The area of the region bounded by the curve y | vedclass

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(A) We have the curve $y=x^{3}$ and the point $A(1,1)$.First, we find the slope of the tangent at $A(1,1)$ by differentiating the curve equation:$\fra

Vedclass · Accepted Answer

$\frac{1}{12} 	ext{ sq unit}$

Vedclass · Answer

(A) We have the curve $y=x^{3}$ and the point $A(1,1)$.First, we find the slope of the tangent at $A(1,1)$ by differentiating the curve equation:$\frac{dy}{dx} = 3x^{2}$.At $x=1$, the slope is $m = 3(1)^{2} = 3$.The equation of the tangent at $(1,1)$ is given by $y-1 = 3(x-1)$, which simplifies to $y = 3x-2$.The tangent intersects the $X$-axis where $y=0$, so $3x-2=0$, which gives $x = \frac{2}{3}$.The required area is the area under the curve $y=x^{3}$ from $x=0$ to $x=1$ minus the area under the tangent line $y=3x-2$ from $x=\frac{2}{3}$ to $x=1$.$	ext{Required Area} = \int_{0}^{1} x^{3} dx - \int_{2/3}^{1} (3x-2) dx$$= \left[ \frac{x^{4}}{4} ight]_{0}^{1} - \left[ \frac{3x^{2}}{2} - 2x ight]_{2/3}^{1}$$= \left( \frac{1}{4} - 0 ight) - \left[ \left( \frac{3}{2} - 2 ight) - \left( \frac{3}{2} \cdot \frac{4}{9} - 2 \cdot \frac{2}{3} ight) ight]$$= \frac{1}{4} - \left[ -\frac{1}{2} - \left( \frac{2}{3} - \frac{4}{3} ight) ight]$$= \frac{1}{4} - \left[ -\frac{1}{2} - \left( -\frac{2}{3} ight) ight]$$= \frac{1}{4} - \left[ -\frac{1}{2} + \frac{2}{3} ight]$$= \frac{1}{4} - \left[ \frac{-3+4}{6} ight] = \frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12} 	ext{ sq unit}$.

The area of the region bounded by the curve $y=x^3$, its tangent at $(1,1)$, and the $X$-axis is

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