The area (in sq. units) of the region bounded | vedclass

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(A) To find the area bounded by the curve $y=1-x-6x^2$ and the $X$-axis,we first find the intersection points with the $X$-axis by setting $y=0$.$1-x-

Vedclass · Accepted Answer

$\frac{125}{216}$

Vedclass · Answer

(A) To find the area bounded by the curve $y=1-x-6x^2$ and the $X$-axis,we first find the intersection points with the $X$-axis by setting $y=0$.$1-x-6x^2=0$$6x^2+x-1=0$$6x^2+3x-2x-1=0$$3x(2x+1)-1(2x+1)=0$$(3x-1)(2x+1)=0$So,$x = \frac{1}{3}$ and $x = -\frac{1}{2}$.The required area is given by the integral:$	ext{Area} = \int_{-1/2}^{1/3} (1-x-6x^2) dx$$= [x - \frac{x^2}{2} - 2x^3]_{-1/2}^{1/3}$$= (\frac{1}{3} - \frac{(1/3)^2}{2} - 2(1/3)^3) - (-\frac{1}{2} - \frac{(-1/2)^2}{2} - 2(-1/2)^3)$$= (\frac{1}{3} - \frac{1}{18} - \frac{2}{27}) - (-\frac{1}{2} - \frac{1}{8} + \frac{1}{4})$$= (\frac{18-3-4}{54}) - (\frac{-4-1+2}{8})$$= \frac{11}{54} - (-\frac{3}{8}) = \frac{11}{54} + \frac{3}{8}$$= \frac{44+81}{216} = \frac{125}{216} 	ext{ sq. units}$.

The area (in sq. units) of the region bounded by the $X$-axis and the curve $y=1-x-6x^2$ is

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