A) \[\frac{19}{6}\]
B) \[\frac{17}{6}\]
C) \[\frac{13}{6}\]
D) Cannot be determined
Correct Answer: A
Solution :
| We have,\[I=\int\limits_{0}^{2}{\sqrt{x+\sqrt{x+\sqrt{x+...\infty }}}}dx\] |
| Let \[y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}\] |
| \[y=\sqrt{x+y}\]\[\Rightarrow \]\[{{y}^{2}}-y-x=0\]\[\Rightarrow \]\[y=\frac{1\pm \,\sqrt{4x+1}}{2}\]\[\Rightarrow \]\[y=\frac{1+\sqrt{4x+1}}{2}y\ge 1\] |
| \[\because \]\[I=\int\limits_{0}^{2}{\frac{1+\sqrt{4x+1}}{2}dx}\]\[\Rightarrow \]\[I=\frac{1}{2}\left[ x+\frac{{{(1+4x)}^{3/2}}}{6} \right]_{0}^{2}\]\[\Rightarrow \]\[I=\frac{1}{2}\left[ \left( 2+\frac{27}{6} \right)-\left( 0+\frac{1}{6} \right) \right]\]\[\Rightarrow \]\[I=\frac{1}{2}\left( \frac{12+27-1}{6} \right)=\frac{19}{6}\] |
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