The approximate value of int_1^3 frac{dx}{2+3 | vedclass

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

Question

(C) Let $f(x) = \frac{1}{2+3x}$. We divide the interval $[1,3]$ into $n=2$ equal parts. $h = \frac{3-1}{2} = 1$. The points are $x_0 = 1, x_1 = 2, x_2

Vedclass · Accepted Answer

$\frac{29}{110}$

Vedclass · Answer

(C) Let $f(x) = \frac{1}{2+3x}$. We divide the interval $[1,3]$ into $n=2$ equal parts. $h = \frac{3-1}{2} = 1$. The points are $x_0 = 1, x_1 = 2, x_2 = 3$. $f(x_0) = f(1) = \frac{1}{2+3(1)} = \frac{1}{5} = 0.2$. $f(x_1) = f(2) = \frac{1}{2+3(2)} = \frac{1}{8} = 0.125$. $f(x_2) = f(3) = \frac{1}{2+3(3)} = \frac{1}{11} \approx 0.0909$. Using Simpson's rule: $\int_a^b f(x) dx \approx \frac{h}{3} [f(x_0) + 4f(x_1) + f(x_2)]$. $\int_1^3 f(x) dx \approx \frac{1}{3} [0.2 + 4(0.125) + 0.0909] = \frac{1}{3} [0.2 + 0.5 + 0.0909] = \frac{0.7909}{3} \approx 0.2636$. Comparing with the options, $\frac{29}{110} \approx 0.2636$.

The approximate value of $\int_1^3 \frac{dx}{2+3x}$ using Simpson's rule and dividing the interval $[1,3]$ into two equal parts is

Similar Questions

$\int_{0}^{\pi /2} \{ x - [\sin x] \} \,dx$ is equal to

If $f(x) = \begin{cases} \sqrt{1-x} & 0 \leqslant x \leqslant 1 \\ (7x-6)^{-1/3} & 1 < x \leqslant 2 \end{cases}$,then $\int_{0}^{2} f(x) dx$ is equal to

$\int_{1 / 2}^{1 / \sqrt{2}} \frac{1}{\left(x+\sqrt{1-x^2}\right)\left(1-x^2\right)} d x=$

$\int_{-2}^3 |1-x^2| dx =$

Let $f(x) = \{x\}$ denote the fractional part of a real number $x$. Then, the value of $\int_{0}^{\sqrt{3}} f(x^2) dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module