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Evaluate:\[\int\limits_0^{16} \frac{x^{1/4}}{1+x^{1/2}}dx\]

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  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)If we substitute a function f(x) as t, then $f'(x)dx=dt$ hence the function $\int f(x)dx=\int t.dt$
  • (iii)$\int \frac {dx}{x^2+a^2}=\frac{1}{a} \tan ^{-1} (x/a)+c$
Given $ I=\int\limits_0^{16} \frac{x^{1/4}}{1+x^{1/2}}dx$
Let $ x^{1/4}=t.$ on differentiating w.r.t x
$\frac{1}{4} x^{-3/4}dx=dt$
$=\frac{1}{4} t^{-3} dx =dt \qquad =\frac{1}{4t^3}dx=dt$
As we substitute for t the limit changes
Therefore $dx=4t^3dt$
when $x=0,t=0, x=16,t=2$
Substituting for t and dt,
$I=\int \limits_0^2 \large\frac{t(4t^3)dt}{1+t^2}$
$=4 \int _0^2 \large\frac{t^4 dt}{1+t^2}$
This is an improper rational function,
hence on dividing we get,
Therefore $\frac{t^4}{1+t^2}=(t^2-1)+\frac{1}{1+t^2}$
Therefore $4 \int \limits_0^2\frac{t^4}{1+t^2}dt=4 \int \limits_0^2[(t^2-1)+\frac{1}{1+t^2}]dt$
On seperating the terms,
$I=\int \limits_0^2 t^2-\int \limits_0^2 dt+\int \limits_0^2 \frac{dt}{1+t^2}$
$\large\frac{dt}{1+t^2}$ is of the form $\frac {dx}{x^2+a^2}=\frac{1}{a} \tan ^{-1} (x/a)$
On integrating we get
Therefore $ I=4 \bigg\{[t^3/3]_0^2-[t]_0^2+[\tan ^{-1}(t)]_0^2 \bigg\}$
On applying the limits,
$=4 \bigg\{(\frac{8}{3}-0)-(2-0)+\tan ^{-1}(2)\bigg\}$
$=\frac {32}{3}-8+4 \tan ^{-1}(2)$
$=\frac{8}{3}+4 \tan ^{-1}(2)$



answered Mar 15, 2013 by meena.p
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