int_{1}^{2} frac{x^{3} - 1}{x^{2}} dx = | vedclass

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(C) We are given the integral $I = \int_{1}^{2} \frac{x^{3} - 1}{x^{2}} dx$. First,simplify the integrand: $\frac{x^{3} - 1}{x^{2}} = \frac{x^{3}}{x^{

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(C) We are given the integral $I = \int_{1}^{2} \frac{x^{3} - 1}{x^{2}} dx$. First,simplify the integrand: $\frac{x^{3} - 1}{x^{2}} = \frac{x^{3}}{x^{2}} - \frac{1}{x^{2}} = x - x^{-2}$. Now,integrate term by term: $I = \int_{1}^{2} (x - x^{-2}) dx = \left[ \frac{x^{2}}{2} - \frac{x^{-1}}{-1} \right]_{1}^{2} = \left[ \frac{x^{2}}{2} + \frac{1}{x} \right]_{1}^{2}$. Substitute the limits: $I = \left( \frac{2^{2}}{2} + \frac{1}{2} \right) - \left( \frac{1^{2}}{2} + \frac{1}{1} \right) = \left( 2 + \frac{1}{2} \right) - \left( \frac{1}{2} + 1 \right) = 2 + \frac{1}{2} - \frac{1}{2} - 1 = 1$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

$\int_{1}^{2} \frac{x^{3} - 1}{x^{2}} dx =$

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