\displaystyle{9}{x}−{29}{x}−{2}\) pounds moves an object along a straight line when it is xx feet from the origin. Calculate the work done in moving the object from \(\displaystyle{x}={1}{f}{t},\to, {x}={19}{f}{t}.\) (Round your answer to two decimal places.)

Answers (1)

2021-03-12

Remember that work along a straight line is defined as
\(\displaystyle{W}=∫{\left[{x}{0},{x}{1}\right]}{F}{\left.{d}{x}\right.}\)
Substituting in our expressions, we thus find that
\(\displaystyle{W}={\int_{{{1}}}^{{{19}}}}{\left({9}{x}-{2}\right)}{\left.{d}{x}\right.}\)
\(=\int_{1}^{19}9xdx-\int_{1}^{19}2dx\)
\(=9\int_{1}^{19}xdx-2\int_{1}^{19}dx\)
\(=9[\frac{1}{2}x^{2}][1,19]-2[x][1,19]\)
\(=9[\frac{1}{2}(19)^{2}-\frac{1}{2})(1)^{2}]-2[19-1]=1584 ft \times lb\)

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