Dividing the interval [0,6] into 6 equal part | vedclass

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

Question

(D) Given the interval $[0, 6]$ is divided into $n = 6$ equal parts.The width of each sub-interval is $h = \frac{b-a}{n} = \frac{6-0}{6} = 1$.The valu

Vedclass · Accepted Answer

$333$

Vedclass · Answer

(D) Given the interval $[0, 6]$ is divided into $n = 6$ equal parts.The width of each sub-interval is $h = \frac{b-a}{n} = \frac{6-0}{6} = 1$.The values of $f(x) = x^3$ at $x = 0, 1, 2, 3, 4, 5, 6$ are:$y_0 = f(0) = 0^3 = 0$$y_1 = f(1) = 1^3 = 1$$y_2 = f(2) = 2^3 = 8$$y_3 = f(3) = 3^3 = 27$$y_4 = f(4) = 4^3 = 64$$y_5 = f(5) = 5^3 = 125$$y_6 = f(6) = 6^3 = 216$By the trapezoidal rule:$\int_0^6 x^3 dx \approx \frac{h}{2} \{y_0 + y_6 + 2(y_1 + y_2 + y_3 + y_4 + y_5)\}$$= \frac{1}{2} \{0 + 216 + 2(1 + 8 + 27 + 64 + 125)\}$$= \frac{1}{2} \{216 + 2(225)\}$$= \frac{1}{2} \{216 + 450\}$$= \frac{666}{2} = 333$.

Dividing the interval $[0,6]$ into $6$ equal parts and by using the trapezoidal rule,the value of $\int_0^6 x^3 dx$ is approximately:

Similar Questions

For $0 < x \leq \pi$,$\sinh ^{-1}(\cot x)$ is equal to

In maize,hybrid vigour is exploited by

An eight-digit number divisible by $9$ is to be formed using digits from $0$ to $9$ without repeating the digits. The number of ways in which this can be done is: (in $(7!)$)

Sporeine was developed in

Given $0 \leq x \leq \frac{1}{2}$,then the value of $\tan \left(\sin ^{-1}\left(\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^2}}{\sqrt{2}}\right)-\sin ^{-1} x\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module