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Question

Evaluate $$\displaystyle\int _{ 1 }^{ 5 }{ \frac { dx }{ \sqrt { 2x-1 }  }  } $$.


Solution

Let $$I = \displaystyle\int _{ 1 }^{ 5 }{ \frac { dx }{ \sqrt { 2x-1 }  }  } $$.

Substitute $$2x-1=t^2\Rightarrow dx=tdt$$
Also upper limit $$\sqrt{2(5)-1}=3$$
and lower limit $$=\sqrt{2(1)-1}=1$$
Thus 
$$I = \displaystyle\int _{ 1 }^{ 3 }{ \frac { tdt }{ t  }  } =\int_1^3dt=[t]_1^3=3-1=2$$.


Mathematics

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