Math Problem Statement

Let �  ⁣ : � → � f:R→R be the function defined by � ( � )

� 2 f(x)=x 2 , �

( 0 , 0.5 , 1 , 1.6 , 2 ) P=(0,0.5,1,1.6,2) and �

( 0 , 0.6 , 1.2 , 1.7 ) T=(0,0.6,1.2,1.7). What is � ( � , � , � ) R(f,P,T)?

Solution

The expression R(f,P,T)R(f, P, T) typically represents a Riemann sum, which is used to approximate the area under a curve defined by a function f(x)f(x) over a certain interval. Here’s a step-by-step solution for calculating R(f,P,T)R(f, P, T) given the information:

  1. Function Definition: The function f(x)=x2f(x) = x^2 is defined over R\mathbb{R}, meaning it maps real numbers to real numbers.

  2. Partition PP: The partition P=(0,0.5,1,1.6,2)P = (0, 0.5, 1, 1.6, 2) divides the interval [0,2][0, 2] into subintervals:

    • [0,0.5][0, 0.5]
    • [0.5,1][0.5, 1]
    • [1,1.6][1, 1.6]
    • [1.6,2][1.6, 2]
  3. Sample Points TT: The sample points T=(0,0.6,1.2,1.7)T = (0, 0.6, 1.2, 1.7) indicate the points in each interval where we evaluate the function f(x)=x2f(x) = x^2:

    • x=0x = 0 in [0,0.5][0, 0.5]
    • x=0.6x = 0.6 in [0.5,1][0.5, 1]
    • x=1.2x = 1.2 in [1,1.6][1, 1.6]
    • x=1.7x = 1.7 in [1.6,2][1.6, 2]
  4. Calculate the Riemann Sum R(f,P,T)R(f, P, T): The Riemann sum for this partition is computed by summing up the products of f(x)f(x) at the sample points TT and the widths of the corresponding subintervals defined by PP: R(f,P,T)=i=1nf(Ti)ΔxiR(f, P, T) = \sum_{i=1}^n f(T_i) \cdot \Delta x_i where Δxi\Delta x_i is the width of each subinterval and TiT_i is the sample point in that subinterval.

    • Interval [0,0.5][0, 0.5]: Δx1=0.50=0.5\Delta x_1 = 0.5 - 0 = 0.5, T1=0T_1 = 0, so f(T1)=f(0)=02=0f(T_1) = f(0) = 0^2 = 0
    • Interval [0.5,1][0.5, 1]: Δx2=10.5=0.5\Delta x_2 = 1 - 0.5 = 0.5, T2=0.6T_2 = 0.6, so f(T2)=f(0.6)=0.62=0.36f(T_2) = f(0.6) = 0.6^2 = 0.36
    • Interval [1,1.6][1, 1.6]: Δx3=1.61=0.6\Delta x_3 = 1.6 - 1 = 0.6, T3=1.2T_3 = 1.2, so f(T3)=f(1.2)=1.22=1.44f(T_3) = f(1.2) = 1.2^2 = 1.44
    • Interval [1.6,2][1.6, 2]: Δx4=21.6=0.4\Delta x_4 = 2 - 1.6 = 0.4, T4=1.7T_4 = 1.7, so f(T4)=f(1.7)=1.72=2.89f(T_4) = f(1.7) = 1.7^2 = 2.89
  5. Compute the Sum: R(f,P,T)=(00.5)+(0.360.5)+(1.440.6)+(2.890.4)R(f, P, T) = (0 \cdot 0.5) + (0.36 \cdot 0.5) + (1.44 \cdot 0.6) + (2.89 \cdot 0.4) =0+0.18+0.864+1.156= 0 + 0.18 + 0.864 + 1.156 =2.2= 2.2

Final Answer:

R(f,P,T)=2.2R(f, P, T) = 2.2

Would you like further details on this calculation?

  1. What if the function f(x)f(x) changes to a different polynomial function?
  2. How would changing the sample points TT affect the Riemann sum?
  3. What does this Riemann sum approximate in terms of area under the curve?
  4. How does the partition size affect the accuracy of the Riemann sum?
  5. Could we use this approach to estimate the integral as the partition becomes finer?

Tip: For more accurate area approximations, increase the number of intervals in the partition PP, which reduces each subinterval’s width.

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Math Problem Analysis

Mathematical Concepts

Riemann Sum
Integration
Partitioning Intervals
Quadratic Functions

Formulas

Riemann Sum formula: R(f, P, T) = Σ f(T_i) * Δx_i

Theorems

Riemann Sum Approximation

Suitable Grade Level

Grades 11-12