Question

Evaluate the integral? : `int_0^2 xsqrt(2x-x^2) dx`

Answer 1

`int_0^2 \ xsqrt(2x-x^2) \ dx = pi/2`

Explanation:

Consider the indefinite integral:

`I = int \ xsqrt(2x-x^2) \ dx`

Which we can write as:

`I = int \ (x-1)sqrt(2x-x^2) + sqrt(2x-x^2) \ dx`
`\ \ = int \ (x-1)sqrt(2x-x^2) \ dx + int \ sqrt(2x-x^2) \ dx`

For the first integral, we have:

`I_1 = int \ (x-1)sqrt(2x-x^2) \ dx`

we can use a substitution, Let:

`u = 2x-x^2 => (du)/dx = 2-2x = -2(x-1)`

And if we perform this substitution then we get:

`I_1 = int \ (-1/2) sqrt(u) \ du`
`\ \ \ = -1/2 u^(3/2)/(3/2)`
`\ \ \ = -1/3 u^(3/2)`

And restoring the substitution we get:

`I_1 = -1/3 (2x-x^2)^(3/2)`

Next we consider the second integral,

`I_2 = int \ sqrt(2x-x^2) \ dx`
`\ \ \ = int \ sqrt(1-(x-1)^2) \ dx`

And here we can perform a substitution; Let

`sin theta=x-1 => (d theta)/(dx)cos theta=1`

And if we perform this substitution then we get:

`I_2 = int \ sqrt(1-sin^2 theta) \ (cos theta) \ d theta`
`\ \ \ = int \ cos^2 theta \ d theta`
`\ \ \ = int \ (cos(2theta)+1)/2 \ d theta`
`\ \ \ = (sin2theta + theta)/2`

And restoring the substitution we find that `theta = arcsin(x-1)` and `sin(2theta)=2(x-1)sqrt(1-(x-1)^2)` and so:

`I_2 = (arcsin(x-1) + (x-1)sqrt((x-1)-(x-1)^2))/2`
`\ \ \ = (arcsin(x-1) + (x-1)sqrt(2x-x^2))/2`

Combining our two results we then gave:

`I = (arcsin(x-1) + (x-1)sqrt(2x-x^2))/2 -1/3 (2x-x^2)^(3/2)`

Given this result we then have:

`int_0^2 \ xsqrt(2x-x^2) \ dx = [(arcsin(x-1) + (x-1)sqrt(2x-x^2))/2 -1/3 (2x-x^2)^(3/2)]_0^2`

`" " = 1/2((arcsin1 - arcsin(-1))`

`" " = 1/2(pi/2-(-pi/2))`

`" " = pi/2`

Answer 2

`int_0^2 xsqrt(2x-x^2)*dx=pi/2`

Explanation:

`int_0^2 xsqrt(2x-x^2)*dx`

=`int_0^2 xsqrt(1^2-(x-1)^2)*dx`

After using `x-1=sinu`, `x=1+sinu` and `dx=cosu*du` transforms, this integral became

`int_(-pi/2)^(pi/2) (1+sinu)*cosu*cosu*du`

=`int_(-pi/2)^(pi/2) (1+sinu)*(cosu)^2*du`

=`1/2int_(-pi/2)^(pi/2) (1+sinu)*(1+cos2u)*du`

=`1/2int_(-pi/2)^(pi/2) (1+cos2u+sinu+cos2u*sinu)*du`

=`1/2int_(-pi/2)^(pi/2) [1+cos2u+sinu+1/2*(sin3u-sinu)]*du`

=`1/4int_(-pi/2)^(pi/2) (2+2cos2u+sin3u+sinu)*du`

=`1/4*[2u+sin2u-1/3*cos3u-cosu]_(-pi/2)^(pi/2)`

=`1/4*2pi`

=`pi/2`