Math Problem Statement

Let Rn be the right endpoint approximation to the area under the graph of f with N intervals, 1.) Compute R4 for f=4-x over the interval [0,4]. 2.) determine the value of limit n approaches infinity of Rn

Solution

Let's tackle each part of the question step by step:

1. Compute R4R_4 for f(x)=4xf(x) = 4 - x over the interval [0,4][0, 4]

The right endpoint approximation (RnR_n) uses the right endpoints of the subintervals to approximate the area under a curve. We are asked to compute R4R_4, which means we will divide the interval [0,4][0, 4] into 44 subintervals.

Step 1: Determine the width of each subinterval.

The formula for the width (Δx\Delta x) of each subinterval is:

Δx=ban\Delta x = \frac{b - a}{n}

Here, a=0a = 0, b=4b = 4, and n=4n = 4. So,

Δx=404=1\Delta x = \frac{4 - 0}{4} = 1

Step 2: Determine the right endpoints.

The right endpoints for the 4 subintervals are:

x1=1,x2=2,x3=3,x4=4x_1 = 1, \quad x_2 = 2, \quad x_3 = 3, \quad x_4 = 4

Step 3: Evaluate the function at the right endpoints.

We are using the function f(x)=4xf(x) = 4 - x. Evaluate the function at the right endpoints:

f(1)=41=3f(1) = 4 - 1 = 3 f(2)=42=2f(2) = 4 - 2 = 2 f(3)=43=1f(3) = 4 - 3 = 1 f(4)=44=0f(4) = 4 - 4 = 0

Step 4: Compute R4R_4.

The formula for the right endpoint approximation is:

R4=Δx(f(x1)+f(x2)+f(x3)+f(x4))R_4 = \Delta x \left( f(x_1) + f(x_2) + f(x_3) + f(x_4) \right)

Substituting the values:

R4=1(3+2+1+0)=1×6=6R_4 = 1 \left( 3 + 2 + 1 + 0 \right) = 1 \times 6 = 6

Thus, R4=6R_4 = 6.

2. Determine the value of limnRn\lim_{n \to \infty} R_n.

As nn \to \infty, the right endpoint approximation converges to the exact value of the definite integral of the function over the interval. So, we need to compute the definite integral of f(x)=4xf(x) = 4 - x over the interval [0,4][0, 4].

The definite integral is:

04(4x)dx\int_0^4 (4 - x) \, dx

Step 1: Find the antiderivative of 4x4 - x.

The antiderivative of 4x4 - x is:

(4x)dx=4xx22\int (4 - x) \, dx = 4x - \frac{x^2}{2}

Step 2: Evaluate the definite integral.

Now, evaluate the antiderivative at the endpoints 44 and 00:

[4xx22]04=(4(4)422)(4(0)022)\left[ 4x - \frac{x^2}{2} \right]_0^4 = \left( 4(4) - \frac{4^2}{2} \right) - \left( 4(0) - \frac{0^2}{2} \right)

Simplifying:

=(16162)0=168=8= \left( 16 - \frac{16}{2} \right) - 0 = 16 - 8 = 8

Thus, the value of limnRn\lim_{n \to \infty} R_n is 88.

Final Answers:

  1. R4=6R_4 = 6
  2. limnRn=8\lim_{n \to \infty} R_n = 8

Would you like any further explanation on these steps?


Relative Questions:

  1. How is the left endpoint approximation different from the right endpoint approximation?
  2. How does the midpoint approximation compare to the right endpoint approximation?
  3. Can the trapezoidal rule provide a more accurate estimate for f(x)=4xf(x) = 4 - x?
  4. What is the formula for the error in a right endpoint approximation?
  5. How would R4R_4 change if the function were f(x)=x2f(x) = x^2?

Tip: To improve approximation accuracy, increasing the number of subintervals nn gives a better estimate of the integral.

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Math Problem Analysis

Mathematical Concepts

Calculus
Definite Integral
Riemann Sums
Limit
Area under a curve

Formulas

Right Endpoint Approximation: R_n = Δx [f(x_1) + f(x_2) + ... + f(x_n)]
Width of subintervals: Δx = (b - a) / n
Definite Integral: ∫[a,b] f(x) dx

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

College Level Calculus