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Chapter 1: Problem 33

In the following exercises, graph the function then use a calculator or a computer program to evaluate the following left and right endpoint sums. Is the area under the curve between the left and right endpoint sums? $$ [\mathrm{T}] L_{100} \text { and } R_{100} \text { for } y=e^{2 x} \text { on the interval }[-1,1] $$

Short Answer

Expert verified
Yes, the area is between \( L_{100} \) and \( R_{100} \).

Step by step solution

01

Understanding the Function and Interval

We are asked to find the left endpoint sum \( L_{100} \) and the right endpoint sum \( R_{100} \) for the function \( y = e^{2x} \) on the interval \([-1, 1]\). This means we need to estimate the area under the curve from \( x = -1 \) to \( x = 1 \) using 100 subintervals.
02

Divide the Interval into Subintervals

The interval \([-1, 1]\) will be divided into 100 equal parts. The width of each subinterval, \( \Delta x \), is calculated as \( \Delta x = \frac{1 - (-1)}{100} = \frac{2}{100} = 0.02 \).
03

Calculate Left Endpoint Sum \(L_{100}\)

The left endpoint sum \( L_{100} \) uses the value of the function at the left-end of each subinterval. The formula for \( L_{100} \) is given by:\[ L_{100} = \sum_{i=0}^{99} e^{2(-1 + i \cdot 0.02)} \cdot 0.02 \].Calculate each term and sum them up using a calculator or a computer program.
04

Calculate Right Endpoint Sum \(R_{100}\)

The right endpoint sum \( R_{100} \) uses the value of the function at the right-end of each subinterval. The formula for \( R_{100} \) is given by:\[ R_{100} = \sum_{i=1}^{100} e^{2(-1 + i \cdot 0.02)} \cdot 0.02 \].Calculate each term and sum them up using a calculator or a computer program.
05

Compare the Sums to the Exact Area

Evaluate both \( L_{100} \) and \( R_{100} \). The true area under the curve \( y = e^{2x} \) from \( x = -1 \) to \( x = 1 \) can be found by computing the definite integral \( \int_{-1}^{1} e^{2x} \, dx \).Compute the integral:\[ \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C \]Evaluate the definite integral:\[ \left[ \frac{1}{2} e^{2x} \right]_{-1}^{1} = \frac{1}{2} e^{2} - \frac{1}{2} e^{-2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Left Endpoint Sum
The left endpoint sum is a method for approximating the area under a curve. In this case, we are exploring the exponential function \( y = e^{2x} \) over the interval \([-1, 1]\). To approximate the area, we divide the interval into 100 subintervals. The width of each subinterval, \( \Delta x \), is \( 0.02 \). This means we're evaluating the function value at the left end of each subinterval and multiplying it by \( \Delta x \) to find the area of each rectangle:
  • The left endpoint at each subinterval starts at \( x = -1 + i \cdot \Delta x \) for \( i = 0 \) to \( 99 \).
  • The formula \( L_{100} = \sum_{i=0}^{99} e^{2(-1 + i \cdot 0.02)} \cdot 0.02 \) sums these rectangular areas.
Using this method, we gather an approximation that tends to underestimate the actual area as it ignores the slope within each interval.
Right Endpoint Sum
The right endpoint sum is another method used to estimate the area under a curve. Similarly, we look at the exponential function \( y = e^{2x} \) over the interval \([-1, 1]\) divided into 100 parts each of width \( \Delta x = 0.02 \). Unlike the left endpoint sum, this method considers the right side of each subinterval to evaluate the height of the rectangle:
  • The right endpoint is calculated at \( x = -1 + i \cdot \Delta x \) for \( i = 1 \) to \( 100 \).
  • The formula \( R_{100} = \sum_{i=1}^{100} e^{2(-1 + i \cdot 0.02)} \cdot 0.02 \) then provides the sum of all rectangular areas.
This technique often results in an overestimation of the area as it also disregards changes within each interval's slope. Depending on the function's growth, the right endpoint sum can be quite different from the left, highlighting how the choice of endpoint affects approximation.
Exponential Function
An exponential function is one of the simplest yet most powerful types of mathematical functions. It involves a constant base raised to a variable exponent, such as \( y = e^{2x} \), where \( e \) is Euler's number, approximately 2.718. This kind of function is known for its rapid growth or decay:
  • In \( e^{2x} \), the exponent \( 2x \) signifies a transformation in the rate of growth based on \( x \).
  • Exponential functions are essential in modeling real-world phenomena like population growth and radioactive decay.
They are continuous and smooth, which makes them intriguing candidates for integral approximation, as their properties influence how sums like the left or right endpoint sum behave.
Definite Integral
The definite integral is a fundamental concept in calculus used to calculate the exact area under a curve over a specific interval. In our case, for \( y = e^{2x} \) from \( x = -1 \) to \( x = 1 \), determining the definite integral involves finding:
  • The antiderivative: \( \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C \).
  • Evaluating it as \( \left[ \frac{1}{2} e^{2x} \right]_{-1}^{1} = \frac{1}{2} e^{2} - \frac{1}{2} e^{-2} \).
This computation reveals the precise area, giving a benchmark against which the approximate areas from endpoint sums can be measured. The definite integral serves as the true measurement of area under the curve, balancing over- and underestimations from approximate Riemann sums.

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Most popular questions from this chapter

In the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer. $$ \text { [T] } y=\frac{x}{\left(x^{2}+1\right)^{2}} \text { over }[-1,1] $$

In the following exercises, use a change of variables to show that each definite integral is equal to zero. $$ \int_{0}^{\pi} \cos ^{2}(2 \theta) \sin (2 \theta) d \theta $$

In the following exercises, \(f(x) \geq 0\) for \(a \leq x \leq b\). Find the area under the graph of \(f(x)\) between the given values \(a\) and \(b\) by integrating. $$ f(x)=2^{-x} ; a=3, b=4 $$

In the following exercises, find the antiderivative using the indicated substitution. $$ \int(x-1)^{5} d x ; u=x-1 $$

The area of a semicircle of radius 1 can be expressed as \(\int_{-1}^{1} \sqrt{1-x^{2}} d x\). Use the substitution \(x=\cos t\) to express the area of a semicircle as the integral of a trigonometric function. You do not need to compute the integral.
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