∫_( 1) ^( 4) (√(((√z)−1)/( (√z)))) dz =?


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 138916 by mathsuji last updated on 20/Apr/21

$\underset{1}{\int^{4}} \sqrt{\frac{\sqrt{z} - 1}{\sqrt{z}}} d z = ?$
	Answered by bramlexs22 last updated on 20/Apr/21
	$(√(((√z)−1)/( (√z)))) = r ; (((√z) −1)/( (√z))) = r^2 → { ((z=1→r=0)),((z=4→r=(√(1/2)))) :} ⇒r^2 (√z) = (√z) −1 ⇒(√z) = (1/(1−r^2 )) ; z = (1−r^2 )^(−2) dz = −2(−2r)(1−r^2 )^(−3) dr dz = ((4r)/((1−r^2 )^3 )) dr I= ∫_0 ^((√2)/2) r(((4r)/((1−r^2 )^3 )))dr I=∫ 2r(((2r)/((1−r^2 )^3 )))dr by parts { ((u=2r⇒du=2dr)),((v=−∫ ((d(1−r^2 ))/((1−r^2 )^3 ))=(1/(2(1−r^2 )^2 )))) :} I= [ (r/((1−r^2 )^2 )) ]_0 ^((√2)/2) +∫_( 0) ^( ((√2)/2)) (dr/((1−r^2 )^2 )) now you can solve it$
	$\sqrt{\frac{\sqrt{z} - 1}{\sqrt{z}}} = r; \frac{\sqrt{z} - 1}{\sqrt{z}} = r^{2} \to {\begin{cases} z = 1 \to r = 0 \\ z = 4 \to r = \sqrt{\frac{1}{2}} \end{cases}$ $\Rightarrow r^{2} \sqrt{z} = \sqrt{z} - 1$ $\Rightarrow \sqrt{z} = \frac{1}{1 - r^{2}}; z = {(1 - r^{2})}^{- 2}$ $dz = - 2 (- 2 r) {(1 - r^{2})}^{- 3} dr$ $dz = \frac{4 r}{{(1 - r^{2})}^{3}} dr$ $I = \underset{0}{\int^{\frac{\sqrt{2}}{2}}} r (\frac{4 r}{{(1 - r^{2})}^{3}}) dr$ $I = \int 2 r (\frac{2 r}{{(1 - r^{2})}^{3}}) dr$ $by parts {\begin{cases} u = 2 r \Rightarrow du = 2 dr \\ v = - \int \frac{d (1 - r^{2})}{{(1 - r^{2})}^{3}} = \frac{1}{2 {(1 - r^{2})}^{2}} \end{cases}$ $I = {[\frac{r}{{(1 - r^{2})}^{2}}]}_{0}^{\frac{\sqrt{2}}{2}} + \int_{0}^{\frac{\sqrt{2}}{2}} \frac{dr}{{(1 - r^{2})}^{2}}$ $now you can solve it$
	Answered by Ankushkumarparcha last updated on 20/Apr/21

	$S o l u t i o n : Ω (s a y) = \int_{1}^{4} \sqrt{\frac{\sqrt{z} - 1}{\sqrt{z}}} d z (L e t I = \int \sqrt{\frac{\sqrt{z} - 1}{\sqrt{z}}} d z)$ $I = \int \sqrt{\frac{\sqrt{z} - 1}{\sqrt{z}}} d z \underset{s e t \sqrt{z} = y}{\underset{⏟}{=}} \int 2 \sqrt{y^{2} - y} d y$ $\frac{I}{2} = \int \sqrt{y^{2} - y} d y => \int \sqrt{{(y - \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} d y (∵ \int \sqrt{x^{2} - a^{2}} d x = \frac{x \sqrt{x^{2} - a^{2}}}{2} - \frac{a^{2}}{2} \log_{e} ∣ x + \sqrt{x^{2} - a^{2}} ∣ + C)$ $\frac{I}{2} = \frac{(2 \sqrt{z} - 1) \sqrt{\sqrt{z} (\sqrt{z} - 1)}}{4} - \frac{1}{8} \log_{e} ∣ \sqrt{z} - \frac{1}{2} + \sqrt{\sqrt{z} (\sqrt{z} - 1)} ∣ + C (N o w t a k i n g l i m i t s b o t h s i d e s f r o m 1 t o 4)$ $Ω = \int_{1}^{4} \sqrt{\frac{\sqrt{z} - 1}{\sqrt{z}}} d z = \frac{3 \sqrt{2}}{2} + \frac{1}{4} \log_{e} (3 - 2 \sqrt{2})$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum