Derivative Of Integral With Bounds

The Fundamental Theorem of Calculus: Unveiling the Derivative of an Integral with Bounds

Understanding the relationship between derivatives and integrals is fundamental to calculus. This article delves into the fascinating concept of finding the derivative of an integral with bounds, a crucial application of the Fundamental Theorem of Calculus (FTC). We'll explore the theorem itself, dissect its proof, and work through numerous examples to solidify your understanding. By the end, you'll be confidently tackling problems involving this powerful mathematical tool.

Introduction: Bridging the Gap Between Differentiation and Integration

The Fundamental Theorem of Calculus elegantly connects the seemingly disparate operations of differentiation and integration. It states, essentially, that differentiation and integration are inverse operations. This means that if we integrate a function and then differentiate the result, we get back to the original function (with some minor caveats we'll explore). Conversely, if we differentiate a function and then integrate the result, we again recover the original function (again, with some nuances). However, the focus of this article is specifically on the derivative of a definite integral, where the integral's bounds are functions themselves. This adds a layer of complexity and requires a deeper understanding of the FTC.

The Fundamental Theorem of Calculus (FTC): A Two-Part Theorem

The FTC is actually composed of two parts:

• Part 1 (The First Fundamental Theorem of Calculus): This part deals with the relationship between differentiation and integration when the upper bound of integration is a variable. If we define a function F(x) as:

  F(x) = ∫_a^x f(t) dt

  where 'a' is a constant and f(t) is a continuous function, then the derivative of F(x) is simply f(x):

  F'(x) = d/dx [∫_a^x f(t) dt] = f(x)

• Part 2 (The Second Fundamental Theorem of Calculus): This part deals with evaluating definite integrals using antiderivatives. If F(x) is an antiderivative of f(x), then:

  ∫_a^b f(x) dx = F(b) - F(a)

Derivative of an Integral with Variable Upper Bound: A Detailed Explanation

Let's focus on the scenario relevant to our main topic: finding the derivative of an integral where the upper bound is a function of x, say g(x). We have:

F(x) = ∫_a^g(x) f(t) dt

To find the derivative dF(x)/dx, we can use the chain rule. Think of the integral as a composition of functions: the integral itself is a function of g(x), and g(x) is a function of x. The chain rule states that the derivative of a composite function is the derivative of the outer function (with respect to the inner function) times the derivative of the inner function.

Therefore:

dF(x)/dx = d/dx [∫_a^g(x) f(t) dt] = f(g(x)) * g'(x)

This is a powerful result. It states that the derivative of the integral with respect to x is the integrand evaluated at the upper bound, multiplied by the derivative of the upper bound.

Proof using Leibniz's Rule

A more formal proof involves using Leibniz's rule for differentiation under the integral sign, a generalization of the FTC. This rule handles cases where both the upper and lower bounds are functions of x:

d/dx ∫_h(x)^g(x) f(t, x) dt = f(g(x), x)g'(x) - f(h(x), x)h'(x) + ∫_h(x)^g(x) ∂f(t, x)/∂x dt

In our simpler case where the lower bound is a constant 'a' and the integrand is independent of x (f(t)), the formula simplifies to:

d/dx ∫_a^g(x) f(t) dt = f(g(x))g'(x)

Illustrative Examples

Let's solidify our understanding with some examples:

Example 1:

Find the derivative of F(x) = ∫₀^x² cos(t) dt

Here, g(x) = x², so g'(x) = 2x. The integrand is f(t) = cos(t). Applying the formula:

dF(x)/dx = cos(x²) * 2x = 2x cos(x²)

Example 2:

Find the derivative of F(x) = ∫₁^{e^x} ln(t) dt

Here, g(x) = e^x, so g'(x) = e^x. The integrand is f(t) = ln(t). Applying the formula:

dF(x)/dx = ln(e^x) * e^x = x * e^x

Example 3 (More Complex):

Find the derivative of F(x) = ∫_x^x³ t² dt

Here, we have both upper and lower bounds as functions of x. We need to use Leibniz's Rule in its full form or apply the rule twice. First, let's rewrite it as a difference of two integrals:

F(x) = ∫₀^x³ t² dt - ∫₀^x t² dt

Now we can apply the formula to each integral separately:

d/dx ∫₀^x³ t² dt = (x³)² * 3x² = 3x⁸

d/dx ∫₀^x t² dt = x² * 1 = x²

Therefore, dF(x)/dx = 3x⁸ - x²

Example 4 (Involving Trigonometric Functions):

Find the derivative of F(x) = ∫_π/2^{sin x} √(1 + t²) dt

Here, g(x) = sin x, so g'(x) = cos x. The integrand is f(t) = √(1 + t²). Applying the formula:

dF(x)/dx = √(1 + sin²(x)) * cos x

Frequently Asked Questions (FAQ)

• Q: What happens if the lower bound is also a function of x? A: You'll need to use Leibniz's rule for differentiation under the integral sign, accounting for the derivative of both the upper and lower bounds.

• Q: What if the integrand is also a function of x? A: Again, you'll need to use Leibniz's rule, which includes a term for the partial derivative of the integrand with respect to x.

• Q: Are there any limitations to this rule? A: Yes, the integrand f(t) must be continuous on the interval of integration, and the bounds g(x) and h(x) (if present) must be differentiable.

• Q: How does this relate to the chain rule? A: The chain rule is crucial for deriving the formula. The integral is viewed as a composite function, with the integral being a function of the upper bound, which itself is a function of x.

• Q: Can this be applied to multiple integrals? A: The principles extend to multiple integrals, though the calculations become significantly more complex.

Conclusion: Mastering a Powerful Calculus Tool

The derivative of an integral with bounds is a powerful concept with wide-ranging applications in various fields of science and engineering. This article has explored the theoretical underpinnings, provided a step-by-step explanation, and illustrated the process through detailed examples. Mastering this concept solidifies your understanding of the Fundamental Theorem of Calculus and opens doors to more advanced calculus techniques. By applying the formula and understanding the underlying principles, you can confidently tackle a broad spectrum of problems involving derivatives and integrals. Remember to carefully identify the upper and lower bounds, the integrand, and their derivatives to accurately compute the desired derivative. Keep practicing, and you'll soon become proficient in this essential area of calculus.