Evaluate the definite integral. 5dxV2x – 1

∫15x2x-1 dx Consider, ∫x2x-1 dx Put, 2x-1=t⇒21-0=dtdx⇒2 dx=dt⇒dx=dt2 Therefore, ∫x2x-1 dx=∫t+12tdt2…

Answered: 5 dx V2x – 1

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015

1st Edition

ISBN:9781680331141

Author:HOUGHTON MIFFLIN HARCOURT

Publisher:HOUGHTON MIFFLIN HARCOURT

Chapter10: Radical Functions And Equations

Section10.3: Solving Radical Equations

Problem 85E

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Question

Evaluate the definite integral.

Expert Solution

Given,

$\int_{1}^{5} \frac{x}{\sqrt{2 x - 1}} d x$

Consider,

$\int \frac{x}{\sqrt{2 x - 1}} d x$

Put,

$2 x - 1 = t \Rightarrow 2 (1) - 0 = \frac{d t}{d x} \Rightarrow 2 d x = d t \Rightarrow d x = \frac{d t}{2}$

Therefore,

$\begin{array}{rcl} \int \frac{x}{\sqrt{2 x - 1}} d x & = & \int \frac{\frac{t + 1}{2}}{\sqrt{t}} (\frac{d t}{2}) [∵ 2 x - 1 = t \Rightarrow 2 x = t + 1 \Rightarrow x = \frac{t + 1}{2}] \\ = & \frac{1}{4} \int \frac{t + 1}{\sqrt{t}} d t \\ = & \frac{1}{4} \int (\frac{t}{\sqrt{t}} + \frac{1}{\sqrt{t}}) d t \\ = & \frac{1}{4} \int (\sqrt{t} + \frac{1}{\sqrt{t}}) d t \\ = & \frac{1}{4} \int (t^{\frac{1}{2}} + t^{- \frac{1}{2}}) d t \\ = & \frac{1}{4} [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C \\ = & \frac{1}{4} [\frac{t^{\frac{3}{2}}}{\frac{3}{2}} + \frac{t^{\frac{1}{2}}}{\frac{1}{2}}] + C \\ = & \frac{1}{2} [\frac{t^{\frac{3}{2}}}{3} + t^{\frac{1}{2}}] + C \\ = & \frac{1}{2} [\frac{{(2 x - 1)}^{\frac{3}{2}}}{3} + {(2 x - 1)}^{\frac{1}{2}}] + C \\ = & \frac{1}{2} [\frac{{(2 x - 1)}^{\frac{3}{2}}}{3} + \sqrt{2 x - 1}] + C \end{array}$

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