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

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(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

Vedclass · Accepted Answer

$333$

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(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 $x_i = 0, 1, 2, 3, 4, 5, 6$ are:$y_0 = f(0) = 0$$y_1 = f(1) = 1$$y_2 = f(2) = 8$$y_3 = f(3) = 27$$y_4 = f(4) = 64$$y_5 = f(5) = 125$$y_6 = f(6) = 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:

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