2) Use the property \[\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx\]
\[I=\int_{1}^{2011}\frac{\sqrt{x}}{\sqrt{2012-x}+\sqrt{x}}dx=\int_{1}^{2011}\frac{\sqrt{2012-x}}{\sqrt{x}+\sqrt{2012-x}}dx\]
Therefore,\[2I=\int_{1}^{2011}\frac{\sqrt{x}}{\sqrt{2012-x}+\sqrt{x}}dx+\int_{1}^{2011}\frac{\sqrt{2012-x}}{\sqrt{2012-x}+\sqrt{x}}dx=\int_{1}^{2011}dx\]
Therefore, \[I=1005\]
–
Aaron Jerry Ninan
·
6 months, 2 weeks ago

Log in to reply

1) By algebric manupilations the given integral can be converted into -\[\int \frac{(1-\frac{1}{x^{2}})}{(x+\frac{1}{x}+2)\sqrt{x+\frac{1}{x}+1}}dx\]
Now by u-substitution method-\[u^{2}=x+\frac{1}{x}+1\]
And \[(1-\frac{1}{x^{2}})dx=2u\cdot du\]
Now the problem can be solved using standard integrals.
–
Aaron Jerry Ninan
·
6 months, 2 weeks ago

## Comments

Sort by:

TopNewest2) Use the property \[\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx\] \[I=\int_{1}^{2011}\frac{\sqrt{x}}{\sqrt{2012-x}+\sqrt{x}}dx=\int_{1}^{2011}\frac{\sqrt{2012-x}}{\sqrt{x}+\sqrt{2012-x}}dx\] Therefore,\[2I=\int_{1}^{2011}\frac{\sqrt{x}}{\sqrt{2012-x}+\sqrt{x}}dx+\int_{1}^{2011}\frac{\sqrt{2012-x}}{\sqrt{2012-x}+\sqrt{x}}dx=\int_{1}^{2011}dx\] Therefore, \[I=1005\] – Aaron Jerry Ninan · 6 months, 2 weeks ago

Log in to reply

1) By algebric manupilations the given integral can be converted into -\[\int \frac{(1-\frac{1}{x^{2}})}{(x+\frac{1}{x}+2)\sqrt{x+\frac{1}{x}+1}}dx\] Now by u-substitution method-\[u^{2}=x+\frac{1}{x}+1\] And \[(1-\frac{1}{x^{2}})dx=2u\cdot du\] Now the problem can be solved using standard integrals. – Aaron Jerry Ninan · 6 months, 2 weeks ago

Log in to reply

– Rishabh Bhatnagar · 6 months, 2 weeks ago

thanx a lotLog in to reply