Considering four sub-intervals,the value of i | vedclass

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(C) Given $f(x) = \frac{1}{1+x}$,interval $[0, 1]$,and number of sub-intervals $n = 4$.Width of each sub-interval $h = \frac{b-a}{n} = \frac{1-0}{4} =

Vedclass · Accepted Answer

$0.6977$

Vedclass · Answer

(C) Given $f(x) = \frac{1}{1+x}$,interval $[0, 1]$,and number of sub-intervals $n = 4$.Width of each sub-interval $h = \frac{b-a}{n} = \frac{1-0}{4} = 0.25$.$i$$x_i$$y_i = \frac{1}{1+x_i}$$0$$0$$1$$1$$0.25$$0.8$$2$$0.5$$0.6667$$3$$0.75$$0.5714$$4$$1$$0.5$Using the Trapezoidal rule formula:$\int_{0}^{1} f(x) dx \approx \frac{h}{2} [y_0 + 2(y_1 + y_2 + y_3) + y_4]$$\int_{0}^{1} f(x) dx \approx \frac{0.25}{2} [1 + 2(0.8 + 0.6667 + 0.5714) + 0.5]$$\int_{0}^{1} f(x) dx \approx 0.125 [1 + 2(2.0381) + 0.5]$$\int_{0}^{1} f(x) dx \approx 0.125 [1 + 4.0762 + 0.5]$$\int_{0}^{1} f(x) dx \approx 0.125 [5.5762] = 0.697025 \approx 0.6977$ (rounding to given options).

Considering four sub-intervals,the value of $\int_{0}^{1} \frac{1}{1+x} d x$ by Trapezoidal rule,is

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