Evaluate the Integral sqrt(1+7s)ds

1. May 12, 2012

needmathhelp!

1.The question is Evaluate the Integral $\sqrt{1+7s}$ds

2. Ok, so i've tried this problem but I'm not coming up with the correct solution. Could you help me.

3. This is how i'm handling the problem.

Evaluate: (1+7s) ^1/2
= $\frac{2(1+7s)^(3/2)}{3}$ +C

The answer is supposed to be 2/21 (1+7s) ^(3/2) + C Could someone please "spell out" how to get this. I dont understand! THank you!

2. May 12, 2012

cjc0117

take the derivative of $\frac{2}{3}(1+7s)^{3/2}$ and then you'll probably see what's going wrong. You're forgetting the chain rule.

EDIT: let $f(s)=s^{3/2}$ and $g(s)=1+7s$. Then $\frac{2}{3}(1+7s)^{3/2}=\frac{2}{3}f(g(s))$ and by the chain rule $\frac{d}{ds}(\frac{2}{3}(1+7s)^{3/2})=\frac{2}{3}f'(g(s))*g'(s)$. Can you tell me what f'(g(s)) and g'(s) are?

Last edited: May 12, 2012
3. May 12, 2012

Infinitum

What you did is apply,

$$\int \sqrt{x} dx = \frac{2x^{\frac{3}{2}}}{3} + C$$

What happens when your given integral is

$\int \sqrt{ax+b}\ dx$

Hint: Use substitution.

4. May 12, 2012

needmathhelp!

AHHHH!! I got it!! THank you guys so much!!! I truly appreciate it!

5. May 13, 2012

needmathhelp!

What happens when you are taking the integral of something like this.....

∫$\frac{6r^2}{\sqrt{6-r^3}}$

I tried:
u = 6-r^3
du = 3r^2

but i cannot get the correct answer. Help would be great!

6. May 13, 2012

HallsofIvy

Staff Emeritus
With $u= 6- r^3$, $du= -3r^2 dr$. You seem to have left out the negative sign.

7. May 13, 2012

sharks

Your integral in post #5 is in this general form:
$$\int \frac{f'(x)}{\sqrt{f(x)}}\,dx=2\sqrt{f(x)}+C$$
In that case, you just rearrange the real constant coefficient of the numerator to match that of f'(x).

Last edited: May 13, 2012
8. May 13, 2012

needmathhelp!

Great.. So if I use that equation then I am getting

-2 (6-r^3)^(1/2) + C

But the book says it should be -4(6-r^3)^(1/2) +C
What am i missing?
(thankyou so much for the help!!)

9. May 13, 2012

sharks

You wrote your integral wrong. It should be:
$$\int \frac{6r^2}{\sqrt{6-r^3}}\,dr$$

10. May 13, 2012

cjc0117

You should write out the steps that you've tried.

You let $u=6-r^{3}$ and $du=-3r^{2}dr$.

Then $3r^2dr=-du$

However, notice that the numerator is actually $6r^{2}$, not $3r^{2}$

Also, can you tell me what is $\int\frac{1}{\sqrt{x}}dx$?

Last edited: May 13, 2012