# Integration Question

• May 21st 2008, 06:21 AM
looi76
Integration Question
Question
A curve is such that $\displaystyle \frac{dy}{dx} = \frac{4}{\sqrt(6 - 2x)}$, and $\displaystyle P(1,8)$ is a point on the curve. Find the equation of the curve.

Attempt:

$\displaystyle \frac{dy}{dx} = \frac{4}{\sqrt(6 - 2x)}$

$\displaystyle \int 4(6-2x)^{-\frac{1}{2}}dx$

I don't know how to integrate it.. need help!
• May 21st 2008, 06:29 AM
wingless
Substitute $\displaystyle u=6-2x$, for example.
• May 21st 2008, 06:29 AM
colby2152
Quote:

Originally Posted by looi76
Question
A curve is such that $\displaystyle \frac{dy}{dx} = \frac{4}{\sqrt(6 - 2x)}$, and $\displaystyle P(1,8)$ is a point on the curve. Find the equation of the curve.

Attempt:

$\displaystyle \frac{dy}{dx} = \frac{4}{\sqrt(6 - 2x)}$

$\displaystyle \int 4(6-2x)^{-\frac{1}{2}}dx$

I don't know how to integrate it.. need help!

Let $\displaystyle u = 6 - 2x$, $\displaystyle du = -2 dx$, $\displaystyle dx = -0.5 du$

$\displaystyle \int 4(6-2x)^{-\frac{1}{2}}dx$

$\displaystyle \int -2(u)^{-\frac{1}{2}}du$

$\displaystyle -4(u)^{\frac{1}{2}}$

$\displaystyle y = -4(6 - 2x)^{\frac{1}{2}} + C$

Solve for C by knowing that $\displaystyle y(1)=8$

The important part is that you knew what to do!