The area of the region enclosed by the curve | vedclass

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(A) Given the curve $y = x^3$. The derivative is $\frac{dy}{dx} = 3x^2$.At the point $(-1, -1)$,the slope of the tangent is $m = 3(-1)^2 = 3$.The equa

Vedclass · Accepted Answer

$\frac{27}{4}$

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(A) Given the curve $y = x^3$. The derivative is $\frac{dy}{dx} = 3x^2$.At the point $(-1, -1)$,the slope of the tangent is $m = 3(-1)^2 = 3$.The equation of the tangent line is $y - (-1) = 3(x - (-1))$,which simplifies to $y = 3x + 2$.To find the point of intersection of the curve $y = x^3$ and the tangent $y = 3x + 2$,we set $x^3 = 3x + 2$,which gives $x^3 - 3x - 2 = 0$.Factoring,we get $(x + 1)^2(x - 2) = 0$,so the intersection points are $x = -1$ and $x = 2$.The area $A$ is given by the integral $\int_{-1}^{2} ((3x + 2) - x^3) dx$.$A = [\frac{3x^2}{2} + 2x - \frac{x^4}{4}]_{-1}^{2}$.$A = (\frac{3(4)}{2} + 2(2) - \frac{16}{4}) - (\frac{3(1)}{2} + 2(-1) - \frac{1}{4})$.$A = (6 + 4 - 4) - (\frac{3}{2} - 2 - \frac{1}{4}) = 6 - (\frac{6 - 8 - 1}{4}) = 6 - (-\frac{3}{4}) = 6 + \frac{3}{4} = \frac{27}{4}$.

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

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