Arclength &

Approximating Integrals

Example: Find the length of the curve from to .

Solution: We plot the graph for convenience.

Now consider a small piece of arc of length

Now

𝑑𝑠=√𝑑𝑥2+𝑑𝑦 2

¿√𝑑𝑥2+(𝑑𝑦𝑑𝑥 )2

𝑑𝑥2

¿√1+(𝑑𝑦𝑑𝑥 )2

𝑑𝑥

We obtain the formula:

𝑠=∫𝑎

𝑏

𝑑𝑠¿∫−1

1

√1+( 𝑑𝑦𝑑𝑥 )2

𝑑𝑥

𝑠=∫−1

1

√1+(𝑑𝑦𝑑𝑥 )2

𝑑𝑥 𝑦=√1−𝑥2𝑑𝑦𝑑𝑥

=−𝑥

√1−𝑥2¿∫−1

1

√1+ 𝑥2

1−𝑥2𝑑𝑥

¿∫−1

11

√1−𝑥2𝑑𝑥

¿ sin− 1𝑥 ]−11

¿ 𝜋2−(− 𝜋2 )

¿𝜋∎

Example 2: Determine the arclength of from to

Solution: 𝑠=∫𝑎

𝑏

√1+( 𝑑𝑦𝑑𝑥 )2

𝑑𝑥

¿∫2

4

√1+(𝑥− 14 𝑥 )

2

𝑑𝑥

¿∫2

4

√1+𝑥2− 12 + 116 𝑥2

¿∫2

414 𝑥

+𝑥

Simplifies to…

¿6+14ln 2 ∎

Unlike the last two examples, most arclength problems cannot be solved exactly because the integrals are too difficult.

Example Find the arclength of from to .

Solution We set up the arclength integral.

Arclength=∫−2

2

√1+(𝑑𝑦𝑑𝑥 )2

𝑑𝑥

¿∫− 2

2

√1+ (3−3𝑥2 )2𝑑𝑥

¿∫− 2

2

√9𝑥4−18𝑥2+10𝑑𝑥

This integral cannot be evaluated for an exact value, even by a computer program.

We need to use an integral approximation technique

New problem Approximate

Solution We draw a picture. Now we break the interval into, say, pieces.

Now we draw rectangles on each of the pieces.

Notice, we consistently draw the rectangles so that the top-left corner lies on the graph

Now we add up the total area of the rectangles (Mathematica)

We find that the integral’s value is

The correct value can be found with Mathematica’s Nintegrate function:

The correct value was .∫−2

2

√9 𝑥4−18 𝑥2+10𝑑𝑥 ≈13.0371∎

Facts about Numerical Integration

• The previous example used left-endpoint numerical integration

• We can also use right endpoint and midpoint numerical integration, though the accuracy is similar

• Better, more advanced techniques exist, such as Simpson’s Rule

• Accuracy of left-endpoint numerical integration is proportional to

• Accuracy of Simpson’s Rule is proportional to • Here is the number of subdivisions used. Thus, the more

subdivisions, the more accurate