How do you evaluate the definite integral $\int \frac{u - 2}{\sqrt{u}}$ $int (u-2)/sqrtu$ from [1,4]?

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How do you evaluate the definite integral $\int \frac{u - 2}{\sqrt{u}}$ $int (u-2)/sqrtu$ from [1,4]?

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Answer 1 · 2018-06-08T04:59:17

$ϕ = \int_{1}^{4} \frac{u - 2}{\sqrt{u}} d u = \frac{2}{3}$ $phi= int_1^4(u-2)/sqrtu du =2/3$

Explanation:

Let,
$ϕ = \int_{1}^{4} \frac{u - 2}{\sqrt{u}} d u$ $phi= int_1^4(u-2)/sqrtu du$

$u = t^{2}$ $u=t^2$ $\Rightarrow$ $=>$ $d u = 2 t d t$ $du=2tdt$

When, $u = 1$ $u=1$ $\Rightarrow$ $=>$ $t = 1$ $t=1$ and $u = 4$ $u=4$ $\Rightarrow$ $=>$ $t = 2$ $t=2$

$ϕ = \int_{1}^{2} \frac{t^{2} - 2}{t} 2 t d t$ $phi= int_1^2(t^2-2)/t 2tdt$

$ϕ = 2 \int_{1}^{2} t^{2} - 2 d t$ $phi=2 int_1^2t^2-2dt$

$ϕ = 2 {[\frac{t^{3}}{3} - 2 t]}_{1}^{2}$ $phi=2 [t^3/3-2t]_1^2$

$ϕ = 2 [\frac{8}{3} - 4 - \frac{1}{3} + 2]$ $phi=2 [8/3-4-1/3+2]$

$ϕ = 2 [\frac{7}{3} - 2]$ $phi=2 [7/3-2]$

$ϕ = 2 [\frac{1}{3}]$ $phi=2 [1/3]$

$ϕ = \frac{2}{3}$ $phi=2/3$

$ϕ = \int_{1}^{4} \frac{u - 2}{\sqrt{u}} d u = \frac{2}{3}$ $phi= int_1^4(u-2)/sqrtu du =2/3$

Answer 2 · 2018-06-08T05:01:17

$I = \int_{1}^{4} \frac{u - 2}{\sqrt{u}} d u = \frac{2}{3}$ $I=int_1^4 (u-2)/sqrtudu=2/3$

Explanation:

Here,

$I = \int_{1}^{4} \frac{u - 2}{\sqrt{u}} d u$ $I=int_1^4 (u-2)/sqrtudu$

$= \int_{1}^{4} [\frac{u}{\sqrt{u}} - \frac{2}{\sqrt{u}}] d u$ $=int_1^4 [u/sqrtu-2/sqrtu]du$

$= \int_{1}^{4} [\sqrt{u} - \frac{2}{\sqrt{u}}] d u$ $=int_1^4 [sqrtu-2/sqrtu]du$

$= \int_{1}^{4} [u^{\frac{1}{2}} - 2 u^{- \frac{1}{2}}] d u$ $=int_1^4 [u^(1/2)-2u^(-1/2)]du$

$= {[\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - 2 \times \frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}]}_{1}^{4}$ $=[u^(1/2+1)/(1/2+1)-2xx(u^(-1/2+1))/(-1/2+1)]_1^4$

$= {[\frac{u^{\frac{3}{2}}}{\frac{3}{2}} - 2 \times \frac{u^{\frac{1}{2}}}{\frac{1}{2}}]}_{1}^{4}$ $=[u^(3/2)/(3/2)-2xxu^(1/2)/(1/2)]_1^4$

$= ⎡ ⎣ \frac{{(4)}^{\frac{3}{2}}}{\frac{3}{2}} - 2 \times \frac{{(4)}^{\frac{1}{2}}}{\frac{1}{2}} ⎤ ⎦ - ⎡ ⎣ \frac{{(1)}^{\frac{3}{2}}}{\frac{3}{2}} - 2 \times \frac{{(1)}^{\frac{1}{2}}}{\frac{1}{2}} ⎤ ⎦$ $=[(4)^(3/2)/(3/2)-2xx(4)^(1/2)/(1/2)]- [(1)^(3/2)/(3/2)-2xx(1)^(1/2)/(1/2)]$

$= [\frac{{(2)}^{3}}{\frac{3}{2}} - 2 \times \frac{2}{\frac{1}{2}}] - [\frac{1}{\frac{3}{2}} - 2 \times \frac{1}{\frac{1}{2}}]$ $=[(2)^3/(3/2)-2xx2/(1/2)]-[1/(3/2)-2xx1/(1/2)]$

$= [\frac{2}{3} \times 8 - 8] - [\frac{2}{3} - 4]$ $=[2/3xx8-8]-[2/3-4]$

$= \frac{16}{3} - 8 - \frac{2}{3} + 4$ $=16/3-8-2/3+4$

$= \frac{14}{3} - 4$ $=14/3-4$

$= \frac{2}{3}$ $=2/3$

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2 Answers

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