What is the average value of the function $u (x) = 10 x sin (x^{2})$ $u(x) = 10xsin(x^2)$ on the interval $[0, \sqrt{π}]$ $[0,sqrt pi]$ ?

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What is the average value of the function $u (x) = 10 x sin (x^{2})$ $u(x) = 10xsin(x^2)$ on the interval $[0, \sqrt{π}]$ $[0,sqrt pi]$ ?

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Answer 1 · 2016-12-20T14:07:35

See below.

Explanation:

The average value is

$\frac{1}{\sqrt{π} - 0} \int_{0}^{\sqrt{π}} 10 x sin (x^{2}) d x = \frac{5}{\sqrt{π}} \int_{0}^{\sqrt{π}} 2 x sin (x^{2}) d x$ $1/(sqrtpi-0) int_0^sqrtpi 10xsin(x^2) dx = 5/sqrtpiint_0^sqrtpi 2xsin(x^2) dx$

$= \frac{5}{\sqrt{π}} {[- cos (x^{2})]}_{0}^{\sqrt{π}}$ $= 5/sqrtpi [-cos(x^2)]_0^sqrtpi$

$= \frac{12}{\sqrt{π}}$ $= 12/sqrtpi$

Pedantic Note

$\frac{12 \sqrt{π}}{π}$ $(12sqrtpi)/pi$ does NOT have a rational denominator.

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What is the average value of the function $u (x) = 10 x sin (x^{2})$ $u(x) = 10xsin(x^2)$ on the interval $[0, \sqrt{π}]$ $[0,sqrt pi]$ ?

1 Answer

Explanation:

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What is the average value of the function $u (x) = 10 x sin (x^{2})$ $u(x) = 10xsin(x^2)$ on the interval $[0, \sqrt{π}]$ $[0,sqrt pi]$ ?