Mathematics, 17.07.2019 12:50, joel53
Using a graph, it can be seen that the curves y=x+ln(x)y=x+ln(x) and y=x3−xy=x3−x intersect at the points (0.447141,−0.357742)(0.447141,−0.35 7742) and (1.507397,1..507397,1.917782). find the coordinates (to three decimal places) of the centroid for the region bounded by these curves.
Answers: 1
and 
, the area bounded by the curves is given by:![Area= \int\limits^{1.507397}_{0.447141} {(x+\ln x)-(x^3-x)} \, dx \\ \\ =\int\limits^{1.507397}_{0.447141} {(2x+\ln x-x^3)} \, dx \\ \\ =\left[x^2+x\ln x-x- \frac{1}{4}x^4 \right]^{1.507397}_{0.447141} \\ \\ =\left[(1.507397)^2+1.507397\ln1.507397-1.507397- \frac{1}{4} (1.507397)^4\right] \\ -\left[(0.447141)^2+0.447141\ln0.447141-0.447141- \frac{1}{4} (0.447141)^4\right] \\ \\ =0.092686-(-0.617095)=0.709781\approx\bold{0.710}](/tpl/images/0100/2049/daf4b.png)
![C_x= \frac{\int\limits^{1.507397}_{0.447141} {x\left[(x+\ln x)-(x^3-x)\right]} \, dx}{Area} \\ \\ =\frac{\int\limits^{1.507397}_{0.447141} {\left(2x^2+x\ln x-x^4\right)} \, dx}{0.709781} \\ \\ = \frac{\left[ \frac{2}{3} x^3+ \frac{1}{2}x^2\ln x- \frac{1}{4} x^2- \frac{1}{5} x^5 \right]^{1.507397}_{0.447141}}{0.709781} \\ \\ = \frac{0.625068-(-0.0744211)}{0.709781} \\ \\ = \frac{0.6994891}{0.709781} \approx0.985](/tpl/images/0100/2049/e275e.png)
![C_y= \frac{\int\limits^{1.507397}_{0.447141} {\left[(x+\ln x)^2-(x^3-x)^2\right]} \, dx}{2(Area)} \\ \\ =\frac{\int\limits^{1.507397}_{0.447141} {\left(2x\ln x+(\ln x)^2-x^6+2x^4\right)} \, dx}{2(0.709781)} \\ \\ = \frac{\left[x^2\ln x- \frac{1}{2} x^2+x(\ln x)^2-2x\ln x+2x- \frac{1}{7} x^7+ \frac{2}{5} x^5 \right]^{1.507397}_{0.447141}}{1.419562} \\ \\ = \frac{2.41459-1.64949}{1.419562} \\ \\ = \frac{0.7651}{1.419562} \approx0.539](/tpl/images/0100/2049/793bd.png)
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Using a graph, it can be seen that the curves y=x+ln(x)y=x+ln(x) and y=x3−xy=x3−x intersect at the...
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