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Q. 1

Expert-verifiedFound in: Page 1095

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Work as an integral of force and distance: Find the work done in moving an object along the *x*-axis from the origin to $x=\frac{\mathrm{\pi}}{2}$ if the force acting on the object at a given value of *x* is role="math" localid="1650297715748" $\mathrm{F}\left(x\right)=x\mathrm{sin}x$.

The work done in moving an object along the *x*-axis from the origin to$x=\frac{\mathrm{\pi}}{2}$ is $0$.

Work as an integral of force and distance: Find the work done in moving an object along the *x*-axis from the origin to $x=\frac{\mathrm{\pi}}{2}$ if the force acting on the object at a given value of *x* is data-custom-editor="chemistry" $\mathrm{F}\left(\mathrm{x}\right)=\mathrm{xsinx}$.

$W={\int}_{0}^{\mathrm{\pi}/2}x\mathrm{sin}xdx$

Firstly solving the integral

$W=\int x\mathrm{sin}xdx\phantom{\rule{0ex}{0ex}}W=\left[x\int \mathrm{sin}x\mathrm{d}x-\left\{\int \frac{d}{dx}\left(x\right)\left(\int \mathrm{sin}xdx\right)\right\}\right]\phantom{\rule{0ex}{0ex}}W=\left[-x\mathrm{cos}x-\left\{-\frac{{x}^{2}}{2}\mathrm{cos}x\right\}\right]\phantom{\rule{0ex}{0ex}}W=\left[-x\mathrm{cos}x+\frac{{x}^{2}}{2}\mathrm{cos}x\right]$

$W={\left[-x\mathrm{cos}x+\frac{{x}^{2}}{2}\mathrm{cos}x\right]}_{0}^{\mathrm{\pi}/2}\phantom{\rule{0ex}{0ex}}W=\left[\left(-\frac{\mathrm{\pi}}{2}\mathrm{cos}\frac{\mathrm{\pi}}{2}+\frac{{\left(\frac{\mathrm{\pi}}{2}\right)}^{2}}{2}\mathrm{cos}\frac{\mathrm{\pi}}{2}\right)-\left(0\mathrm{cos}0+\frac{{\left(0\right)}^{2}}{2}\mathrm{cos}0\right)\right]\phantom{\rule{0ex}{0ex}}W=\left[\left(-\frac{\mathrm{\pi}}{2}\times 0+\frac{{\mathrm{\pi}}^{2}}{4\xb72}\times 0\right)-\left(0\mathrm{cos}0+\frac{{\left(0\right)}^{2}}{2}\mathrm{cos}0\right)\right]\phantom{\rule{0ex}{0ex}}W=\left[\left(0+0\right)-\left(0+0\right)\right]\phantom{\rule{0ex}{0ex}}W=0$

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