# Find the following definite integral by using intergration by parts. `int_(-4)^0 x sqrt (1-2x) dx`

lemjay | Certified Educator

`int_(-4)^0 x sqrt(1-2x)dx`

Using integration by parts, apply the formula `int udv = uv - int vdu` .

So let,

`u = x`              and            `dv= sqrt(1-2x) dx`

`du=dx`                              `v= int sqrt(1-2x)dx = int (1-2x)^(1/2) dx`

To determine v, use substitution method. Let,

`y=1-2x`         `dy=-2dx`          `-(dy)/2 = dx`

So we have,

`v= int y^1/2 (-(dy)/2) = -1/2 int y^(1/2) =- y^(3/2)/3`

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

Then, substitute u, v, du and dv to the formula of integration by parts.

`int xsqrt(1-2x)dx = -(x(1-2x)^(3/2))/3 - int -(1-2x)^(3/2)/3 dx`

`=-(x(1-2x)^(3/2))/3 + 1/3int(1-2x)^(3/2)dx`

To evaluate the integral part, use substitution method. Let,

z=1-2x           dz = -2dx          -(dz)/2 = dx

`int(1-2x)^(3/2)dx = -1/2 int z^(3/2)dz = -z^(5/2)/5 = -(1-2x)^(5/2)/5`

We then have,

`=-(x(1-2x)^(3/2))/3 +1/3*(-(1-2x)^(5/2)/5) = -(x(1-2x)^(3/2))/3-(1-2x)^(5/2)/15`

Next, evaluate the limits of the integral.

`int_(-4)^0 x sqrt(1-2x)dx = -(x(1-2x)^(3/2))/3 - (1-2x)^(5/2)/15 `   `|_(-4)^0`

`= [0-1/15] - [(4*9^(3/2))/3 - 9^(5/2)/15]=-1/15-[(4*3^3)/3 - 3^5/15]=-1/15-99/5`

`= -298/15=-19 13/15`

Hence,  `int_(-4)^0 xsqrt(1-2x)dx = -19 13/15` .