Evaluate the Integral sqrt(1+7s)ds

[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 don't understand! THank you!

 

[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?

 

[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.

 

[needmathhelp!]

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

 

[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!

 

[HallsofIvy]

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

 

[DryRun]

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).

 

[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!)

 

[DryRun]

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

 

[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?