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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integral\[\int\limits_1^2(4x^3-5x^2+6x+9)\;dx\]

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Toolbox:
  • $\int \limits_a^b f(x)dx=F(b)-F(a)$
  • $\int x^ndx=\Large\frac{x^{n+1}}{n+1}$
Given $I=\int \limits_1^2(4x^3-5x^2+6x+9)dx$
 
On integrating we get,
 
$ \bigg[4(\Large\frac{x^4}{4})-5(\frac{x^3}{3})+6(\frac{x^2}{2})+9x\bigg]^2_1$
 
=$\bigg[x^4-\Large\frac{5x^3}{3}+3x^2+9x\bigg]^2_1$
 
Now applying limits we get
 
$\bigg[2^4-\large\frac{5 \times 2^3}{3}+3(2)^2+9 \times 2\bigg]-\bigg[1^4-\large\frac{5 \times 1^3}{3}+3(1)^2+9 \times 1\bigg]$
 
$\bigg[16-\frac{40}{3}+12+18 \bigg]-\bigg[1-\frac{5}{3}+3+9 \bigg]$
 
$\large\frac{98}{3}-\frac{34}{3}=\frac{64}{3}$

 

answered Feb 10, 2013 by meena.p
 
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