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Home  >>  CBSE XII  >>  Math  >>  Integrals

Find an anti derivative(or integral)of the function by the method of inspection $(ax+b)^2$

1 Answer

Toolbox:
  • $\frac{d}{dx}[F(x)+c]=f(x),$
  • $\int f(x)\;dx=F(x)+c$
We know that $\frac{d}{dx}(ax+b)^3=3a(ax+b)^2$
 
$\Rightarrow (ax+b)^2=\frac{1}{3a}\frac{d}{dx}(ax+b)^3.$
 
$\qquad\;\;=\frac{d}{dx}\begin{bmatrix}\frac{1}{3a}(ax+b)^3\end{bmatrix}.$
 
Hence $\frac{1}{3a}(ax+b)^3$ is an anti derivative of $(ax+b)^2$.

 

answered Jan 25, 2013 by sreemathi.v
 
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