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

The value of $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sin x)^{-4} \,dx$ is

Evaluate the definite integral $\int_{0}^{\frac{\pi}{4}} \sin 2x \,dx$.

Let the domain of the function $f(x) = \log_2 \log_4 \log_6(3 + 4x - x^2)$ be $(a, b)$. If $\int_0^{b-a} [x^2] dx = p - \sqrt{q} - \sqrt{r}$,where $p, q, r \in \mathbb{N}$,$\gcd(p, q, r) = 1$,and $[\cdot]$ is the greatest integer function,then $p + q + r$ is equal to

Evaluate the following definite integral as the limit of a sum:
$\int_{1}^{4}(x^{2}-x) dx$

$\int_0^1 {(1 - x)^9} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module