Question

Find first moment of area about x-axis for area under curve y=1+x+x^2 from x=o to x=2. Your answer should be correct to one decimal.​

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

Let us consider a curve y = f(x) from x = a to x = b, then first moment of area about x - axis is given by

$\bf \: M_y \: = \:\: \int_a^b \:\dfrac{ \{{f(x) \}}^{2} }{2} \: dx$

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

Given that

• f(x) = 1 + x + x²

• a = 0

• b = 2

Now,

First moment of area about x - axis is given by

$\rm :\longmapsto\: \: M_y \: = \:\: \int_a^b \:\dfrac{ \{{f(x) \}}^{2} }{2} \: dx$

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

$\rm :\longmapsto\: \: M_y \: = \:\: \int_0^2 \:\dfrac{1 + {x}^{2} + {x}^{4} +2x + 2 {x}^{3} + 2{x}^{2}}{2} \: dx$

$\rm :\longmapsto\: \: M_y \: = \:\: \int_0^2 \:\dfrac{ 1 + 2x + {3x}^{2} + 2 {x}^{3} + {x}^{4}}{2} \: dx$

$\rm :\longmapsto\:M_y = \dfrac{1}{2} \bigg( x + \dfrac{2 {x}^{2} }{2} + \dfrac{3 {x}^{3} }{3} + \dfrac{2 {x}^{4} }{4} + \dfrac{ {x}^{5} }{5}\bigg) _0^2$

$\rm :\longmapsto\:M_y = \dfrac{1}{2} \bigg(2 + 4 + 8 + 8 + \dfrac{32}{5} \bigg)$

$\rm :\longmapsto\:M_y = \dfrac{1}{2} \bigg(22+ \dfrac{32}{5} \bigg)$

$\rm :\longmapsto\:M_y = \dfrac{1}{2} \bigg( \dfrac{110 + 32}{5} \bigg)$

$\rm :\longmapsto\:M_y = \dfrac{1}{2} \times \bigg(\dfrac{142}{5} \bigg)$

$\bf\implies \:M_y = 14.2$

Additional Information :-

Let us consider a curve y = f(x) from x = a to x = b, then

1. First moment of area about y - axis is given by

$\rm :\longmapsto\:M_x = \: \int_a^b \:x y \: dx$

and

$2. \: \: \sf \: Centroid \: is \: given \: by \: ({\overline x }, \: {\overline y})$

where,

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

and

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

and

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