Question

Integrate (x) (1-x)^(1/2) dx

not sure where to even start....thanks

Answer 1

try u = 1-x

Answer 2

making x = 1-u

Answer 3

\[\int\limits\limits\limits_{?}^{?} x \sqrt{1-x} dx\]

Answer 4

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

Answer 5

using equation thing is not very intuitive is it? trying u = 1-x

Answer 6

well no, but it is a standard trick.

Answer 7

you have to deal with the annoying 1-x inside the radical somehow.

Answer 8

so even though the derivative of 1-x would be -1 I just keep marching?

Answer 9

yes. you have \[u=1-x\]
\[du=-dx\]
or
\[dx=-du\]
and since \[u=1-x\] we know \[x=1-u\] to get
\[\int x sqrt{1-x}dx=-\int (1-u) \sqrt{u}du = \int u^{\frac{3}{2}}du - \int u^{\frac{1}{2}}du\]

Answer 10

i assume it is ok from there. take the antiderivative using the power rule and then replace u by 1-x

Answer 11

- 2/3 (1-x)^ 3/2 + 2/5 (1-x)^5/2 +C

Answer 12

that is it!