Help please. Find the centroid of the area. Where the area bounded by the curve y=2(1+x^3) and the coordinate axes

$dA = y \, dx$

$\displaystyle A = \int_{x_1}^{x_2} y \, dx$

$\displaystyle A = 2 \int_{-1}^0 (1 + x^3) \, dx$

$A = 1.5 ~ \text{unit}^2$

$\displaystyle A\,Xg = \int x_c \, dA$

$\displaystyle 1.5Xg = \int_{x_1}^{x_2} (-x) \, (y \, dx)$

$\displaystyle 1.5Xg = -2\int_{-1}^{0} x(1 + x^3) \, dx$

$1.5Xg = 3/5$

$Xg = 2/5 ~ \text{unit}$

$\displaystyle A\,Yg = \int y_c \, dA$

$\displaystyle 1.5Yg = \int_{x_1}^{x_2} (y / 2) \, (y \, dx)$

$\displaystyle 1.5Yg = \frac{1}{2}\int_{x_1}^{x_2} y^2 \, dx$

$\displaystyle 1.5Yg = 2\int_{-1}^{0} (1 + x^3)^2 \, dx$

$1.5Yg = 9/7$

$Yg = 6/7 ~ \text{unit}$

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$dA = y \, dx$

$\displaystyle A = \int_{x_1}^{x_2} y \, dx$

$\displaystyle A = 2 \int_{-1}^0 (1 + x^3) \, dx$

$A = 1.5 ~ \text{unit}^2$

$\displaystyle A\,Xg = \int x_c \, dA$

$\displaystyle 1.5Xg = \int_{x_1}^{x_2} (-x) \, (y \, dx)$

$\displaystyle 1.5Xg = -2\int_{-1}^{0} x(1 + x^3) \, dx$

$1.5Xg = 3/5$

$Xg = 2/5 ~ \text{unit}$

$\displaystyle A\,Yg = \int y_c \, dA$

$\displaystyle 1.5Yg = \int_{x_1}^{x_2} (y / 2) \, (y \, dx)$

$\displaystyle 1.5Yg = \frac{1}{2}\int_{x_1}^{x_2} y^2 \, dx$

$\displaystyle 1.5Yg = 2\int_{-1}^{0} (1 + x^3)^2 \, dx$

$1.5Yg = 9/7$

$Yg = 6/7 ~ \text{unit}$