Math Problem Statement

Use the definition of area as a limit to evaluate the integral: ∫(2x - x^2)dx from 0 to 2.

Solution

The given problem is to evaluate the integral of 02(2xx2)dx\int_0^2 (2x - x^2) \, dx using the definition of area as a limit.

Step-by-Step Solution:

  1. Definition of Area as a Limit: The integral abf(x)dx\int_a^b f(x) \, dx can be interpreted as the limit of a Riemann sum: abf(x)dx=limni=1nf(xi)Δx\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x where Δx=ban\Delta x = \frac{b - a}{n} and xix_i^* is a sample point in each subinterval.

  2. Riemann Sum Setup: For 02(2xx2)dx\int_0^2 (2x - x^2) \, dx, we divide the interval [0,2][0, 2] into nn equal subintervals. Thus, Δx=20n=2n\Delta x = \frac{2 - 0}{n} = \frac{2}{n}. The sample points xix_i^* are chosen as xi=2inx_i^* = \frac{2i}{n} for i=1,2,...,ni = 1, 2, ..., n.

  3. Evaluate the Riemann Sum: The function to be integrated is f(x)=2xx2f(x) = 2x - x^2. Thus, the Riemann sum becomes: i=1nf(xi)Δx=i=1n(2(2in)(2in)2)2n\sum_{i=1}^n f(x_i^*) \Delta x = \sum_{i=1}^n \left( 2 \left( \frac{2i}{n} \right) - \left( \frac{2i}{n} \right)^2 \right) \cdot \frac{2}{n} Simplifying the terms inside the sum:

    = \frac{2}{n} \sum_{i=1}^n \left( \frac{8i}{n} - \frac{8i^2}{n^2} \right)$$
  4. Split the Summation: Now, we can split the sum into two parts: 2n(8ni=1ni8n2i=1ni2)\frac{2}{n} \left( \frac{8}{n} \sum_{i=1}^n i - \frac{8}{n^2} \sum_{i=1}^n i^2 \right) Using the known formulas for the sums of integers and squares of integers: i=1ni=n(n+1)2,i=1ni2=n(n+1)(2n+1)6\sum_{i=1}^n i = \frac{n(n+1)}{2}, \quad \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} Substituting these into the expression: 2n(8nn(n+1)28n2n(n+1)(2n+1)6)\frac{2}{n} \left( \frac{8}{n} \cdot \frac{n(n+1)}{2} - \frac{8}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} \right)

  5. Simplify and Take the Limit: Simplifying the first term: 2n8nn(n+1)2=8(n+1n)\frac{2}{n} \cdot \frac{8}{n} \cdot \frac{n(n+1)}{2} = 8 \cdot \left( \frac{n+1}{n} \right) And simplifying the second term: 2n8n2n(n+1)(2n+1)6=16(n+1)(2n+1)3n2\frac{2}{n} \cdot \frac{8}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{16(n+1)(2n+1)}{3n^2} As nn \to \infty, n+1n1\frac{n+1}{n} \to 1 and 2n+1n2\frac{2n+1}{n} \to 2. Therefore, the terms become: 8116(1)(2)3=8323=243323=838 \cdot 1 - \frac{16(1)(2)}{3} = 8 - \frac{32}{3} = \frac{24}{3} - \frac{32}{3} = -\frac{8}{3}

Final Answer:

The value of the integral is: 83\boxed{\frac{8}{3}}


Do you have any questions, or would you like further details?

Here are some related questions you might consider:

  1. How can you visualize the area under the curve geometrically?
  2. How does the choice of sample points in the Riemann sum affect the result?
  3. What are other methods of evaluating integrals aside from the limit definition?
  4. How does the Fundamental Theorem of Calculus apply to this integral?
  5. What is the significance of negative area in definite integrals?

Tip: When using the Riemann sum, splitting the summation and using known formulas for sum of integers can greatly simplify the calculation process.

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Riemann Sum
Limit

Formulas

∫_a^b f(x) dx = lim (n -> ∞) Σ f(x_i*) Δx
Sum of integers: Σ i = n(n+1)/2
Sum of squares: Σ i^2 = n(n+1)(2n+1)/6

Theorems

Definition of Integral as a Limit
Summation Formulas

Suitable Grade Level

Grades 11-12 or College Calculus