The value of the integral int_0^4 frac{d x}{1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the integral $\int_0^4 f(x) d x$ where $f(x) = \frac{1}{1+x^2}$ and $h=1$.The values of $f(x)$ at $x = 0, 1, 2, 3, 4$ are:$y_0 = f(0) = 1$$y

Vedclass · Accepted Answer

$\frac{113}{85}$

Vedclass · Answer

(D) Given the integral $\int_0^4 f(x) d x$ where $f(x) = \frac{1}{1+x^2}$ and $h=1$.The values of $f(x)$ at $x = 0, 1, 2, 3, 4$ are:$y_0 = f(0) = 1$$y_1 = f(1) = \frac{1}{2}$$y_2 = f(2) = \frac{1}{5}$$y_3 = f(3) = \frac{1}{10}$$y_4 = f(4) = \frac{1}{17}$By using the Trapezoidal rule:$\int_0^4 f(x) d x = \frac{h}{2} [ (y_0 + y_4) + 2(y_1 + y_2 + y_3) ]$$= \frac{1}{2} [ (1 + \frac{1}{17}) + 2(\frac{1}{2} + \frac{1}{5} + \frac{1}{10}) ]$$= \frac{1}{2} [ \frac{18}{17} + 2(\frac{5+2+1}{10}) ]$$= \frac{1}{2} [ \frac{18}{17} + 2(\frac{8}{10}) ]$$= \frac{1}{2} [ \frac{18}{17} + \frac{8}{5} ]$$= \frac{1}{2} [ \frac{90 + 136}{85} ]$$= \frac{1}{2} [ \frac{226}{85} ] = \frac{113}{85}$

The value of the integral $\int_0^4 \frac{d x}{1+x^2}$ obtained by using the Trapezoidal rule with $h=1$ is

Similar Questions

$\int\limits_{ - 1}^{\frac{3}{2}} {|x\sin \pi x|dx} $ equals

If $\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$,then $b$ is equal to

The value of $\int_0^{\frac{\pi}{2}}|\sin x-\cos x| d x$ is

$\int_0^1 {{e^{2\ln x}}dx} = $

The value of the integral $\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module