# Length of Curve by Integrals

#### alane1994

##### Active member
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

##### Well-known member
MHB Math Helper
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.