Class notes
AP Calculus BC Unit 6 Applications of Differentiation
- Course
- Institution
This document goes over area between curves as well as volumes of curves rotated various ways, covering both disk and washer methods.
[Show more]
Seller
Follow
Content preview
Unit 6 | Applications of IntegrationSection 6.1 | Areas Between Curves
The area Aof the region bounded by the curves y = f(x), y = g(x), and the lines x = a, x = b, where f and gare
continuous and f(x) ⩾ g(x)for all xin [a, b], is
b
A = ∫ [f(x) − g(x)] dx
a
Example 1
Find the area of the region bounded above by y = ex , bounded below by y = x, and bounded on the sides by x = 0and
x = 1
Solution The region is shown below. The upper boundary curve is y = ex and the lower boundary curve is y = x. So we
use the area formula with f(x) = e , g(x) = x, a = 0, and b = 1:
x
1 1
1
A = ∫ (ex − x) dx = ex − x2 ]
2
0 0
1
= e − − 1 = e − 1.5
2
https://www.desmos.com/calculator/bveib1tzkf?embed
Example 2
Find the area of the region enclosed by the parabolas y = x2 and y = 2x − x2 .
Solution We first find the points of intersection of the parabolas by solving their equations simultaneously. This gives x2 =
2 2
2x − x or 2x − 2x = 0. Thus 2x(x − 1) = 0, so x = 0or 1. The points of intersection are (0, 0)and (1, 1).
= 2x − x2 and yB = x2 . The area of a typical
We see from the figure below that the top and bottom boundaries are yT
2 2
rectangle is (yT − yB )Δx = (2x − x − x )Δxand the region lies between x = 0and x = 1. So the total area is
1 1
Unit 6 | Applications of Integration 1
, 1 1
A=∫ (2x − 2x2 ) dx = 2 ∫ (x − x2 ) dx
0 0
1
x2 x3 1 1 1
= 2[ − ] = 2( − ) =
2 3 0 2 3 3
https://www.desmos.com/calculator/zitmmdtvhe?embed
Example 3
Find the approximate area of the region bounded by the curves y = x
x2 +1
and
y = x4 − x.
Solution If we were to try to find the exact intersection points, we would have to solve the equation
x
= x4 − x
x2 + 1
This looks like a very difficult equation to solve exactly (in fact it’s impossible), so instead we use a graphing device to draw
the graphs of the two curves below. One intersection point is the origin. We zoom in toward the other point of intersection
and find that x ≈ 1.18. So an approximation to the area between the curve is
1.18
[
x
A≈∫ − (x4 − x)] dx
x2 +1
0
To integrate the first term we use substitution u = x2 + 1. Then du = 2x dx, and when x = 1.18, we have u ≈ 2.39;
when x = 0, u = 1. So
1 2.39 du 1.18
A≈ ∫ −∫ (x4 − x) dx
2 1
u 0
1.18
x5 x2
− ] −[
2.39
= u ]1
5 2 0
(1.18)5 (1.18)2
= 2.39 − 1 − +
5 2
≈ 0.785
https://www.desmos.com/calculator/iz1ayfe2j1?embed
Example 4
Unit 6 | Applications of Integration 2
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller clegge4656. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $2.99. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
79650 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now