Math 1A: introduction to functions and calculus Oliver Knill, 2011Lecture 21: Area computationIf f(x) ≥ 0, t henRbaf(x) dx is the area under the graph of f(x) and above theinterval [a, b] on the x axes.As you have seen in a homework, any function can be written as f(x) = f+(x) − f−(x), wheref+(x) ≥ 0 and f−(x) ≥ 0. This means that we can write any integralRbaf(x) dx as the differenceof the area above the graph minus the area below the graph.Zbaf(x) dx =Zbaf+(x) − f−(x) dx .Here is the most common situation:If a region is enclosed by two g r aphs f ≤ g and x is also enclosed between a and bthen its area isRbag(x) − f(x) dx.1 Find the area of the region enclosed by the x-axes, the y-axes and the graph of the cosfunction. Solution:Rπ/20cos(x) dx = 1.2 Find the area of the region enclosed by the gr aphs f(x) = x2and f(x) = x4.3 Find the area of the region enclosed by the gr aphs f(x) = 1 − x2and g(x) = x4.124 Find the area of the region enclosed by a half circle of radius 1. Solution: The half circleis the graph of the function f(x) =√1 − x2. The area under the graph isZ1−1√1 − x2dx .Finding the anti-derivative is not so easy. We will find techniques to do so, for now wepop it together: we know that arcsin(x) has the derivative 1/√1 − x2and x√1 − x2hasthe derivative√1 − x2− x2/√1 − x2. The sum of these two functions has the derivative√1 − x2− (1 − x2)/√1 − x2= 2√1 − x2. We find the anti derivative to be (x√1 − x2+arcsin(x))/2. The area is thereforex√1 − x2+ arcsin(x)2|1−1=π2.5 Find the area of the region between the graphs of f (x) = 1−|x|1/4and g(x) = −1+|x|1/4.36 Find the area under the curve of f (x) = 1/x2between −6 and 6. Solution.R6−6x−2dx =−x−1|6−6= −1/6 − 1/6 = −1/3. There is something fishy with this computation becausef(x) is nonnegative so that the area should be positive. But we obta ined a negative answer.What is going on?7 Find the area between the curves x = 0 and x = 2 + sin(y), y = 2π and y = 0. Solution:We turn the picture by 90 degrees so t hat we compute the area under the curve y = 0, y =2 + sin(x) and x = 2π and x = 0.48 The grass problem. Find the area between the curves |x|1/3and |x|1/2. Solution. Thisexample illustrates how important it is to have a picture. This is good advise for any”word problem” in mathematics.Use a picture of the situation while doing the computation.Homework1 Find the area of the region enclosed by the graphs f(x) = x3and g(x) =p|x|.2 Find the area of the region enclosed by the four lines y = x, y = 3 − x, x = 1.3 Find the area of the region enclosed by the curves y = 4π, y = 2π, x = −3 + sin(3y), x =2 + sin(2y).4 Write down an integral which gives the area of the elliptical region 4x4+ y2≤ 1. Evaluatethe integral numerically using Wolfram alpha, Mathematica or any other software.5 The graphs sin(x) and cos(x) − 1 intersect at x = 0, 2π and a point between. They definea humming bird region,consisting of a larger region and a tail region. Find the a rea ofeach and assume the bird has its eye
View Full Document
Unlocking...