December 8, 2022

Definite integration involving modulus function

Definite  Integration Involving Modulus Function

There is no anti-derivative for a modulus function; however we know it’s definition

\left | x \right |= \left\{\begin{matrix} x\, \, if \, \, x\geq 0\\ -x \, \, otherwise \end{matrix}\right.

Thus we can split up our integral  in two parts. One  part must be completely negative and the another  must be completely positive. This strategy is known as splitting .  To split correctly it is better to  draw the modulus graph so that we can get the correct equation to represent each portion of the modulus graph .

Example 1 Find the value of \int_{\pi }^{10\pi }\left | sin x \right |dx .

Sol.  For a periodic function f(x) with period T\int_{a}^{a+nT}f(x)dx=n\int_{0}^{T}f(x)dx, where n is any natural number.

Here  f(x)=\left | sinx \right |  whose  period  is \pi.

Hence \int_{\pi }^{10\pi }\left | sin x \right |dx=\int_{\pi }^{\pi+9\pi }\left | sin x \right |dx= 9\int_{0}^{\pi}\left | sin x \right | dx

= 9\int_{0}^{\pi }sin x \, dx         (  since sinx \geq 0, \forall \, \, x\in [0,\, \pi ]  )

=9{\left [ -cosx \right]}_{0}^{\pi }   = 9 x 2 =18 .

Thus  \int_{\pi }^{10\pi }\left | sin x \right |dx= 18

graph of |sinx|

Example 2.  Evaluate \int_{-1}^{2}\left | x^{3} -x\right |dx

Sol.  We note that x^{3}-x\geq 0 on   \left [ -1, \, 0 \right ] and x^{3}-x\leq 0 on \left [ 0, \, 1\right ]   and that x^{3}-x\geq 0 on \left [ 1, 2 \right ] .

graph 0f x^3−x

Graph of  x^{3}-x

So \int_{-1}^{2}\left |x^{3} -x\right|dx=\int_{-1}^{0}(x^{3}-x)dx+\int_{0}^{1}-(x^{3}-x)dx+\int_{1}^{2}(x^{3}-x)dx

=\int_{-1}^{0}(x^{3}-x)dx+ \int_{0}^{1}(x-x^{3})dx+ \int_{1}^{2}(x^{3}-x)dx

=\left [ \frac{x^{4}}{4}-\frac{x^{2}}{2} \right ]_{-1}^{0}+\left [ \frac{x^{2}}{2}-\frac{x^{4}}{4} \right ]_{0}^{1}+\left [ \frac{x^{4}}{4}-\frac{x^{2}}{2}\right ]_{1}^{2}

= -(\frac{1}{4}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{4})+(4-2)-(\frac{1}{4}-\frac{1}{2})

=\frac{11}{4}

Example 3. Find the value \int_{0}^{\frac{\pi }{2}}\left | cosx-sinx \right |dx

Sol.  We know that  sin\, x   and  cos\, \, x has one common value at x=\frac{\pi }{4}.   In the interval \left [ 0,\, \frac{\pi}{4} \right ] ,   cos\, x\geq sin\, x and in \left [ \frac{\pi }{4} , \, \frac{\pi }{2}\right ]  , sin\, x\geq cos\, x.  Thus  cosx-sinx \geq 0 in \left [ 0,\, \frac{\pi}{4} \right ]  and cosx-sinx \leq 0 in \left [ \frac{\pi }{4} , \, \frac{\pi }{2}\right ].

So \int_{0}^{\frac{\pi }{2}}\left | cosx-sinx \right |dx=\int_{0}^{\frac{\pi }{4}}\left ( cosx-sinx \right )dx+\int_{\frac{\pi }{4}}^{\frac{\pi }{2}}-\left ( cosx-sinx \right )dx

=[sin\, x+cos\, x]_{0}^{\frac{\pi }{4}}+[-cos\, x -sin\, x]_{\frac{\pi }{4}}^{\frac{\pi }{2}}

=(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1)+\left \{- 1-(-\frac{1}{\sqrt{2}} -\frac{1}{\sqrt{2}})\right\}

=2\sqrt{2}-2

https://youtu.be/EAYq3ySAy9k

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