Question

Integrating ∫√(4x - x^2) dx.

Could someone please tell me how to start this one?

Answer 1

complete the square is my first guess

Answer 2

then trig sub

Answer 3

follow @TuringTest you will get it

Answer 4

Trig sub is fun, we'll help you! Just give it your best try. @s3a

Answer 5

well;\[\int\limits_{}^{}\sqrt{x}= \frac{ 2x ^{3/2} }{ 3 }\]
so:\[\int\limits_{}^{}\sqrt{4x-x^2}=\frac{ (8x-2x^2)^{3/2} }{ 3 }\]

and remember:
\[x ^{3/2}=\sqrt{x^3}\]
so:\[(8x-2x^2)^{3/2}=\sqrt{-8 x^6+96 x^5-384 x^4+512 x^3}\]

going back to the original integral:

\[\int\limits_{}^{}\sqrt{4x-x^2}=\frac{ \sqrt{-8 x^6+96 x^5-384 x^4+512 x^3} }{ 3 }\]

Answer 6

i like using formulas for integration i hate u substitution and trig sub

Answer 7

Seems to me that trig subst. is the way to go ... and that after completing the square.

Let's focus on the quantity under the square root operator:

4x - x^2 = - (x^2 - 4x)

Completing the square: - (x^2 - 4x + 4 - 4)

Rewriting x^2 - 4x + 4, we get - ([x-2)^2 - 4), or 4 - (x-2)^2, or 2^2 - (x-2)^2.

Then the integral becomes Int Sqrt( 2^2 - (x_2)^2 ) dx.

Hint: if you have the situation Sqrt( a^2 - x^2), what is an appropriate substitution for x?

Answer 8

Actually, thank you everyone, but TuringTest's advice helped me get on the path I wanted. My problem now is that I am making a mistake somewhere in that path.

Here is my work:
http://www.tiikoni.com/tis/view/?id=ac56dfe

Could someone please tell me where I went wrong?

Answer 9

you should get it to be \(\sqrt{1+....}\) under the radical before substitution, so factor out that 4

Answer 10

Thanks, TuringTest (and everyone else). :)

Here's my correct work:
http://www.tiikoni.com/tis/view/?id=090cade