Question

Find (x, y) position of centroid of the area under curve y=1+x+x^2 from x=0 to x=2. your answer should be correct to two decimals.​

Answer 1

$\large\underline{\bold{Solution-}}$

We have to first find the area bounded between the curve

$\sf \: y \: = \: 1 + x + {x}^{2} \: from \: x = 0 \: and \: x = 2$

Required area is given by

$\rm :\longmapsto\:A \: = \: \int_a^b \: y \: dx$

Here,

y = 1 + x + x²

a = 0

b = 2

So,

$\rm :\longmapsto\:A \: = \: \int_0^2 \: (1 + x + {x}^{2}) \: dx$

$\rm :\longmapsto\:A = \bigg(x + \dfrac{ {x}^{2} }{2} + \dfrac{ {x}^{3} }{3} \bigg) _0^2$

$\rm :\longmapsto\:A = 2 + \dfrac{4}{2} + \dfrac{8}{3} = \dfrac{12 + 12 + 16}{6} = \dfrac{20}{3}$

Now, using the formula for the x-coordinate of the centroid we have:

$\rm :\longmapsto\:{\overline x } = \dfrac{1}{A} \: \int_a^b \:x y \: dx$

$\rm :\longmapsto\:{\overline x } = \dfrac{3}{20} \: \int_0^2 \:x (1 + x + {x}^{2}) \: dx$

$\rm :\longmapsto\:{\overline x } = \dfrac{3}{20} \: \int_0^2 \:(x + {x}^{2} + {x}^{3} ) \: dx$

$\rm :\longmapsto\:{\overline x } = \dfrac{3}{20} \bigg(\dfrac{ {x}^{2} }{2} + \dfrac{ {x}^{3} }{3} + \dfrac{ {x}^{4} }{4} \bigg) _0^2$

$\rm :\longmapsto\:{\overline x } = \dfrac{3}{20}\bigg( \dfrac{4}{2} + \dfrac{8}{3} + \dfrac{16}{4} \bigg)$

$\rm :\longmapsto\:{\overline x } = \dfrac{3}{20} \bigg( 6 + \dfrac{8}{3} \bigg)$

$\rm :\longmapsto\:{\overline x } = \dfrac{3}{20}\bigg( \dfrac{18 + 8 }{3} \bigg)$

$\bf :\implies\:{\overline x } = \dfrac{26}{20} \: = \: 1.3$

Again,

using the formula for the y-coordinate of the centroid we have:

$\rm :\longmapsto\:{\overline y } = \dfrac{1}{A} \: \int_a^b \:\dfrac{ {y}^{2} }{2} \: dx$

$\rm :\longmapsto\:{\overline y } = \dfrac{3}{20} \: \int_0^2 \:\dfrac{ {(1 + x + {x}^{2} )}^{2} }{2} \: dx$

$\rm :\longmapsto\:{\overline y } = \dfrac{3}{40} \: \int_0^2 \:(1 + {x}^{2} + {x}^{4} + 2x + {2x}^{3} + {2x}^{2})\: dx$

$\ = \dfrac{3}{40} \bigg(x + \dfrac{ {x}^{3} }{3} + \dfrac{ {x}^{5} }{5} + {x}^{2} + \dfrac{ {x}^{4} }{2} + \dfrac{2 {x}^{3} }{3} \bigg) _0^2$

$\rm :\longmapsto\:{\overline y } = \dfrac{3}{40}\bigg(2 + \dfrac{8}{3} + \dfrac{32}{5} + 4 + \dfrac{16}{2} + \dfrac{16}{3}\bigg)$

$\rm :\longmapsto\:{\overline y } = \dfrac{3}{40}\bigg(22 + \dfrac{32}{5} \bigg)$

$\rm :\longmapsto\:{\overline y } = \dfrac{3}{40}\bigg( \dfrac{110 + 32}{5} \bigg)$

$\rm :\longmapsto\:{\overline y } = \dfrac{426}{200}$

$\bf\implies \:{\overline y } = 2.13$

$\bf \: Hence, \: position \: of \: centroid \: is \: (1.3,2.13)$