The area bounded by the curve x(x^2 + p) = y | vedclass

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(C) The given curve is $y = x^3 + px + 1$. To find the area bounded by the curve and the line $y = 1$,we set $x^3 + px + 1 = 1$,which gives $x^3 + px

Vedclass · Accepted Answer

$\frac{p^2}{2}$

Vedclass · Answer

(C) The given curve is $y = x^3 + px + 1$. To find the area bounded by the curve and the line $y = 1$,we set $x^3 + px + 1 = 1$,which gives $x^3 + px = 0$. This implies $x(x^2 + p) = 0$. Assuming $p  0$. The intersection points are $x = 0, x = a, x = -a$. The area $A$ is given by the integral of the absolute difference between the curve and the line: $A = \int_{-a}^{0} (x^3 - a^2x + 1 - 1) dx + \int_{0}^{a} (1 - (x^3 - a^2x + 1)) dx$. $A = \int_{-a}^{0} (x^3 - a^2x) dx + \int_{0}^{a} (-x^3 + a^2x) dx$. Evaluating the first integral: $[\frac{x^4}{4} - \frac{a^2x^2}{2}]_{-a}^{0} = 0 - (\frac{a^4}{4} - \frac{a^4}{2}) = \frac{a^4}{4}$. Evaluating the second integral: $[-\frac{x^4}{4} + \frac{a^2x^2}{2}]_{0}^{a} = (-\frac{a^4}{4} + \frac{a^4}{2}) - 0 = \frac{a^4}{4}$. Total area $A = \frac{a^4}{4} + \frac{a^4}{4} = \frac{a^4}{2}$. Since $p = -a^2$,then $p^2 = a^4$. Therefore,the area is $\frac{p^2}{2}$.

The area bounded by the curve $x(x^2 + p) = y - 1$ and the line $y = 1$ is:

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