The approximate value of int_2^{10} x^2 dx by | vedclass

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Question

(A) Given the integral $\int_2^{10} x^2 dx$ with $n = 4$ intervals.Here, $a = 2$, $b = 10$, and the step size $h = \frac{b - a}{n} = \frac{10 - 2}{4}

Vedclass · Accepted Answer

$336$

Vedclass · Answer

(A) Given the integral $\int_2^{10} x^2 dx$ with $n = 4$ intervals.Here, $a = 2$, $b = 10$, and the step size $h = \frac{b - a}{n} = \frac{10 - 2}{4} = 2$.The points are $x_0 = 2, x_1 = 4, x_2 = 6, x_3 = 8, x_4 = 10$.The corresponding values of $f(x) = x^2$ are:$y_0 = f(2) = 4$$y_1 = f(4) = 16$$y_2 = f(6) = 36$$y_3 = f(8) = 64$$y_4 = f(10) = 100$Using the trapezoidal rule:$\int_a^b f(x) dx \approx \frac{h}{2} [y_0 + 2(y_1 + y_2 + y_3) + y_4]$$\int_2^{10} x^2 dx \approx \frac{2}{2} [4 + 2(16 + 36 + 64) + 100]$$= 1 \cdot [4 + 2(116) + 100] = 4 + 232 + 100 = 336$.

The approximate value of $\int_2^{10} x^2 dx$ by using the trapezoidal rule with $4$ equal intervals is:

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