# Thread: y = sqrt{1 + (9x/4)} dx...Arc Length

1. ## y = sqrt{1 + (9x/4)} dx...Arc Length

Find arc length given
y = x^(3/2) - 1 over [0,1].

Let INT be the integral symbol.

I was able to construct the integral but after plugging the
limits x = 0 to x = 1, I get a different answer every time.

Here is the integral:

INT sqrt{1 + (9x/4)} dx from
x = 0 to x = 1.

The correct answer is 1.43971 according to the wolfram website.

Can someone do this one for me step by step?

2. ## Re: y = sqrt{1 + (9x/4)} dx...Arc Length

well the method in general is just

$\ell = \displaystyle{\int_0^1}\sqrt{1+(f^\prime(x))^2}~dx$

here $f(x)=x^{3/2}-1$

$f^\prime(x)=\dfrac 3 2 x^{1/2}$

$(f^\prime(x))^2= \dfrac 9 4 x$

so we must find the integral

$\ell = \displaystyle{\int_0^1}\sqrt{1+\dfrac 9 4 x}~dx$

$u=\dfrac 9 4 x$

$du=\dfrac 9 4 ~dx$

$\ell = \displaystyle{\int_0^1}\sqrt{1+\dfrac 9 4 x}~dx\Rightarrow \dfrac 4 9 \displaystyle{\int_0^{9/4}}\sqrt{1+u}~du$

$\ell = \dfrac 4 9 \dfrac 2 3 (1+u)^{3/2}|^{9/4}_0=\dfrac 8 {27} \left(\left(1+\dfrac 9 4\right)^{3/2} - 1\right)\approx 1.43971$

3. ## Re: y = sqrt{1 + (9x/4)} dx...Arc Length

I let u = 1 + (9x/4). I should have let u be simply (9x/4). This made the difference.
I have trouble knowing what to substitute for u in some of these problems.

I am having fun with calculus. Today, I started learning integration by parts. It is such a cool skill to know. Expect questions in terms of integration by parts in the next couple of days. I also forgot to change the limits of integration. Good night.

4. ## Re: y = sqrt{1 + (9x/4)} dx...Arc Length

Originally Posted by nycmath
I let u = 1 + (9x/4). I should have let u be simply (9x/4). This made the difference.
I have trouble knowing what to substitute for u in some of these problems.

I am having fun with calculus. Today, I started learning integration by parts. It is such a cool skill to know. Expect questions in terms of integration by parts in the next couple of days. I also forgot to change the limits of integration. Good night.
You can use that substitution, in fact, it would be easier

5. ## Re: y = sqrt{1 + (9x/4)} dx...Arc Length

Originally Posted by Prove It
You can use that substitution, in fact, it would be easier
Can you show me by letting u be [1+(9x/4)]?
I find indefinite integrals a lot easier to play with. It is so easy to make mistakes with definite integrals.

6. ## Re: y = sqrt{1 + (9x/4)} dx...Arc Length

Originally Posted by nycmath
I let u = 1 + (9x/4). I should have let u be simply (9x/4). This made the difference.
I have trouble knowing what to substitute for u in some of these problems.

I am having fun with calculus. Today, I started learning integration by parts. It is such a cool skill to know. Expect questions in terms of integration by parts in the next couple of days. I also forgot to change the limits of integration. Good night.
that should have worked just as well. I suspect you didn't modify the integration limits properly.