Evaluate-1-4-x-2-x-2x-1-dx-Question-ID-53-How-does-the-limits-change-in-the-solution-of-Q-No-53-

Question Number 15022 by Tinkutara last updated on 07/Jun/17

Evaluate : \int_{1}^{4} \frac{x^{2} + x}{\sqrt{2 x + 1}} d x (Question ID :

53) How does the limits change in the

solution of Q . No . 53 ?

Answered by Joel577 last updated on 07/Jun/17

Let u = 2 x + 1 \Leftrightarrow x = \frac{u - 1}{2}

d u = 2 d x

\underset{1}{\int^{4}} \frac{x (x + 1)}{\sqrt{2 x + 1}} d x

= \underset{1}{\int^{4}} \frac{\frac{u - 1}{2} . \frac{u + 1}{2}}{2 \sqrt{u}} d u

= \frac{1}{8} \underset{1}{\int^{4}} \frac{u^{2} - 1}{\sqrt{u}} d u

= \frac{1}{8} \underset{1}{\int^{4}} (u^{2} - 1) (u^{- 1 / 2}) d u

= \frac{1}{8} \underset{1}{\int^{4}} u^{3 / 2} - u^{- 1 / 2} d u

= \frac{1}{8} {[\frac{2}{5} {(2 x + 1)}^{5 / 2} - 2 {(2 x + 1)}^{1 / 2}]}_{1}^{4}

c o n t i n u e \dots

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