How do you find the average value of the function $u (x) = 10 x sin (x^{2})$ $u(x) = 10xsin(x^2)$ on the interval $[0, \sqrt{π}]$ $[ 0, sqrtpi ]$ ?

Question

How do you find the average value of the function $u (x) = 10 x sin (x^{2})$ $u(x) = 10xsin(x^2)$ on the interval $[0, \sqrt{π}]$ $[ 0, sqrtpi ]$ ?

1 Answer

Impact of this question

2651 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-31T12:33:26

The average value is $\frac{10}{\sqrt{π}}$ $10/sqrtpi$ . [Average rate of change is $0$ $0$ ]

Explanation:

The question asks for the average value which is:

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

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

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

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

$= \frac{5}{\sqrt{π}} [2] = \frac{10}{\sqrt{π}}$ $= 5/sqrtpi [2] = 10/sqrtpi$

A different question

The question was posted under the topic "Average Rate of Change . . . " which is different from the average value.

The average rate of change of this function on this interval is

$\frac{f (\sqrt{π}) - f (0)}{\sqrt{π} - 0} = \frac{10 \sqrt{π} sin ({(\sqrt{π})}^{2}) - 10 (0) sin (0^{2})}{\sqrt{π} - 0}$ $(f(sqrtpi)-f(0))/(sqrtpi-0) = (10sqrtpisin((sqrtpi)^2) - 10(0)sin(0^2))/(sqrtpi - 0)$

$= \frac{0}{\sqrt{π}} = 0$ $= 0/sqrtpi =0$

Science

Math

Humanities

... and beyond

How do you find the average value of the function $u (x) = 10 x sin (x^{2})$ $u(x) = 10xsin(x^2)$ on the interval $[0, \sqrt{π}]$ $[ 0, sqrtpi ]$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the average value of the function u(x)=10xsin(x2)u(x) = 10xsin(x^2) on the interval [0,√π][ 0, sqrtpi ]?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the average value of the function $u (x) = 10 x sin (x^{2})$ $u(x) = 10xsin(x^2)$ on the interval $[0, \sqrt{π}]$ $[ 0, sqrtpi ]$ ?