# Comparing dy/dx with another dy/dx to find Y

• Aug 21st 2010, 03:58 AM
Punch
Comparing dy/dx with another dy/dx to find Y
Given that $y=(3x+2)\sqrt{x+2}$ and $\frac{dy}{dx}=\frac{3x+8}{2\sqrt{x+2}}$.

Integrate $\int^7_{2}\frac{x}{2\sqrt{x+2}}$ with respect to x
• Aug 21st 2010, 04:07 AM
HallsofIvy
$\frac{x}{2\sqrt{x+2}}= \frac{1}{3}\frac{3x+ 8}{2\sqrt{x+ 2}}- \frac{4}{3\sqrt{x+2}}$
You are "given" how to integrate the first and the second can be integrated by letting u= x+ 2.
• Aug 21st 2010, 08:52 AM
Jester
Quote:

Originally Posted by Punch
Given that $y=(3x+2)\sqrt{x+2}$ and $\frac{dy}{dx}=\frac{3x+8}{2\sqrt{x+2}}$.

Integrate $\int^7_{2}\frac{x}{2\sqrt{x+2}}$ with respect to x

Something's not right here. It's either your $y$ or your $\frac{dy}{dx}$ b/c

$\dfrac{d}{dx} (3x+2)\sqrt{x+2} = \dfrac{9x+14}{2\sqrt{x+2}}$.
• Aug 28th 2010, 08:16 AM
Punch
$dy/dx=\frac{(3x+2)}{2\sqrt{x+2}}+3\sqrt{x+2}$
$=\frac{9x+14}{2\sqrt{x+2}}$

you are absolutely right but i still dont get how i can compare $\frac{9x+14}{2\sqrt{x+2}}$ with $\frac{x}{2\sqrt{x+2}}$
• Aug 28th 2010, 10:35 AM
Jester
First off, as HallsofIvy point out, the substitution u = x+2 or x = u - 2 will give you two integrals that you can easily integrate (and the way I personally would have done this problem). In what follows, the constants of integration will be supressed. First, using the fact that

$\displaystyle \int \dfrac{9x+14}{2\sqrt{x+2}}dx = (3x+2)\sqrt{x+2}$

then

$\displaystyle 9 \int \dfrac{x}{\sqrt{x+2}}dx + 14 \int \dfrac{1}{\sqrt{x+2}}dx = (6x+4)\sqrt{x+2},\;\;(1)$

noting that what you want is the first integral. Also $9x+14 = 9(x+2) - 4$ so your integral can be written as

$\displaystyle \int \dfrac{9(x+2)-4}{\sqrt{x+2}}dx = 9 \int \sqrt{x+2}\,dx - 4\int \dfrac{1}{\sqrt{x+2}}dx$

so

$\displaystyle 9 \int \sqrt{x+2}\,dx - 4\int \dfrac{1}{\sqrt{x+2}}dx = (6x+4)\sqrt{x+2},\;\;\;(2)$

Now use integration by parts $\int u dv = uv - \int v du$ on the first integral in (2)

$u = \sqrt{x+2}, \;\; du = \dfrac{dx}{2\sqrt{x+2}}$

$dv = dx,\;\;\; v = x$

gives

$\displaystyle 9 x \sqrt{x+2} - \dfrac{9}{2}\int \dfrac{x}{\sqrt{x+2}}dx - 4\int \dfrac{1}{\sqrt{x+2}}dx = (6x+4)\sqrt{x+2}$

or

$\displaystyle - \dfrac{9}{2}\int \dfrac{x}{\sqrt{x+2}}dx - 4\int \dfrac{1}{\sqrt{x+2}}dx = (-3x+4)\sqrt{x+2}$

or

$\displaystyle 9\int \dfrac{x}{\sqrt{x+2}}dx + 8\int \dfrac{1}{\sqrt{x+2}}dx = (6x-8)\sqrt{x+2}.\;\;(3)$

Now multiply $7 \times (3) - 4 \times (1)$ gives

$\displaystyle 27 \int \dfrac{x}{\sqrt{x+2}}dx = 18(x-4)\sqrt{x+2}$

from which it follows that

$\displaystyle \int \dfrac{x}{\sqrt{x+2}}dx = \frac{2}{3} (x-4)\sqrt{x+2}$.

Kinda long and again, not the way I would have done this problem.