Calculus : U-Substitution

**Afterme** · December 9th 2008, 05:35 PM

The problem is x(16-3x)^1/2 dx

u = 16 - 3x
du = -3 dx

My problem is how do I get rid of the X in the original problem? I can divide -1/3 to get rid of the -3 and that'll take care of the dx but as for the X I have no idea how to go about it. Help =)

**skeeter** · December 9th 2008, 05:42 PM

Originally Posted by Afterme

The problem is x(16-3x)^1/2 dx

u = 16 - 3x
du = -3 dx

My problem is how do I get rid of the X in the original problem? I can divide -1/3 to get rid of the -3 and that'll take care of the dx but as for the X I have no idea how to go about it. Help =)

$u = 16-3x$

$3x = 16-u$

$x = \frac{1}{3}(16-u)$

substitute ...

$\int \frac{1}{3}(16-u) \cdot u^{\frac{1}{2}} \cdot \frac{du}{-3}$

$-\frac{1}{9} \int 16u^{\frac{1}{2}} - u^{\frac{3}{2}} \, du$

**Afterme** · December 9th 2008, 06:06 PM

Originally Posted by skeeter

$u = 16-3x$

$3x = 16-u$

$x = \frac{1}{3}(16-u)$

substitute ...

$\int \frac{1}{3}(16-u) \cdot u^{\frac{1}{2}} \cdot \frac{du}{-3}$

$-\frac{1}{9} \int 16u^{\frac{1}{2}} - u^{\frac{3}{2}} \, du$

Thanks alot =)

**skeeter** · December 9th 2008, 06:10 PM

distributive property of multiplication ...

$(16 - u)u^{\frac{1}{2}} = 16 \cdot u^{\frac{1}{2}} - u \cdot u^{\frac{1}{2}} = 16u^{\frac{1}{2}} - u^{\frac{3}{2}}$

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