How do you integrate by substitution $\int (9 - y) \sqrt{y} d y$ $int (9-y)sqrty dy$ ?

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How do you integrate by substitution $\int (9 - y) \sqrt{y} d y$ $int (9-y)sqrty dy$ ?

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Answer 1 · 2018-03-19T17:04:47

The answer is $= 6 y^{\frac{3}{2}} - \frac{2}{5} y^{\frac{5}{2}} + C$ $=6y^(3/2)-2/5y^(5/2)+C$

Explanation:

We need

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C (x \neq - 1)$ $intx^ndx=x^(n+1)/(n+1)+C(x!=-1)$

Let $u = \sqrt{y}$ $u=sqrty$ , $\Rightarrow$ $=>$ , $d u = \frac{d y}{2 \sqrt{y}}$ $du=(dy)/(2sqrty)$

Therefore,

$\int (9 - y) \sqrt{y} d y = \int (9 - u^{2}) \sqrt{y} \cdot 2 \sqrt{y} d u$ $int(9-y)sqrtydy=int(9-u^2)sqrty*2sqrtydu$

$= 2 \int (9 u^{2} - u^{4}) d u$ $=2int(9u^2-u^4)du$

$= 2 \cdot 9 \frac{u^{3}}{3} - \frac{2}{5} u^{5}$ $=2*9u^3/3-2/5u^5$

$= 6 y^{\frac{3}{2}} - \frac{2}{5} y^{\frac{5}{2}} + C$ $=6y^(3/2)-2/5y^(5/2)+C$

Answer 2 · 2018-03-19T17:05:07

$\frac{2}{5} y^{\frac{3}{2}} (15 - y) + c$ $2/5y^(3/2)(15-y)+c$

Explanation:

$I = \int (9 - y) \sqrt{y} d y$ $I=int(9-y)sqrtydy$

to do this by substitution:

$u = \sqrt{y} \Rightarrow u = y^{\frac{1}{2}}$ $u=sqrty=>u=y^(1/2$

$\Rightarrow d u = \frac{1}{2} y^{- \frac{1}{2}} d y$ $=>du=1/2y^(-1/2)dy$

$d y = 2 y^{\frac{1}{2}} d u$ $dy=2y^(1/2)du$

$I = \int (9 - u^{2}) y^{\frac{1}{2}} 2 y^{\frac{1}{2}} d u$ $I=int(9-u^2)y^(1/2)2y^(1/2)du$

$= \int (9 - u^{2}) y d u$ $=int(9-u^2)ydu$

$= 2 \int (9 - u^{2}) u^{2} d u$ $=2int(9-u^2)u^2du$

$= 2 \int (9 u^{2} - u^{4}) d u$ $=2int(9u^2-u^4)du$

$= 2 (3 u^{3} - \frac{u^{5}}{5}) + c$ $=2(3u^3-u^5/5)+c$

$= \frac{2}{5} u^{3} (15 - u^{2}) = c$ $=2/5u^3(15-u^2)=c$

$= \frac{2}{5} y^{\frac{3}{2}} (15 - y) + c$ $=2/5y^(3/2)(15-y)+c$

note for this integral it would be more efficient to multiply the brackets out and integrate directly using the power rule

ie

$\int (9 - y) y^{\frac{1}{2}} d y = \int (9 y^{\frac{1}{2}} - y^{\frac{3}{2}}) d y$ $int(9-y)y^(1/2)dy=int(9y^(1/2)-y^(3/2))dy$

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How do you integrate by substitution $\int (9 - y) \sqrt{y} d y$ $int (9-y)sqrty dy$ ?

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How do you integrate by substitution $\int (9 - y) \sqrt{y} d y$ $int (9-y)sqrty dy$ ?