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Differentiation of a Determinant

∫ 133 x-2 d x

Question

$\int_{1}^{3} (3 x - 2) d x$

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Solution

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$
$Here a = 1, b = 3, f (x) = 3 x - 2, h = \frac{3 - 1}{n} = \frac{2}{n} Therefore, I = \int_{1}^{3} (3 x - 2) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [(3 - 2) + (3 + 3 h - 2) + (3 + 6 h - 2) . . . . . . . . . . . . . . . + (3 (n - 1) h + 3 - 2)] = \lim_{h \to 0} h [n + 3 h (1 + 2 + 3 . . . . . . . . . + (n - 1))] = \lim_{h \to 0} h [n + 3 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [n + 3 n - 3] = \lim_{n \to \infty} 2 (4 - \frac{3}{n}) = 8$

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