Question

Integrate 2xSqrt{1-5x} dx

Answer 1

\[\int\limits_{}^{}2x \sqrt{1-5x} ~~dx\]

Answer 2

\[u=1-5x\]

Answer 3

i tried but i couldnt get the answer

Answer 4

heres my work

Answer 5

u^2=1-5x , x=(u^2-1) / (-5)
2udu=-5dx

\[∫2(\frac{u^2-1}{5})u(-\frac{2}{5}u)du\]
\[-\frac{2}{5}∫(u^2-1)u^2du\]
\[-\frac{2}{5}∫(u^4-u^2)du\]
\[-\frac{2}{5}[\frac{u^5}{5}-\frac{u^3}{3}]\]
\[\huge -\frac{2}{75}[3u^{5}-5u^3]\]
\\[\huge -\frac{2}{75}u^3[3u^{2}-5]\]\]
\[\huge -\frac{2}{75}(1-5x)^{3/2}[3(1-5x)-5]\]
\[\huge -\frac{2}{75}(1-5x)^{3/2}[-2-15x]\]

Answer 6

@zar

Answer 7

@Zarkon

Answer 8

@LagrangeSon678

Answer 9

?

Answer 10

can you check my working...

Answer 11

\[∫2(\frac{u^2-1}{5})u(-\frac{2}{5}u)du=-\frac{4}{25}\int\left(u^2-1\right)u^2du\]

Answer 12

ahh i found my mistake but , is it wrong to substitute like ,
e.g. u^2=1-5x

Answer 13

you can do that...but I think the substitution \(u=1-5x\) is better

Answer 14

1 more thing, is
u^2=1-5x ---------> x=(u^2-1) / (-5)

correct?

Answer 15

\[\int2x \sqrt{1-5x} dx\]\[u=1-5x\implies x={1-u\over5}\]\[du=-5dx\implies dx=-\frac15du\]\[-\frac2{25}\int(1-u)u^{1/2}du\]is probably a little more direct

Answer 16

@Calculator yes, but you forgot the - sign when you did the integration

Answer 17

but I don't see anything illegal with your rather complicated approach

Answer 18

\[∫2(\frac{u^2-1}{5})u(-\frac{2}{5}u)du\]

should be

\[∫2(\frac{u^2-1}{-5})u(-\frac{2}{5}u)du\]

Answer 19

its a typo ,my actual working is okay, but you caught my mistakes at multiplying fractons,
thanks!!!!!!!!
@Zarkon @TuringTest