problem 5.2 - locate the centroid of the plan area shown by integration. y 1 in y = 4x^5 - 3x^2 + 12x + 1 13…

Explanation:

Step1: Define area formula

The area (A) of the planar region is given by (A=\int_{a}^{b}y\mathrm{d}x). Here, (a = 0), (b = 1), and (y=4x^{5}-3x^{2}+12x + 1). So (A=\int_{0}^{1}(4x^{5}-3x^{2}+12x + 1)\mathrm{d}x). Using the power - rule of integration (\int x^{n}\mathrm{d}x=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have: [ \begin{align*} A&=\left[\frac{4x^{6}}{6}-\frac{3x^{3}}{3}+\frac{12x^{2}}{2}+x\right]_{0}^{1}\ &=\frac{4}{6}-1 + 6+1\ &=\frac{2}{3}+6\ &=\frac{2 + 18}{3}=\frac{20}{3}\text{ in}^2 \end{align*} ]

Step2: Define (x) - coordinate of centroid formula

The (x) - coordinate of the centroid (\bar{x}) is (\bar{x}=\frac{1}{A}\int_{a}^{b}x\cdot y\mathrm{d}x). [ \begin{align*} \int_{0}^{1}x(4x^{5}-3x^{2}+12x + 1)\mathrm{d}x&=\int_{0}^{1}(4x^{6}-3x^{3}+12x^{2}+x)\mathrm{d}x\ &=\left[\frac{4x^{7}}{7}-\frac{3x^{4}}{4}+\frac{12x^{3}}{3}+\frac{x^{2}}{2}\right]_{0}^{1}\ &=\frac{4}{7}-\frac{3}{4}+4+\frac{1}{2}\ &=\frac{16 - 21+112 + 14}{28}\ &=\frac{121}{28} \end{align*} ] (\bar{x}=\frac{3}{20}\times\frac{121}{28}=\frac{363}{560}\text{ in})

Step3: Define (y) - coordinate of centroid formula

The (y) - coordinate of the centroid (\bar{y}) is (\bar{y}=\frac{1}{2A}\int_{a}^{b}y^{2}\mathrm{d}x). [ \begin{align*} y^{2}&=(4x^{5}-3x^{2}+12x + 1)^{2}\ &=16x^{10}-24x^{7}+9x^{4}+96x^{6}-72x^{3}+144x^{2}+8x^{5}-6x^{2}+24x + 1\ &=16x^{10}-24x^{7}+96x^{6}+8x^{5}+9x^{4}-72x^{3}+138x^{2}+24x + 1 \end{align*} ] [ \begin{align*} \int_{0}^{1}y^{2}\mathrm{d}x&=\int_{0}^{1}(16x^{10}-24x^{7}+96x^{6}+8x^{5}+9x^{4}-72x^{3}+138x^{2}+24x + 1)\mathrm{d}x\ &=\left[\frac{16x^{11}}{11}-\frac{24x^{8}}{8}+\frac{96x^{7}}{7}+\frac{8x^{6}}{6}+\frac{9x^{5}}{5}-\frac{72x^{4}}{4}+\frac{138x^{3}}{3}+\frac{24x^{2}}{2}+x\right]{0}^{1}\ &=\frac{16}{11}-3+\frac{96}{7}+\frac{4}{3}+\frac{9}{5}-18 + 46+12 + 1\ \end{align*} ] [ \begin{align*} \int{0}^{1}y^{2}\mathrm{d}x&=\frac{16\times105 - 3\times1155+96\times165+4\times385+9\times231-18\times1265+46\times1155+12\times1155+1\times1155}{11\times105}\ &=\frac{1680-3465 + 15840+1540+2079-22770+53130+13860+1155}{1155}\ &=\frac{(1680 + 1540+2079+53130+13860+1155)-(3465 + 22770)}{1155}\ &=\frac{73444 - 26235}{1155}\ &=\frac{47209}{1155} \end{align*} ] (\bar{y}=\frac{3}{40}\times\frac{47209}{1155}=\frac{47209}{15400}\text{ in})

Answer:

(\bar{x}=\frac{363}{560}\text{ in},\bar{y}=\frac{47209}{15400}\text{ in})