Express The Limit As A Definite Integral

The concept of expressing a limit as a definite integral bridges the gap between discrete sums and continuous functions, providing a powerful tool for evaluating limits that might otherwise be intractable. This transformation leverages the fundamental theorem of calculus and the definition of the definite integral as the limit of a Riemann sum. Understanding this connection not only enhances your problem-solving capabilities in calculus but also provides a deeper appreciation for the relationship between integration and summation.

Understanding Riemann Sums and Definite Integrals

At the heart of expressing a limit as a definite integral lies the concept of the Riemann sum. A Riemann sum approximates the area under a curve by dividing the area into a series of rectangles and summing their areas.

• Partition: Divide the interval [a, b] into n subintervals, not necessarily of equal width. Let the endpoints of these subintervals be denoted by x₀, x₁, x₂, ..., xₙ, where a = x₀ < x₁ < x₂ < ... < xₙ = b.

• Sample Points: Choose a sample point xᵢ** within each subinterval [xᵢ₋₁, xᵢ]. This point will determine the height of the rectangle in that subinterval.

• Riemann Sum: The Riemann sum is then defined as:

  ∑ᵢ₌₁ⁿ f(xᵢ**) Δxᵢ*

  where Δxᵢ = xᵢ - xᵢ₋₁ is the width of the i-th subinterval.

As n approaches infinity and the width of the widest subinterval (the norm of the partition, denoted by ||P||) approaches zero, the Riemann sum converges to the definite integral of f(x) from a to b, provided the limit exists. This is formally written as:

∫ₐᵇ f(x) dx = lim(||P||→₀) ∑ᵢ₌₁ⁿ f(xᵢ**) Δxᵢ*

This limit represents the exact area under the curve f(x) from x = a to x = x = b.

Transforming a Limit into a Definite Integral

The key to expressing a limit as a definite integral lies in recognizing patterns within the limit that resemble a Riemann sum. Often, the limit will involve a summation with terms that depend on n, the number of subintervals. By manipulating the expression algebraically, we can identify the following components:

• The function f(x): This is the function being evaluated in the summation. You'll need to express the terms in the sum in terms of a function of a variable x that varies between a lower and upper bound.
• The interval [a, b]*: This determines the limits of integration. The interval is often implicitly defined by the range of indices in the summation.
• The width of the subinterval Δx: This corresponds to the term that approaches zero as n approaches infinity. It's often of the form 1/n or a multiple thereof.
• xᵢ: This is the point at which the function f(x) is evaluated in each subinterval. It often has the form a + iΔx, where a is the lower limit of integration.

General Strategies:

1. Identify the Summation: The limit should contain a summation (∑) with an index that runs from a lower limit (usually 1) to an upper limit (usually n).

2. Look for i/n: Try to rewrite the terms inside the summation to include the term i/n. This is crucial because i/n will become the variable of integration, x, in the definite integral.

3. Determine f(x): Once you have i/n, identify the function f(x) that is being evaluated. This may require some algebraic manipulation.

4. Find the Interval [a, b]*: The interval of integration is usually determined by the limits of the summation. If the summation runs from i = 1 to n, and you have expressed the terms as functions of i/n, then the interval is usually [0, 1]. However, this is not always the case. You might need to rescale the variable or adjust the interval based on the specific problem. If the summation runs from i = 1 to 2n, the interval will be [0, 2].

5. Express as a Definite Integral: After identifying f(x) and the interval [a, b], you can write the limit as a definite integral:

   lim (n→∞) ∑ᵢ₌₁ⁿ f(xᵢ) Δx = ∫ₐᵇ f(x) dx

Examples

Let's illustrate this process with several examples:

Example 1:

lim (n→∞) ∑ᵢ₌₁ⁿ (i/n)² (1/n)

• Summation: We have a summation from i = 1 to n.
• i/n: The term (i/n)² already contains i/n.
• f(x): We can identify f(x) = x².
• Interval: The summation runs from i = 1 to n, and we have the term (i/n). This suggests the interval [0, 1].
• Δx: The term (1/n) corresponds to Δx.

Therefore, the limit can be expressed as the definite integral:

∫₀¹ x² dx

Evaluating this integral:

∫₀¹ x² dx = [x³/3]₀¹ = (1/3) - (0/3) = 1/3

Example 2:

lim (n→∞) ∑ᵢ₌₁ⁿ sin(πi/n) (1/n)

• Summation: We have a summation from i = 1 to n.
• i/n: The term sin(πi/n) contains i/n.
• f(x): We can identify f(x) = sin(πx).
• Interval: The summation runs from i = 1 to n, and we have the term (i/n). This suggests the interval [0, 1].
• Δx: The term (1/n) corresponds to Δx.

Therefore, the limit can be expressed as the definite integral:

∫₀¹ sin(πx) dx

Evaluating this integral:

∫₀¹ sin(πx) dx = [-cos(πx)/π]₀¹ = (-cos(π)/π) - (-cos(0)/π) = (1/π) + (1/π) = 2/π

Example 3 (Slightly More Complex):

lim (n→∞) ∑ᵢ₌₁ⁿ (3 + 2i/n)⁵ (2/n)

• Summation: We have a summation from i = 1 to n.
• i/n: The term (3 + 2i/n)⁵ contains i/n.
• f(x): We can identify f(x) = (3 + 2x)⁵.
• Interval: The summation runs from i = 1 to n. Since the expression is in the form 3 + 2(i/n), let's analyze it: when i=1, we have 3 + 2/n. When i=n, we have 3 + 2n/n = 5. When n approaches infinity, 3 + 2/n approach 3. Therefore, the interval is not [0, 1]. It is necessary to use the properties of the integral. Let x = i/n, then, x goes from 0 to 1 as i goes from 1 to n. Therefore the interval is [0, 1]
• Δx: The term (2/n) corresponds to 2Δx. This suggests Δx = 1/n.

Therefore, the limit can be expressed as the definite integral:

∫₀¹ (3 + 2x)⁵ * 2 dx*

Let u = 3 + 2x, then du = 2 dx. When x = 0, u = 3. When x = 1, u = 5. The integral becomes:

∫₃⁵ u⁵ du = [u⁶/6]₃⁵ = (5⁶/6) - (3⁶/6) = (15625/6) - (729/6) = 14896/6 = 7448/3

Example 4 (Non-Zero Lower Limit):

lim (n→∞) ∑ᵢ₌₁ⁿ √(1 + i/n) (1/n)

• Summation: We have a summation from i = 1 to n.
• i/n: The term √(1 + i/n) contains i/n.
• f(x): We can identify f(x) = √(1 + x).
• Interval: As i varies from 1 to n, i/n varies from 1/n to 1. As n approaches infinity, 1/n approaches 0. Thus x varies from 0 to 1. Thus, the limits of integration are 0 to 1. The interval of integration is [0, 1]
• Δx: The term (1/n) corresponds to Δx.

The limit becomes the definite integral:

∫₀¹ √(1 + x) dx

Evaluating the integral:

Let u = 1+x, then du = dx. When x = 0, u = 1. When x = 1, u = 2. The integral becomes

∫₁² √u du = ∫₁² u¹/² du = [(2/3)u³/²]₁² = (2/3)(2³/²) - (2/3)(1³/²) = (2/3)(2√2) - (2/3) = (4√2)/3 - 2/3

Example 5:

lim (n→∞) ∑ᵢ₌₁ⁿ (n/(n² + i²))

• Summation: We have a summation from i = 1 to n.

• i/n: We need to manipulate the expression to find i/n. Divide both the numerator and denominator by n²:

  lim (n→∞) ∑ᵢ₌₁ⁿ (1/(1 + (i/n)²)) (1/n)

• f(x): Now we can identify f(x) = 1/(1 + x²).

• Interval: The summation runs from i = 1 to n. Since we have i/n, the interval is [0, 1].

• Δx: The term (1/n) corresponds to Δx.

The limit becomes the definite integral:

∫₀¹ (1/(1 + x²)) dx

Evaluating the integral:

∫₀¹ (1/(1 + x²)) dx = [arctan(x)]₀¹ = arctan(1) - arctan(0) = π/4 - 0 = π/4

Common Mistakes and Pitfalls

• Incorrectly Identifying f(x): This is a common source of error. Make sure you have correctly isolated the function that depends on x (i/n).
• Incorrect Interval: The interval of integration is not always [0, 1]. Pay close attention to the limits of the summation and how they translate to the variable x. Sometimes a substitution is needed to adjust the limits.
• Forgetting Δx: Make sure you correctly identify and include the Δx term (usually a multiple of 1/n) in the definite integral.
• Assuming Equal Widths: While the examples often use equal widths (Δx = (b-a)/n), the definition of the Riemann integral allows for unequal widths as long as the norm of the partition approaches zero.
• Algebraic Errors: Carefully check your algebraic manipulations when rewriting the summation terms.

Generalization and Further Applications

The ability to express a limit as a definite integral extends beyond basic Riemann sums. It's a valuable tool in various areas, including:

• Approximating Areas and Volumes: Definite integrals are fundamental for calculating areas between curves and volumes of solids.
• Probability and Statistics: Continuous probability distributions are defined using integrals.
• Physics: Many physical quantities, such as work, energy, and center of mass, are calculated using integration.
• Numerical Analysis: Numerical integration techniques are used to approximate definite integrals when an analytical solution is not possible.
• Solving Differential Equations: Integration is a key operation in solving differential equations.
• Evaluating Infinite Series: In some instances, understanding the connection between definite integrals and limits of sums enables the approximation or calculation of certain types of infinite series.

Conclusion

Expressing a limit as a definite integral is a powerful technique rooted in the fundamental connection between Riemann sums and definite integrals. By mastering this concept, you gain a deeper understanding of calculus and unlock a valuable tool for solving a wider range of problems. The ability to recognize patterns within limits, manipulate algebraic expressions, and correctly identify the function and interval of integration is essential for success. Through practice and careful attention to detail, you can confidently transform limits into definite integrals and leverage the power of integration to solve complex problems. The relationship between discrete sums and continuous integrals is a cornerstone of calculus, and understanding it unlocks a deeper appreciation for the beauty and power of mathematical analysis.