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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\sqrt{ax+b}$

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Toolbox:
  • Method of substitution:
  • Given $\int f(x)dx$ can be transformed into another form by changing independent variable x to t by substituting x=g(t).
  • Consider $I=\int f(x)dx.$
  • Put x=g(t) so that $\frac{dx}{dx}=g'(t).
  • dx=g'(t)dt.
  • Thus $ I=\int f(g(t).g'(t))dt.$
Given $I=\int\sqrt {ax+b}.dx.$
 
Let ax+b=t.
 
$\;\;$adx=dt
 
$\;\;\;dx=\frac{1}{a}dt.$
 
Substituting for t and dt we get,
 
$I=\int\sqrt t.\frac{dt}{a}.$
 
$\;\;\;=\frac{1}{a}\int\sqrt t.dt=\frac{1}{a}\int t^\frac{1}{2}dt.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{a}\begin{bmatrix}\frac{t^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}+c.$
 
Substituting back for t we get,
 
$\frac{2}{3a}(ax+b)^\frac{3}{2}+c.$
 

 

answered Jan 28, 2013 by sreemathi.v
 
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