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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