Question

find the RMS value of XE^{X} between X=0andx=1​

Answer 1

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Answer 2

RMS value of $\bf f(x) = x{e}^{x}$ is $\bf \frac{1}{2} \sqrt{e^2-1}$

Given:

• $f(x) = x{e}^{x}$

To find:

• Find the RMS value of the given function between x=0 and x=1

Solution:

Formula to be used:

• $\bf MSV = \frac{1}{b - a} \int _a ^{b} {(f(x)})^{2} dx$ and
• $\bf RMS = \sqrt{ MSV}$
• Integration by parts: $\int UV \: dx =U \int Vdx - \int \left (\: \frac{d U}{dx} \int Vdx \right)dx$

Step 1:

Calculate MSV (Mean square value ) of function.

Here

$f(x) = x{e}^{x}$

$x = 0 \: and \: x = 1$

so,

$a = 0 \: and \: b = 1$

$MSV = \frac{1}{1 - 0} \int _0 ^{1} (xe^x)^{2} dx$

$MSV = \frac{1}{1} \int _0 ^{1} (x^2 {e}^{2x}) dx$

Solve integration by parts.

$\int {x}^{2} {e}^{2x} \: dx = {x}^{2} \int {e}^{2x} dx - \int \left (\: \frac{d {x}^{2} }{dx} \int {e}^{2x} dx \right)dx$

$\int {x}^{2} {e}^{2x} \: dx = \frac{ {x}^{2} { {e}^{2x} } }{2} - \int \left (\: 2x \times \frac{1}{2} {e}^{2x} \right)dx$

$\int {x}^{2} {e}^{2x} \: dx = \frac{ {x}^{2} { {e}^{2x} } }{2} - \int \: x {e}^{2x} dx$

again apply integration by parts

$\int {x}^{2} {e}^{2x} \: dx = \frac{ {x}^{2} { {e}^{2x} } }{2} - {x} \int {e}^{2x} dx + \int \left (\: \frac{d {x}}{dx} \int {e}^{2x} dx \right)dx$

$\int {x}^{2} {e}^{2x} \: dx = \frac{ {x}^{2} { {e}^{2x} } }{2} - \frac{ {x} { {e}^{2x} } }{2} + \frac{ { {e}^{2x} } }{4}$

Apply limit

$\int_0 ^{1} {x}^{2} {e}^{2x} \: dx = \left(\frac{ {x}^{2} { {e}^{2x} } }{2} - \frac{ {x} { {e}^{2x} } }{2} + \frac{ { {e}^{2x} } }{4} \right) \bigg | _0 ^{1}$

$\int_0 ^{1} {x}^{2} {e}^{2x} \: dx = \left(\frac{ {1}^{2} { {e}^{2} } }{2} - \frac{ {1.} { {e}^{2} } }{2} + \frac{ { {e}^{2} } }{4} - 0 + 0 - \frac{ {e}^{0} }{4} \right)$

Cancel the common terms with opposite sign.

$\int_0 ^{1} {x}^{2} {e}^{2x} \: dx = \left( \frac{ { {e}^{2} } }{4} - \frac{1}{4} \right)$

$\bf MSV = \frac{ { {e}^{2} -1} }{4}$

Thus,

Mean square value of f(x) is $\bf \frac{ { {e}^{2} -1} }{4}$

Step 2:

Find the RMS value of the function.

$RMS= \sqrt{MSV}$

$RMS= \sqrt{\frac{ { {e}^{2} -1} }{4} }$

$\bf RMS= \frac{1}{2} \sqrt{e^2-1}$

Thus,

RMS value of $\bf f(x) = x{e}^{x}$is $\bf\frac{1}{2} \sqrt{e^2-1}$

Learn more:

1) find the RMS value of √8-4x square between x =0 and x= 2

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2) what is the peak and average values of 500 volts RMS?

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