Question

\[\int (x+1)\sqrt{2-x}\ dx\]

Answer 1

im planning to distribute it thru first; or ibp it

Answer 2

u=2-x and x=2-u

Answer 3

\[\int\limits_{}^{}(2-u)u^{1/2}\]

Answer 4

3-u instead of 2-u.

Answer 5

\[\implies du = -dx\]\[-\int(3-u)\sqrt{u}\;du\]

Answer 6

yes! thank you :)

Answer 7

do it by parts: u = sqrt(2-x), dv = (x+1)dx; du = -1/2sqrt(2-x), v = x^2/2 + x

Answer 8

eww, no need for parts here.

Answer 9

wow,i had to refresh just to see all this

Answer 10

you will get \[\frac{2(x-2)(x+3)\sqrt{2-x}}{5}+C\]

Answer 11

Yeah don't do it by parts; do it using polpak's approach

Answer 12

\[\int (u-3)\sqrt{u}\;du = \int u^\frac{3}{2}\;du - 3\int u^\frac{1}{2}du\]

Answer 13

I got my stats book yesterday!

Answer 14

Oh! But I have to leave now.

Answer 15

\begin{array}
&&&(2-x)^{1/2}\\
+&(x+1)&\frac{2}{3}(2-x)^{3/2}\\
-&1&\frac{2}{3}\frac{5}{3}(2-x)^{5/3}\\
+&0&doesnt\ matter\ here
\end{array}

by parts is fine, a little messy, but somethig for the teacher to wonder about lol

and stats! yay, i like stats. cant do stats, but i like stats

Answer 16

.... and thats spose to be 3/5 in front there

Answer 17

\begin{array}
&&&(2-x)^{1/2}\\
+&(x+1)&\frac{2}{3}(2-x)^{3/2}\\
-&1&\frac{2}{3}\frac{2}{5}(2-x)^{5/2}\\
+&0&doesnt\ matter\ here
\end{array}

\[(x+1)\frac{2}{3}(2-x)^{3/2}-\frac{2}{3}\frac{2}{5}(2-x)^{5/2}\]

\[\frac{2}{3}[(x+1)(2-x)^{3/2}-\frac{2}{5}(2-x)^{5/2}]+C\]

shed prolly kill me

Answer 18

\[\frac{2}{3}[(x+1)(2-x)^{3/2}-\frac{2}{5}(2-x)^{5/2}]+C\]
\[\frac{2(2-x)^{3/2}}{3}[(x+1)-\frac{2}{5}(2-x)]+C\]
\[\frac{2(2-x)^{3/2}}{3}[\frac{5x+5-4+2x}{5}]+C\]
\[\frac{2(2-x)^{3/2}}{3}[\frac{7x+1}{5}]+C\]
\[\frac{2(7x+1)(2-x)^{3/2}}{15}+C\]

the wolf gives me:
\[\frac{-2 (2 - x)^{3/2} (3 + x)}{5}\]
i wonder where i went astray

Answer 19

\[\int (x+1)\sqrt{2-x}\ dx\]
\[\int x(2-x)^{1/2}\ dx+\int(2-x)^{1/2}\ dx\]
\[\frac{2}{3}x(2-x)^{3/2}-\frac{2}{3}\int(2-x)^{3/2}\ dx+\frac{2}{3}(2-x)^{3/2}\]
\[\frac{2}{3}x(2-x)^{3/2}-\frac{2}{3}\frac{2}{5}(2-x)^{5/2}+\frac{2}{3}(2-x)^{3/2}\]

but thats what i ended up with before ..... now i gotta wonder if a remember IBPs

Answer 20

\[\int u\ dv=uv-\int v\ du\]

Answer 21

u = x ; du = dx

\[dv = (2-x)^{1/2} dx;\ v = (2/3)(2-x)^{3/2}\]

Answer 22

Good luck with IBP; the integral expression will just become more complex :-(

Answer 23

Don't mind me - just bookmarking this page
:)

Answer 24

I determined where my fauxpaux was; i forgot to pull out a -1 when i integrated my "v" choice.

\begin{array}l
&&\underline{\int v}\\
&\underline{du}&(2-x)^{1/2}\\
+&(x+1)&-\frac{2}{3}(2-x)^{3/2}\\
-&1&\frac{2}{3}\frac{2}{5}(2-x)^{5/2}\\
+&0&doesnt\ matter\ here
\end{array}

\[(x+1)-\frac{2}{3}(2-x)^{3/2}-\frac{2}{3}\frac{2}{5}(2-x)^{5/3}\]
\[-\frac{2}{3}(2-x)^{3/2}\left((x+1)+\frac{2}{5}(2-x)\right)\]
\[-\frac{2}{3}(2-x)^{3/2}\left(\frac{5(x+1)+2(2-x)}{5}\right)\]\[-\frac{2}{3}(2-x)^{3/2}\left(\frac{5x+5+4-2x}{5}\right)\]
\[-\frac{2}{3}(2-x)^{3/2}\left(\frac{3x+9}{5}\right)\]
\[-\frac{2}{\cancel{3}}(2-x)^{3/2}\frac{\cancel{3}(x+3)}{5}\]
\[\frac{-2}{5}\ (2-x)^{3/2}\ (x+3)\]

Answer 25

now it matches the wolfs

Answer 26

that was a lot of work!

Answer 27

much easier to shift a little yes?