slolve the definite integral∫_1^ ^2 (1/x)dxusingg> ng trapezoidal rule with 4 sub intervals hencefind an approximate value of ln 2.


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Question Number 25472 by rita1608 last updated on 10/Dec/17

$s l o l v e t h e d e f i n i t e i n t e g r a l \int_{1^{}}^{2} \frac{1}{x} d x u s i n g g > n g$ $t r a p e z o i d a l r u l e w i t h 4 s u b i n t e r v a l s$ $h e n c e f i n d a n a p p r o x i m a t e v a l u e o f$ $l n 2 .$
	Answered by prakash jain last updated on 10/Dec/17

	$Four subintervals$ $Δ x = \frac{2 - 1}{4} = \frac{1}{4}$ $x_{0} = 1, x_{1} = 1.25, x_{2} = 1.5, x_{3} = 1.75, x_{4} = 2$ $Δ x = 0.25$ $I = \frac{Δ x}{2} [f (x_{0}) + 2 f (x_{1}) + 2 f (x_{2}) + 2 f (x_{3}) + f (x_{4})]$ $= \frac{0.25}{2} [1 + \frac{2}{1.25} + \frac{2}{1.5} + \frac{2}{1.75} + \frac{1}{2}]$ $= \frac{1}{2 \times 4} [1 + \frac{2 \times 4}{5} + \frac{2 \times 2}{3} + \frac{2 \times 4}{7} + \frac{1}{2}]$ $= \frac{1}{8} [1 + 1.6 + 1.33 + 1.14 + . 5]$ $= \frac{5.57}{8} = 0.69$ $\int_{1}^{2} \frac{1}{x} d x = {[\ln x]}_{1}^{2} = \ln 2$ $\ln 2 \approx 0.69$
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