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**Thursday 8/30/12 "Estimating the Area Between a Curve and the x-axis Using The Derivative and Three Forms of Rieman Sums"**
math f'(x)=\frac{d}{dx}f(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} math
 * Definition of Derivative:**

math \frac{d}{dx}C=0 math math \frac{d}{dx} x^r=rx^{r-1} math math \frac{d}{dx}e^x=e^x math math \frac{d}{dx}a=(lna)a^x math math \frac{d}{dx}sinx=cosx math math \frac{d}{dx}tanx=sec^2x math math \frac{d}{dx}secx=secxtanx math math \frac{d}{dx}lnx=\frac{1}{x} math
 * Common Derivatives:**

Class Example: Find equation of line tangent to... math y=\frac{e^{x^2}}{1+x} ... math using the derivative when ... math x=0 math Soultion: math $ \frac{dy}{dx}=slope $ math math \frac{dy}{dx}=\frac{2xe^{x^2}(1+x)-(1)(e^{x^2})}{(1+x)^2}=\frac{-1}{1}=-1 math math f(0)=1 math math $ using: (y-y_0)=m(x-x_0) $ math math $ equation: y-1=-1(x-0) $ math
 * Using the derivative:**

Class Example: Find... math \int_{0}^{2}(2x+1)dx math We know that "Area 1" is a triangle and so its area can be found using the know geometric equation...
 * Find the area between a curve and the x-axis using known geometric formulas:**

math A_1=\frac{1}{2}bh=\frac{1}{2}(2*4)=4 math

Similarly we know that "Area 2" is a rectangle and that its area can be found using the equation... math A_2=lw=1*2=2 math

By adding these two areas together we can evaluate the total area between the curve and the x-axis. math A=2+4=6 math

Goal: Approximate the area between a curve f(x), and the x-axis over the interval [a,b], when f(x)__>__0...
 * General Functions**

Step 1: Split interval [a,b] into "n" number of subintervals of equal length. This length is referred to as (∆x). math a=x_0<x_1<x_2< ...<x_{n-1}<x_n=b math

Step 2: Approximate area under y=f(x) over the interval math [x_i,x_{i+1}] math by a rectangle Step 3: Add area of all rectangles for an approximation of the area under the curve

**Finding the Area Between a Curve and the x-axis Using the Three Rieman Sum Methods**
Consider the following graph of a function, f(x) We will approximate the area between the curve and the x-axis over the interval [1,3], with 4 subintervals (n=4), using the three different Rieman Sum methods. **1-**The Left-hand Endpoint Approximation **2-**The Right-hand Endpoint Approximation **3-**The Midpoint Endpoint Approximation The ∆x, or length of each interval, can be found using the formula... math \frac{b-a}{n}=\frac{3-1}{4}=\frac{1}{2} math and will remain the same throughout each method as the values of a,b, and n do not change based on which you use. A≈R1+R2+R3+R4 To approximate the area between the curve and the x-axis using a Left-Hand Rieman Sum we must first evaluate f(x) at the left-hand point of each subinterval and multiply each evaluation by ∆x. Finally you add each of the four products together to find an approximation of the total area between the curve and the x-axis over the interval [1,3]. Mathematically this looks like... math \frac{1}{2}[f(1)+f(\frac{3}{2})+f(2)+f(\frac{5}{2})] math As you can see this approximation is an underestimation because there is area left below that curve that is not included in the total area of the rectangles. Solving for area under the curve using Left Hand Rule: math f(x)=\frac{1}{2}x^2+2 math First find ∆x by... math \frac{b-a}{n}=\frac{2-0}{4}=\frac{1}{2} math Now multiply ∆x by the y value on the curve that represents the height of each rectangle: math A=\frac{1}{2}[2+2.125+2.5+3.125]=4.875 math A≈R1+R2+R3+R4 To approximate the area between the curve and the x-axis using a Right-Hand Rieman Sum we must first evaluate f(x) at the right-hand point of each subinterval and multiply each evaluation by ∆x. Finally you add each of the four products together to find an approximation of the total area between the curve and the x-axis over the interval [1,3]. As you can see this method is very similar to that of the Left-Hand Rieman Sum. Mathematically this looks like... math \frac{1}{2}[f(\frac{3}{2})+f(2)+f(\frac{5}{2})+f(3)] math Opposite of the previous method this approximation is an overestimation because there is area included in the rectangles that does not lie between the curve and the x-axis. Solving for area under the curve using Right Hand Rule: math f(x)=\frac{1}{2}x^2+2 math First find ∆x by... math \frac{b-a}{n}=\frac{2-0}{4}=\frac{1}{2} math Now multiply ∆x by the y value on the curve that represents the height of each rectangle: math A=\frac{1}{2}[2.125+2.5+3.125+4]=5.875 math A≈R1+R2+R3+R4 To approximate the area between the curve and the x-axis using a Mid-Point Rieman Sum we must first evaluate f(x) at the mid-point of each subinterval and multiply each evaluation by ∆x. Finally you add each of the four products together to find an approximation of the total area between the curve and the x-axis over the interval [1,3]. As you can see all three methods are very similar and vary only by the points at which f(x) is evaluated. Mathematically this looks like... math \frac{1}{2}[f(\frac{5}{4})+f(\frac{7}{4})+f(\frac{9}{4})+f(\frac{11}{4}) math This method can often produce the most accurate approximation of the area between and curve and the x-axis and does so in this hypothetical case. Solving for area under the curve using Mid-Point Rule: [[image:MPR.jpg]] math f(x)=\frac{1}{2}x^2+2 math First find ∆x by... math \frac{b-a}{n}=\frac{2-0}{4}=\frac{1}{2} math Now multiply ∆x by the y value on the curve that represents the height of each rectangle: math A=\frac{1}{2}[2.0313+2.2813+2.7813+3.5313]=5.3126 math
 * Left-Hand Rieman Sum Approximation:**
 * Real Example:**
 * Right-Hand Rieman Sum Approximation:**
 * Real Example:**
 * Mid-Point Rieman Sum Approximation:**
 * Real Example:**