Integration of y=(1+0.25x^2)^0.5

[gaywaiha]

Homework Statement

I am told to find out the length of he curve y=¼x^2

Homework Equations

The length of the curve should be

∫√(1+¼x^2)dx

The Attempt at a Solution

I substituted tanx and got ∫2/cos^3x dx

So how could I continue and work out the solution? Should I use integration by part, or I had already done wrong?

 

[SammyS]

Homework Statement

I am told to find out the length of he curve y=¼x^2

Homework Equations

The length of the curve should be

∫√(1+¼x^2)dx

The Attempt at a Solution

I substituted tanx and got ∫2/cos^3x dx

So how could I continue and work out the solution? Should I use integration by part, or I had already done wrong?

Hello gaywaiha. Welcome to PF !

Change $\displaystyle \ \frac{1}{\cos^3(x)} \$ to $\ \sec^3(x) \ .$

 

[Qwertywerty]

I substituted tanx and got ∫2/cos^3x dx

How ? Instead substitute x = 2*tan(θ) . You will however have to use by-parts later .

 

[Chestermiller]

You've already done a couple of things wrong.

1. What does the word Precalculus mean to you? I'm moving this to the Calculus and Beyond forum

2. Your substitution should be x = 2 tan θ

3. Show us your work

Chet

 

[Ray Vickson]

You've already done a couple of things wrong.

1. What does the word Precalculus mean to you? I'm moving this to the Calculus and Beyond forum

2. Your substitution should be x = 2 tan θ

3. Show us your work

Chet

As stated, the problem has an easy solution that I am willing to repeat here: the length of the curve $y = (1/4)x^2$ is infinite. No detailed calculations needed!

 

[gaywaiha]

Guys thank you very much. I am sorry that I substituted 2tanx and forgot to type the 2. But now I finally found the answer by separate 1/cos^3x into (1/cosx)(tanx)' and work it out using integration by part.

Thank you

 

[HallsofIvy]

However, you still haven't addressed Chestermiller's point- until you have put some bounds on it, the length is infinite!

 

[gaywaiha]

Sorry about that. The interval is (0≤x≤2).