Find-X-if-0-e-x-0-and-xr-r-e-e-n-or-n-e-0-and-x-y-y-x-1-

Question Number 4807 by Dnilka228 last updated on 14/Mar/16

F i n d X i f \dots

\underset{0}{\int^{e}} x \neq 0 a n d \frac{x r}{r^{e}} = \sqrt{e + n}

o r

\sum_{n!} e = 0 a n d \frac{℧ x \neq y}{℧ y \neq x} = - 1

Commented by Dnilka228 last updated on 14/Mar/16

I^{'} m know, im crazy)) 0)

Commented by prakash jain last updated on 15/Mar/16

What is question ?

Answered by FilupSmith last updated on 16/Mar/16

\frac{x r}{r^{e}} = \sqrt{e + n}

{x r}^{1 - e} = {(e + n)}^{\frac{1}{2}}

x = r^{e - 1} {(e + n)}^{\frac{1}{2}}

l e t A = \int_{0}^{e} x d r (1) d r o r d n i s u n s p e c i f i e d

∴ A = \int_{0}^{e} r^{e - 1} {(e + n)}^{1 / 2} d r

= {(e + n)}^{\frac{1}{2}} \int_{0}^{e} r^{1 - e} d r

= {(e + n)}^{\frac{1}{2}} \frac{1}{2 - e} {[r^{2 - e}]}_{0}^{e}

A = {(e + n)}^{\frac{1}{2}} \frac{1}{2 - e} e^{2 - e}

A = \frac{e^{2} \sqrt{e + n}}{(2 - e) e^{e}}

l e t A = \int_{0}^{e} x d n (2)

∴ A = \int_{0}^{e} r^{e - 1} {(e + n)}^{1 / 2} d n

A = r^{e - 1} \int_{0}^{e} \sqrt{e + n} d n

l e t u = e + n

d u = 1 d n

∴ A = r^{e - 1} \int_{0}^{e} \sqrt{u} d u

∴ A = r^{e - 1} {[\frac{2}{3} u^{\frac{3}{2}}]}_{0}^{e}

A = r^{e - 1} {[\frac{2}{3} {(e + n)}^{\frac{3}{2}}]}_{0}^{e}

A = r^{e - 1} \frac{2}{3} [{(2 e)}^{\frac{3}{2}} - {(e)}^{\frac{3}{2}}]

A = r^{e - 1} \frac{2}{3} (2^{\frac{3}{2}} e^{\frac{3}{2}} - e^{\frac{3}{2}})

A = \frac{2}{3} r^{e - 1} \sqrt{e^{3}} (\sqrt{8} - 1)