# Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.3 Center of Mass - Exercises - Page 483: 16

$$\left(\frac{2 \ln (2)-(3 / 4)}{2 \ln (2)-1}, \frac{\ln ^{2}(2)-2 \ln (2)+1}{2 \ln (2)-1}\right)$$

#### Work Step by Step

Given $$f(x)=\ln x, \quad[1,2]$$ Since \begin{aligned} M_{y} &=\rho \int_{a}^{b} x f(x) d x \\ &=\int_{1}^{2} x\ln x d x \end{aligned} Let \begin{align*} u&=\ln x \ \ \ \ \ \ \ \ \ \ \ dv= xdx\\ du&=\frac{1}{x}dx\ \ \ \ \ \ \ \ \ v=\frac{1}{2}x^2 \end{align*} Then \begin{aligned} \int_{1}^{2} x\ln x d x &=\frac{x^{2} \ln x}{2}\bigg|_{1}^{2}-\int_{1}^{2} \frac{x}{2} d x \\ &=\frac{x^{2} \ln x}{2}-\frac{x^{2}}{4}\bigg|_{1}^{2}\\ &= 2 \ln (2)-\frac{3}{4} \end{aligned} and \begin{aligned} M_{x} &=\frac{1}{2} \rho \int_{a}^{b} f(x)^{2} d x \\ &=\frac{1}{2} \int_{1}^{2}\left(\ln x\right)^{2} d x \end{aligned} Let \begin{array}{rlrl} u & =\ln ^{2} x & d v & =1 \\ d u & =\frac{2 \ln x}{x} & & v=x \end{array} Then $$\int \left(\ln x\right)^{2} d x= x \ln ^{2} x-2 \int \ln x \, d x$$ Let \begin{array}{rlrl} u & =\ln x & d v & =1 \\ d u & =\frac{1}{x} & & v=x \end{array} then $\int \ln x=x\ln x-x$ and $$\int \left(\ln x\right)^{2} d x=x \ln ^{2} x-2 x \ln x+2 x$$ hence \begin{aligned} M_{x} &=\frac{1}{2} \int_{1}^{2} \ln ^{2} x d x \\ &=\left.\frac{1}{2}\left(x \ln ^{2} x-2 x \ln x+2 x\right)\right|_{1} ^{2} \\ &=\frac{1}{2}\left(2 \ln ^{2} 2-4 \ln 2+2\right) \\ &=\ln ^{2}(2)-2 \ln (2)+1 \end{aligned} and \begin{align*} M&=\rho A\\ &=\int_{1}^{2} \ln x d x\\ &=\left.(x \ln x-x)\right|_{1} ^{2}\\ &=2 \ln (2)-1 \end{align*} then \begin{align*} \left(\bar{x}, \bar{y}\right)&=\left(\frac{M_{y}}{M},\frac{M_{x}}{M} \right)\\ &=\left(\frac{2 \ln (2)-(3 / 4)}{2 \ln (2)-1}, \frac{\ln ^{2}(2)-2 \ln (2)+1}{2 \ln (2)-1}\right) \end{align*}

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