Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • (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
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App