# Find y in terms of x

• Sep 13th 2009, 01:52 AM
Oasis1993
Find y in terms of x
Hello,

Ive been having trouble solving this problem.
I could really use help :)

In this case find y in terms of x

dy/dx= (√x+5)^2

Rob
• Sep 13th 2009, 02:54 AM
pickslides
$\frac{dy}{dx}= (\sqrt{x}+5)^2$

$y= \int(\sqrt{x}+5)^2~dx$

I suggest expanding the brackets and integrating each term individually.

$y= \int x+10\sqrt{x}+25~dx$

Can you take it from here?
• Sep 13th 2009, 03:39 AM
Oasis1993
Quote:

Originally Posted by pickslides
$\frac{dy}{dx}= (\sqrt{x}+5)^2$

$y= \int(\sqrt{x}+5)^2~dx$

I suggest expanding the brackets and integrating each term individually.

$y= \int x+10\sqrt{x}+25~dx$

Can you take it from here?

Dear Pickslides thans for the reply but can you show me how its solved?

Ive got the answers but dont know how to get that answer.

Thanks
• Sep 13th 2009, 03:50 AM
pickslides
$y= \int x+10\sqrt{x}+25~dx$

$y= \int x~dx+\int 10\sqrt{x}~dx+\int 25~dx$

Now use the following rule for each part.

$\int x^n~dx = \frac{x^{n+1}}{n+1}+c$

for each part.

Spoiler:
$y= \frac{x^2}{2}+\frac{20x^{\frac{3}{2}}}{3}+25x+c$
• Sep 13th 2009, 05:32 AM
mr fantastic
Quote:

Originally Posted by Oasis1993
Dear Pickslides thans for the reply but can you show me how its solved?

Ive got the answers but dont know how to get that answer.

Thanks

What more is needed? You're expected to realise that $\sqrt{x} = x^{1/2}$ and then use the usual rule for integrating $x^n$.