Math Problem Statement

Evaluate the integral

3

x2  3x 10 dx as limit of sums

Solution

To evaluate the integral 03(x23x+10)dx\int_0^3 (x^2 - 3x + 10) \, dx as a limit of sums, we use the concept of Riemann sums. Here are the steps:

Step 1: Partition the interval

Let's partition the interval [0,3][0, 3] into nn subintervals of equal length. The length of each subinterval, Δx\Delta x, is: Δx=30n=3n\Delta x = \frac{3 - 0}{n} = \frac{3}{n} The ii-th subinterval will be [xi1,xi][x_{i-1}, x_i] where: xi=0+iΔx=3inx_i = 0 + i\Delta x = \frac{3i}{n}

Step 2: Choose sample points

We'll choose the right endpoint of each subinterval as the sample point. So, the sample point xix_i^* in the ii-th subinterval is: xi=3inx_i^* = \frac{3i}{n}

Step 3: Write the Riemann sum

The Riemann sum for the integral is given by: i=1nf(xi)Δx\sum_{i=1}^n f(x_i^*) \Delta x Substituting f(x)=x23x+10f(x) = x^2 - 3x + 10, xi=3inx_i^* = \frac{3i}{n}, and Δx=3n\Delta x = \frac{3}{n}, the Riemann sum becomes: i=1n((3in)23(3in)+10)3n\sum_{i=1}^n \left(\left(\frac{3i}{n}\right)^2 - 3\left(\frac{3i}{n}\right) + 10\right) \cdot \frac{3}{n} Simplifying the expression inside the sum: i=1n(9i2n29in+10)3n\sum_{i=1}^n \left(\frac{9i^2}{n^2} - \frac{9i}{n} + 10\right) \cdot \frac{3}{n} Distribute 3n\frac{3}{n}: i=1n(27i2n327in2+30n)\sum_{i=1}^n \left(\frac{27i^2}{n^3} - \frac{27i}{n^2} + \frac{30}{n}\right)

Step 4: Take the limit as nn \to \infty

The integral is the limit of the Riemann sum as nn approaches infinity: 03(x23x+10)dx=limni=1n(27i2n327in2+30n)\int_0^3 (x^2 - 3x + 10) \, dx = \lim_{n \to \infty} \sum_{i=1}^n \left(\frac{27i^2}{n^3} - \frac{27i}{n^2} + \frac{30}{n}\right) We can evaluate the sum term by term:

  1. First term: limn27n3i=1ni2=27n3n(n+1)(2n+1)6\lim_{n \to \infty} \frac{27}{n^3} \sum_{i=1}^n i^2 = \frac{27}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} Simplifying: limn27n3+27n2+9n6n3=limn(276+276n+96n2)=276=4.5\lim_{n \to \infty} \frac{27n^3 + 27n^2 + 9n}{6n^3} = \lim_{n \to \infty} \left(\frac{27}{6} + \frac{27}{6n} + \frac{9}{6n^2}\right) = \frac{27}{6} = 4.5

  2. Second term: limn27n2i=1ni=27n2n(n+1)2\lim_{n \to \infty} \frac{-27}{n^2} \sum_{i=1}^n i = \frac{-27}{n^2} \cdot \frac{n(n+1)}{2} Simplifying: limn27n227n2n2=limn(272272n)=272=13.5\lim_{n \to \infty} \frac{-27n^2 - 27n}{2n^2} = \lim_{n \to \infty} \left(\frac{-27}{2} - \frac{27}{2n}\right) = \frac{-27}{2} = -13.5

  3. Third term: limni=1n30n=301=30\lim_{n \to \infty} \sum_{i=1}^n \frac{30}{n} = 30 \cdot 1 = 30

Final Result:

Summing up the results of these limits: 03(x23x+10)dx=4.513.5+30=21\int_0^3 (x^2 - 3x + 10) \, dx = 4.5 - 13.5 + 30 = 21

So, the value of the integral is 21\boxed{21}.

Would you like to discuss any part of this calculation in more detail?

Here are 5 related questions to explore:

  1. How does the choice of sample points (left, right, or midpoint) affect the Riemann sum?
  2. Can you apply the same Riemann sum approach to an integral with a different function?
  3. How can we generalize this method for non-uniform partitions?
  4. What is the geometric interpretation of the integral as the area under the curve?
  5. How does the limit process work in Riemann sums, and why does it lead to the exact value of the integral?

Tip: Practicing the evaluation of integrals using Riemann sums helps reinforce the concept of integration as the sum of areas of infinitesimally small rectangles.

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Riemann Sums

Formulas

Riemann sum formula

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

College