Let a, b and c be positive constants. The val | vedclass

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Question

$I = \int\limits_0^1 {(ac{x^{b + 1}} + {a^3}b{x^{3b + 5}})\,dx} $; $ac&middot;= \left. {\frac{{{x^{b + 2}}}}{{b + 2}} + \frac{{{a^3}b{x^{3b + 6}}}}{{3

Vedclass · Accepted Answer

$\sqrt {\frac{{3c}}{2}} $

Vedclass · Answer

$I = \int\limits_0^1 {(ac{x^{b + 1}} + {a^3}b{x^{3b + 5}})\,dx} $; $ac&middot;= \left. {\frac{{{x^{b + 2}}}}{{b + 2}} + \frac{{{a^3}b{x^{3b + 6}}}}{{3b + 6}}} \right]_{\,0}^{\,1} $

$= \frac{{ac}}{{b + 2}} + \frac{{{a^3}b}}{{3(b + 2)}}$
$I(b) = \frac{1}{{3(b + 2)}} [3ac + a^3b]$

$\Rightarrow \, \frac{{{a^3}\left( {b + \frac{{3c}}{{{a^2}}}} \right)}}{{3(b + 2)}}$
If this is independent of $b$, then $\frac{{3c}}{{{a^2}}} = 2$
Alternatively :

If the integral is independent of $b$, then $I&rsquo;(b) = 0$
$\Rightarrow I&rsquo;(b) = \frac{{ - ac}}{{{{(b + 2)}^2}}}$ $+D\left( {\frac{{{a^3}}}{3}\,\cdot\,\left( {\frac{{b + 2 - 2}}{{b + 2}}} \right)} \right)$
$= \frac{{ - ac}}{{{{(b + 2)}^2}}}$ $+ \frac{{{a^3}}}{3}\left( {\,\frac{2}{{{{(b + 2)}^2}}}} \right)$ $= \frac{1}{{{{(b + 2)}^2}}}\,\,\,\left( {\frac{{2{a^3}}}{3} - ac} \right)$

$= \frac{{2{a^3}}}{3}$

 $= ac \Rightarrow a = \sqrt {\frac{{3c}}{2}} $

Let $a, b$ and $c$ be positive constants. The value of $‘a’$ in terms of $‘c’$ if the value of integral $\int\limits_0^1 {(ac{x^{b + 1}} + {a^3}b{x^{3b + 5}})\,dx} $ is independent of $b$ equals

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