The Correct Answer and Explanation is:
Here is the step-by-step solution to the problem.
Part (a): Determine Δx and xᵢ.
To find Δx and xᵢ, we first identify the components of the given integral, ∫¹₂ (2x³) dx, in the context of the definition of a definite integral.
- The function is f(x) = 2x³.
- The lower limit of integration is a = 1.
- The upper limit of integration is b = 2.
1. Calculate Δx:
The width of each subinterval, Δx, is given by the formula:
Δx = (b – a) / n
Substituting the values of a and b:
Δx = (2 – 1) / n = 1/n
2. Determine xᵢ:
The problem specifies using right-hand endpoints. The formula for the right-hand endpoint of the i-th subinterval is:
xᵢ = a + iΔx
Substituting the values for a and Δx:
xᵢ = 1 + i(1/n) = 1 + i/n
Answers for Part (a):
Δx = 1/n
xᵢ = 1 + i/n
Part (b): Using the definition mentioned above, evaluate the integral.
We use the definition of the definite integral as the limit of a Riemann sum:
∫ₐᵇ f(x) dx = lim(n→∞) Σᵢ₌₁ⁿ f(xᵢ)Δx
1. Set up the sum:
Substitute f(xᵢ) and Δx into the formula:
lim(n→∞) Σᵢ₌₁ⁿ [2(1 + i/n)³] * (1/n)
2. Expand and simplify the expression:
First, expand the cubic term (1 + i/n)³:
(1 + i/n)³ = 1³ + 3(1)²(i/n) + 3(1)(i/n)² + (i/n)³ = 1 + 3i/n + 3i²/n² + i³/n³
Now, substitute this back into the sum:
lim(n→∞) Σᵢ₌₁ⁿ [2(1 + 3i/n + 3i²/n² + i³/n³)] * (1/n)
Distribute the 2 and the 1/n:
lim(n→∞) Σᵢ₌₁ⁿ (2/n + 6i/n² + 6i²/n³ + 2i³/n⁴)
3. Apply summation properties and formulas:
Separate the summation term by term and factor out constants:
lim(n→∞) [ (2/n)Σᵢ₌₁ⁿ 1 + (6/n²)Σᵢ₌₁ⁿ i + (6/n³)Σᵢ₌₁ⁿ i² + (2/n⁴)Σᵢ₌₁ⁿ i³ ]
Use the standard summation formulas:
- Σᵢ₌₁ⁿ 1 = n
- Σᵢ₌₁ⁿ i = n(n+1)/2
- Σᵢ₌₁ⁿ i² = n(n+1)(2n+1)/6
- Σᵢ₌₁ⁿ i³ = [n(n+1)/2]² = n²(n+1)²/4
Substitute these formulas into the expression:
lim(n→∞) [ (2/n)(n) + (6/n²)(n(n+1)/2) + (6/n³)(n(n+1)(2n+1)/6) + (2/n⁴)(n²(n+1)²/4) ]
4. Simplify and evaluate the limit:
Simplify each term:
lim(n→∞) [ 2 + 3(n+1)/n + (n+1)(2n+1)/n² + (n+1)²/(2n²) ]
lim(n→∞) [ 2 + 3(1 + 1/n) + (2n²+3n+1)/n² + (n²+2n+1)/(2n²) ]
lim(n→∞) [ 2 + 3(1 + 1/n) + (2 + 3/n + 1/n²) + (1/2 + 1/n + 1/(2n²)) ]
As n approaches infinity, any term with n in the denominator (like 1/n or 1/n²) approaches 0.
So, the limit becomes:
2 + 3(1 + 0) + (2 + 0 + 0) + (1/2 + 0 + 0) = 2 + 3 + 2 + 0.5 = 7.5
Answer for Part (b):
Value of integral: 7.5
