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Use the error formula to find $n$ such that the error is less than or equal to $0.00001$ using the Trapezoidal Rule. $\newline$ $\int_{0}^{\frac{\pi}{2}}\sin x\,dx$

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Conjugate root theorems

Full solution

Q. Use the error formula to find

n

such that the error is less than or equal to

0.00001

using the Trapezoidal Rule.

\newline

\int_{0}^{\frac{\pi}{2}}\sin x\,dx

Error Bound Calculation: The error bound for the Trapezoidal Rule is given by: $\newline$ $E_t \leq \frac{(b-a)^3}{12n^2} \cdot \max|f''(x)|$ on $[a, b]$ $\newline$ Here, $a = 0$ , $b = \frac{\pi}{2}$ , and we need to find $\max|f''(x)|$ for $f(x) = \sin(x)$ .
Second Derivative of $\sin(x)$ : The second derivative of $f(x) = \sin(x)$ is $f''(x) = -\sin(x)$ . The maximum value of $|f''(x)|$ on $[0, \frac{\pi}{2}]$ is $1$ since $|\sin(x)| \leq 1$ for all $x$ .
Error Formula Substitution: Now, plug in the values into the error formula: $\newline$ $E_t \leq (\frac{\pi}{2} - 0)^3 / (12n^2) \times 1$ $\newline$ We want $E_t \leq 0.00001$ .
Solving for n: Solve for n: $\newline$ $0.00001 \geq \left(\frac{\pi}{2}\right)^3 / (12n^2)$ $\newline$ $n^2 \geq \left(\frac{\pi}{2}\right)^3 / (12 \times 0.00001)$ $\newline$ $n \geq \sqrt{\left(\frac{\pi}{2}\right)^3 / (12 \times 0.00001)}$
Calculating n: Calculate n: $\newline$ $n \geq \sqrt{\left(\frac{\pi}{2}\right)^3 / (0.00012)}$ $\newline$ $n \geq \sqrt{\left(\frac{\pi^3}{8 \times 0.00012}\right)}$ $\newline$ $n \geq \sqrt{\left(\frac{\pi^3}{0.00096}\right)}$

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