March 3, 2024

# 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

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

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

$\int_{\pi&space;}^{10\pi&space;}\left&space;|&space;sin&space;x&space;\right&space;|dx$
.

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  $f(x)=\left&space;|&space;sinx&space;\right&space;|$  whose  period  is $\pi$

.

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

= $9\int_{0}^{\pi&space;}sin&space;x&space;\,&space;dx$

(  since $sinx&space;\geq&space;0,&space;\forall&space;\,&space;\,&space;x\in&space;[0,\,&space;\pi&space;]$  )

=$9{\left&space;[&space;-cosx&space;\right]}_{0}^{\pi&space;}$

= 9 x 2 =18 .

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

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

Sol.  We note that $x^{3}-x\geq&space;0$

on   $\left&space;[&space;-1,&space;\,&space;0&space;\right&space;]$ and $x^{3}-x\leq&space;0$
on $\left&space;[&space;0,&space;\,&space;1\right&space;]$   and that $x^{3}-x\geq&space;0$
on $\left&space;[&space;1,&space;2&space;\right&space;]$ .

Graph of  $x^{3}-x$

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

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

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

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

=$\frac{11}{4}$

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

Sol.  We know that  $sin\,&space;x$

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

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

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

=$(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1)+\left&space;\{-&space;1-(-\frac{1}{\sqrt{2}}&space;-\frac{1}{\sqrt{2}})\right\}$

=$2\sqrt{2}-2$

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#### Bina singh

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