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Integrals

Long Answer Type

271.

Evaluate: $\int e^{x} (\frac{\sin 4 x - 4}{1 - \cos 4 x}) dx$

272.

Evaluate: $\int \frac{1 - x^{2}}{x (1 - 2 x)} dx$

273.
Evaluate $\int_{1}^{3} (3 x^{2} + 2 x) dx$ as limit of sums.

$I = \int_{1}^{3} (3 x^{2} + 2 x) dx$

Here a = 1, b = 3

$f (x) = 3 x^{2} + 2 x h = \frac{b - a}{n} = \frac{2}{n} Since, \int_{a}^{b} f (x) dx = \lim_{h \to 0} h [f (a) + F (a + h) . . . . . . + f (+ (n - 1) h)] so, \int_{1}^{3} (3 x^{2} + 2 x) = \lim_{h \to 0} h [(3 (1)^{2} + 2 (1)) + (3 (1 + h)^{2} + 2 (1 + h)) + 3 (1 + 2 h)^{2} + 2 (1 + 2 h)) . . . . . . . . . . + 3 (1 + (n - 1) h)^{2} + 2 (1 + (n - 1) h)] = \lim_{h \to 0} h [3 (n) + 3 (h^{2} + 4 h^{2} + . . . . . . . . . + (n - 1)^{2} h^{2}) + 3 (2 h + 4 h + . . . . . . . + 2 (n - 1) h) + 2 n + 2 (h + 2 h + . . . . . . . . . . + (n - 1) h)] = \lim_{h \to 0} h [5 n h + 3 h^{3} (1^{2} + 2^{2} + . . . . . + (n - 1)^{2} + 6 h^{2} (1 + 2 + . . . . . . . . . . . + (n - 1)) + 2 h^{2} (1 + 2 + (n - 1))]$

$= \lim_{h \to 0} [5 n h + 3 h^{3} \times \frac{(n - 1) n (2 n - 1)}{6} + \frac{8 h^{2} (n) (n - 1)}{2}] = \lim_{h \to 0} [10 + \frac{(n h - h) n h (2 n h - h)}{2} + 4 (n h) (n h - h)] = [10 + \frac{2 \times 2 \times 4}{2} + 4 x 2 x 2] = 10 + 8 + 16 = 34$