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Find the area of the region bounded by the curve $y=3x^{2}-x$ and the $x$-axis between $x=-1$ and $x=1$

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Toolbox:
  • Area bounded by the curve $t=f(x),$ the x-axis and the ordinates $x=a,x=b$ is $\int \limits_a^b f(x) dx $ or $ \int \limits _a^b y dx $
  • If the curve lies below the x-axis for $a \leq x \leq b,$ then the area is $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
Given $y=3x^2 -x$
Step 1:
$y= 3(x^2-\large\frac{x}{3}+\frac{1}{36})-\frac{1}{12}$
$(y+\large\frac{1}{12})$$=3 (x- \large\frac{1}{6})^2$
This is a parabola passing through $(0,0)$ vector at $\bigg(\large\frac{1}{6},\frac{-1}{12}\bigg)$ and opening upwards
Step 2:
Area bounded by $y=3x^2 -x$
$A=-1,x=1$
$A=A_1 (between \; x=-1,x=0)$
$+A_2 (between \; x=0,x =\large\frac{1}{3})$
$+A_3 (between \; x =\large\frac{1}{3}$$x=1)$
$A_2$ lies below the x-axis (on the -ve side of y-axis)
Step 3:
$\therefore A=A_1+A_2+A_3=\int \limits_{-1}^0 y dx-\int \limits_0^{\large\frac{1}{3}} y dx+\int \limits _{\large\frac{1}{3}}^1 ydx$
$\qquad=\int \limits_{-1}^0 (3x^2-x)dx-\int \limits_0^{\large\frac{1}{3}} (3x^2-x)dx+\int \limits _{\large\frac{1}{3}}^1 (3x^2-x)dx$
$\qquad=\bigg(x^3-\large\frac{x^2}{2}\bigg]_{-1}^0-\bigg(x^3-\large\frac{x^2}{2}\bigg]_0^{\large\frac{1}{3}}+\bigg(x^3-\large\frac{x^2}{2}\bigg]_{\large\frac{1}{3}}^1$
$\qquad= -(-1-\large\frac{1}{2})-\bigg[\large\frac{1}{27}-\frac{1}{18}\bigg]+\bigg[1-\large\frac{1}{2}-\bigg(\large\frac{1}{27}-\frac{1}{18}\bigg)\bigg]$
$\qquad= \large\frac{3}{2}+\large\frac{1}{54}+\large\frac{1}{2}+\frac{1}{54}$
$\qquad=2 + \large\frac{1}{27}$
$\qquad= \large\frac{55}{27}$$sq.units$
answered Aug 15, 2013 by meena.p
 

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