Answer to Question #139566 in Quantitative Methods for Moel Tariburu

Question

Find a value for n to ensure that the absolute error in approximating the integral
∫_0^2▒〖x cos⁡〖x dx〗 〗by the midpoint rule will be less than 10^-3

Answer 1

Absolute error in midpoint approximation is calculated as $\frac{(b-a)^3k}{24n^2}$ , where k is a max value of |f''(x)| on [a,b].

In this care a=0, b=2, f(x)=xcos(x), f'(x)=cos(x)-xsin(x), f''(x)=-sin(x)-sin(x)-xcos(x)=-2sin(x)-xcos(x), f'''(x)=-2cos(x)-cos(x)+xsin(x)=-3cos(x)+xsin(x), f''''(x)=3sin(x)+sin(x)-xcos(x)=4sin(x)+xcos(x).

-3cos(x)+xsin(x)=0

xsin(x)=3cos(x)

x=3cot(x). Using a numerical method, we get that x=1.1925 on [0,2].

|f''(0)|=|-2sin(0)-0|=0.

|f''(1.1925)|=|-2sin(1.1925)-1.1925cos(1.1925)|=2.299

|f''(2)||=|-2sin(2)-2cos(2)|=0.9863.

Therefore, a max value of |f''(x)| on [0,2] is 2.299

So, if absolute error is less than 0.001 then:

$\frac{2^3\cdot2.299}{24n^2}\leq0.001 \newline \newline 24n^2\cdot0.001\geq2^3\cdot2.299 \newline n^2\geq\frac{2^3\cdot2.299}{24\cdot0.001} \newline n^2\geq 766.33 \newline n\geq \sqrt{766.33} \newline n\geq 27.6827$

So n has to be at least 28.

Answer: n=28