Definite Integration Set 5

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Example -9

Evaluate \({I_n} = \int\limits_0^{\pi {\rm{/2}}} {{{\sin }^n}x\,dx} \)

Solution: Let \({f_n}(x) = \int {{{\sin }^n}x\,dx} \)

\[\begin{align}& \Rightarrow\quad {f_n}(x)dx = \int {{{\sin }^n}x\,dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\quad\;\;\;= \int {\mathop {{{\sin }^{n - 1}}}\limits_{\scriptstyle \downarrow \atop{\scriptstyle{\rm{Ist}}\atop\scriptstyle{\rm{function}}}} x \cdot \mathop {\sin x}\limits_{\scriptstyle \downarrow \atop{\scriptstyle{\rm{IInd}}\atop\scriptstyle{\rm{function}}}} \,dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\quad\;\;\;= - {\sin ^{n - 1}}x\cos x + (n - 1)\int {{{\sin }^{n - 2}}x \cdot {{\cos }^2}x\,dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\quad\;\;\;= - {\sin ^{n - 1}}x\cos x + (n - 1)\int {{{\sin }^{n - 2}}x(1 - {{\sin }^2}x)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\quad\;\;\;= - {\sin ^{n - 1}}x\cos x + (n - 1){f_{n - 2}}(x) - (n - 1){f_n}(x)\,\,\,\\ &\Rightarrow\quad n{f_n}(x) = - {\sin ^{n - 1}}x\cos x + (n - 1){f_{n - 2}}(x)\\ &\Rightarrow\quad \,\,\,\,\,{f_n}(x) = \frac{{ - {{\sin }^{n - 1}}x\cos x}}{n} + \left( {\frac{{n - 1}}{n}} \right){f_{n - 2}}(x)\\ &\Rightarrow\quad \left. {{I_n} = {f_n}(x)} \right|_0^{\pi {\rm{/2}}} = 0 + \left( {\frac{{n - 1}}{n}} \right){I_{n - 2}}\end{align}\]

Thus, our recursive relation is

\[{I_n} = \left( {\frac{{n - 1}}{n}} \right){I_{n - 2}}\]

We now use this repeatedly:

\[\begin{align}&\,\,\,\,\,\,\,\,{I_n} = \left( {\frac{{n - 1}}{n}} \right){I_{n - 2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;= \frac{{(n - 1)(n - 3)}}{{n(n - 2)}}{I_{n - 4}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\quad\vdots \\\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;= \left\{ \begin{gathered}\frac{{(n - 1)(n - 3).............(2)}}{{n(n - 2)....................(3)}}{I_1} & ({\rm{if\; }}n\;{\rm{ is \; odd)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{or}}\\\frac{{(n - 1)(n - 3).............(1)}}{{n(n - 2)....................(2)}}{I_0} & ({\rm{if \;}}n\;{\rm{ is\; even)}}\end{gathered} \right\}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\end{align}\]

Now, \[\,\,\,\,\,\,\,\,{I_1} = \int\limits_0^{\pi {\rm{/2}}} {\sin x\,dx = 1} \]

and \(\,\,\,\,\,\,\,\,{I_0} = \int\limits_0^{\pi {\rm{/2}}} {dx = \frac{\pi }{2}} \)

Thus, \({I_n}\) is now obtainable from (1).

Notice that \({J_n} = \int\limits_0^{\pi {\rm{/2}}} {{{\cos }^n}x\,dx} \) will be the same as \({I_n}\) , by virtue of property – 9.

Example -10

Suppose that \(a + b = 4\) where \(a \in (0,2)\) and \(g(x)\) is a differentiable function such that \(g'(x) > 0\,\,\,\,\rlap{-} V\,\,\, x\in \mathbb{R}\) .

Show that \(\int\limits_0^a {g(x)dx} + \int\limits_0^b {g(x)dx} \) increases as (ba) increases:

Solution: Let us first intuitively try to justify the stated assertion.

Since \(g'(x) > 0\,\,\,\,\rlap{-} V\,\,\,x \in \mathbb{R}\) , \(g(x)\) is an increasing function on ¡ .

Now, since 0 < a < 2 and \(a + b = 4,\) a and b will lie symmetrically about the point x = 2. As a increases from 0 to 2, b decreases from 4 to 2. Assume an arbitrary (increasing) configuration for \(g(x)\) :

Now we interpret the area given by \(\int\limits_0^a {g(x)dx} + \int\limits_0^b {g(x)dx} \) from the figure below

We have \(A = \int\limits_0^a {g(x)dx} + \int\limits_0^b {g(x)dx} \)

\[\begin{array}{l}\,\,\,\, = {A_1} + ({A_1} + {A_2})\\\,\,\,\, = 2{A_1} + {A_2}\\\,\,\,\, = ({A_1} + {A_2} + {A_3}) - ({A_3} - {A_1}) & & \left( \begin{array}{l}{\rm{We\; introduced \;}}{\;A_{\rm{3}}}{\rm{\; into\; }}\\{\rm{the \;expression}}\end{array} \right)\\\,\,\,\, = \int\limits_0^4 {g(x)dx - ({A_3} - {A_1})} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\end{array}\]

Now consider what will happen as (ba) increases. The area \({A_2}\) expands, while \({A_1}\) and \({A_3}\) will shrink (visualise this ‘expansion’ and ‘shrinking’ of areas from Fig.21). But the important point is that the shrink in \({A_3}\) and \({A_1}\) will not be the same. Since \(g(x)\) is increasing \({A_3}\) , will shrink more than \({A_1}\) will. Thus, as \((b - a)\) increases, \(({A_3} - {A_1})\) will decrease.

From (1), therefore, A will increase!

The maximum A is obtained when \({A_3} - {A_1} = 0,\) i.e.,

When a = 0 and b = 4:

\[{A_{\max }} = \int\limits_0^4 {g(x)dx} \]

Now we redo this example analytically. We treat a as a variable which can lie in (0, 2). Also ,Thus,

\[\begin{align}&A = \int\limits_0^a {g(x)dx} + \int\limits_0^b {g(x)} \,dx\\\,\,\,\, &\;\;= \int\limits_0^a {g(x)dx} + \int\limits_0^{4 - a} {g(x)} \,dx\\ \Rightarrow\qquad \frac{{dA}}{{da}} &= g(a) - g(4 - a)\end{align}\]
Since \(g(x)\) is an increasing function and a < 4 – a, we have \(g(a) < g(4 - a)\) so that

\[\frac{{dA}}{{da}} < 0\]

Thus, A decreases with respect to a or equivalently, increases with respect to ba.

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