How Do You Calculate the Length of a Curve Using Integrals?

[alane1994]

I had a question on a quiz that I missed... I am unsure how they got this answer. If someone could explain it would be great!

Write the integral that gives the length of the curve.

$$y=f(x)=\int_{0}^{4.5x} \sin{t} dt$$

It was multiple-choice(multiple-guess;)).

$$\text{Choice A } L=\int_{0}^{\pi} \sqrt{1+4.5(\sin(x))^2}dx$$

$$\text{Choice B } L=\int_{0}^{\pi} \sqrt{1+20.25(\sin(4.5x))^2}dx$$

$$\text{Choice C } L=\int_{0}^{\pi} \sqrt{1+20.25(\sin(x))^2}dx$$

$$\text{Choice D } L=\int_{0}^{\pi} \sqrt{1+4.5(\sin(4.5x))^2}dx$$

The correct answer is B... any way to explain in everyday people speak?

 

[Sudharaka]

I had a question on a quiz that I missed... I am unsure how they got this answer. If someone could explain it would be great!

Write the integral that gives the length of the curve.

$$y=f(x)=\int_{0}^{4.5x} \sin{t} dt$$

It was multiple-choice(multiple-guess;)).

$$\text{Choice A } L=\int_{0}^{\pi} \sqrt{1+4.5(\sin(x))^2}dx$$

$$\text{Choice B } L=\int_{0}^{\pi} \sqrt{1+20.25(\sin(4.5x))^2}dx$$

$$\text{Choice C } L=\int_{0}^{\pi} \sqrt{1+20.25(\sin(x))^2}dx$$

$$\text{Choice D } L=\int_{0}^{\pi} \sqrt{1+4.5(\sin(4.5x))^2}dx$$

The correct answer is B... any way to explain in everyday people speak?

Hi alane1994, :)

The arc length of the graph of the function \(f\) between the points \(x=a\) and \(x=b\) is given by,

\[s = \int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }\, dx\]

Refer: Arc length - Wikipedia, the free encyclopedia

So first you'll have to find \(f'(x)\). Can you give it a try? :)

Kind Regards,
Sudharaka.