Math Problem Statement

Consider the function f(x) = x over [0, 4]. (a) Sketch the function. Hence, find the area under the graph over [0, 4]. (b) Calculate the right-endpoint approximation R_N where N is the number of rectangles. Hence, find the area under the graph over [0, 4] by calculating lim(N -> ∞) R_N. (c) Calculate the left-endpoint approximation L_N where N is the number of rectangles. Hence, find the area under the graph over [0, 4] by calculating lim(N -> ∞) L_N.

Solution

Let's break this problem into steps to solve each part systematically.


(a) Sketch the function and find the area under the graph over [0, 4]

The function f(x)=xf(x) = x is a straight line that passes through the origin. Over the interval [0,4][0, 4], the graph is a line starting at (0,0)(0, 0) and ending at (4,4)(4, 4).

The area under the graph from [0,4][0, 4] is the area of the triangle formed by the x-axis, the line x=4x = 4, and the function. The formula for the area of a triangle is: Area=12×base×height.\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}. Here, the base is 44 and the height is also 44. Thus: Area=12×4×4=8.\text{Area} = \frac{1}{2} \times 4 \times 4 = 8.

Result for part (a): The area under the graph is 8.


(b) Right-endpoint approximation RNR_N

For the right-endpoint approximation, divide the interval [0,4][0, 4] into NN rectangles. The width of each rectangle is: Δx=4N.\Delta x = \frac{4}{N}. The right endpoints are: xi=4iN,for i=1,2,,N.x_i = \frac{4i}{N}, \quad \text{for } i = 1, 2, \dots, N. The area of each rectangle is f(xi)Δx=(4iN)4Nf(x_i) \Delta x = \left(\frac{4i}{N}\right) \cdot \frac{4}{N}. The total area is: RN=i=1Nf(xi)Δx=i=1N(4iN)4N.R_N = \sum_{i=1}^N f(x_i) \Delta x = \sum_{i=1}^N \left(\frac{4i}{N}\right) \cdot \frac{4}{N}. Simplify: RN=16N2i=1Ni.R_N = \frac{16}{N^2} \sum_{i=1}^N i. Using the formula for the sum of the first NN integers: i=1Ni=N(N+1)2,\sum_{i=1}^N i = \frac{N(N+1)}{2}, we get: RN=16N2N(N+1)2=16(N+1)2N=8+8N.R_N = \frac{16}{N^2} \cdot \frac{N(N+1)}{2} = \frac{16(N+1)}{2N} = 8 + \frac{8}{N}. Take the limit as NN \to \infty: limNRN=8.\lim_{N \to \infty} R_N = 8.

Result for part (b): The area under the graph using right-endpoint approximation is 8.


(c) Left-endpoint approximation LNL_N

For the left-endpoint approximation, divide the interval [0,4][0, 4] into NN rectangles. The width of each rectangle is: Δx=4N.\Delta x = \frac{4}{N}. The left endpoints are: xi=4(i1)N,for i=1,2,,N.x_i = \frac{4(i-1)}{N}, \quad \text{for } i = 1, 2, \dots, N. The area of each rectangle is f(xi)Δx=(4(i1)N)4Nf(x_i) \Delta x = \left(\frac{4(i-1)}{N}\right) \cdot \frac{4}{N}. The total area is: LN=i=1Nf(xi)Δx=i=1N(4(i1)N)4N.L_N = \sum_{i=1}^N f(x_i) \Delta x = \sum_{i=1}^N \left(\frac{4(i-1)}{N}\right) \cdot \frac{4}{N}. Simplify: LN=16N2i=1N(i1).L_N = \frac{16}{N^2} \sum_{i=1}^N (i-1). Using the formula for the sum of the first N1N-1 integers: i=1N(i1)=j=0N1j=(N1)N2,\sum_{i=1}^N (i-1) = \sum_{j=0}^{N-1} j = \frac{(N-1)N}{2}, we get: LN=16N2(N1)N2=16(N1)2N=88N.L_N = \frac{16}{N^2} \cdot \frac{(N-1)N}{2} = \frac{16(N-1)}{2N} = 8 - \frac{8}{N}. Take the limit as NN \to \infty: limNLN=8.\lim_{N \to \infty} L_N = 8.

Result for part (c): The area under the graph using left-endpoint approximation is 8.


Final Results:

  • (a) Exact area: 88.
  • (b) Right-endpoint approximation: limNRN=8\lim_{N \to \infty} R_N = 8.
  • (c) Left-endpoint approximation: limNLN=8\lim_{N \to \infty} L_N = 8.

Do you want me to clarify any part of this?

Here are five related questions to consider:

  1. How does the midpoint approximation compare to the left- and right-endpoint approximations?
  2. Can you apply the trapezoidal rule to this problem? What result would you get?
  3. Why does the exact area match the limit of RNR_N and LNL_N?
  4. How does increasing NN affect the accuracy of these approximations?
  5. Can you generalize this method to find the area under f(x)=x2f(x) = x^2 over any interval [a,b][a, b]?

Tip: When solving for approximations using sums, always simplify the summation expressions step by step to avoid errors!

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Riemann Sums
Limits

Formulas

Area of a triangle: A = (1/2) * base * height
Sum of the first N integers: ∑(j=1 to N) j = N(N+1)/2

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate Calculus