Still Confused?

Try reviewing these fundamentals first

- Home
- Integral Calculus
- Integration Applications

Still Confused?

Try reviewing these fundamentals first

Nope, got it.

That's the last lesson

Start now and get better math marks!

Get Started NowStart now and get better math marks!

Get Started NowStart now and get better math marks!

Get Started NowStart now and get better math marks!

Get Started Now- Intro Lesson11:25
- Lesson: 112:57
- Lesson: 25:12
- Lesson: 39:59

Instead of finding the area under the curve, we are going to be the length of the curve between a and b. Now we know we can find the length of a line using the distance formula, but what about the length of the curve? In the intro video, we will learn that we can find the length of the curve by manipulating the distance formula into an integral. After, we will be applying the formula that we've found by finding the arc length of functions in terms of x. We will then look at some advanced questions where we will find the arc length of functions in terms of y, as well as finding the arc length function with an initial point.

The arc length of a curve from a to b:

$Arc\; Length = \int_{a}^{b} \sqrt{1+[f'(x)]^2}dx$

Arc length as a function of a variable, x:

$s(x)=\int_{intial}^{x} \sqrt{1+[f'(u)]^2}du$

Arc length as a function of a variable, x:

- Introduction
- 1.Find the length of the arc of $y^2=(1+x)^3$ from (-1,0) to (3,8)
- 2.Find the length of the arc of $x=\ln(sin y)$ on $\frac{\pi}{4}\leq y \leq \frac{\pi}{2}$
- 3.Find the arc length function for the curve $y=\frac{1}{x}(\frac{4}{3}x^4+\frac{1}{16})$ with initial point at $x=\frac{1}{2}$

We have over 170 practice questions in Integral Calculus for you to master.

Get Started Now