An extremum value of y = int_{0}^{x} (t - 1)( | vedclass

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(D) Given $y = \int_{0}^{x} (t^2 - 3t + 2) dt$. Applying the Fundamental Theorem of Calculus,we find the derivative: $\frac{dy}{dx} = (x - 1)(x - 2)$.

Vedclass · Accepted Answer

Both $(A)$ and $(B)$

Vedclass · Answer

(D) Given $y = \int_{0}^{x} (t^2 - 3t + 2) dt$. Applying the Fundamental Theorem of Calculus,we find the derivative: $\frac{dy}{dx} = (x - 1)(x - 2)$. For extremum values,set $\frac{dy}{dx} = 0$,which gives $x = 1$ and $x = 2$. Now,find the second derivative: $\frac{d^2y}{dx^2} = 2x - 3$. At $x = 1$,$\frac{d^2y}{dx^2} = 2(1) - 3 = -1 $y_{max} = \int_{0}^{1} (t^2 - 3t + 2) dt = [\frac{t^3}{3} - \frac{3t^2}{2} + 2t]_{0}^{1} = \frac{1}{3} - \frac{3}{2} + 2 = \frac{2 - 9 + 12}{6} = \frac{5}{6}$. At $x = 2$,$\frac{d^2y}{dx^2} = 2(2) - 3 = 1 > 0$,so $y$ has a minimum at $x = 2$. $y_{min} = \int_{0}^{2} (t^2 - 3t + 2) dt = [\frac{t^3}{3} - \frac{3t^2}{2} + 2t]_{0}^{2} = \frac{8}{3} - 6 + 4 = \frac{8}{3} - 2 = \frac{2}{3}$. Thus,the extremum values are $5/6$ and $2/3$. Therefore,the correct option is $(D)$.

An extremum value of $y = \int_{0}^{x} (t - 1)(t - 2) dt$ is

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