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 $\int\frac{1}{\sqrt{x}}dx$?