Find The Derivative Of An Integral

Finding the Derivative of an Integral: The Bridge Between Calculus Pillars

At the heart of calculus lies a profound and elegant connection between its two main operations: differentiation and integration. The process of finding the derivative of an integral is not a mere algebraic trick but a cornerstone concept that reveals the deep, symmetric relationship between these processes. This connection is formally established by the Fundamental Theorem of Calculus (FTC), a theorem so pivotal it essentially unifies the subject. Understanding how to differentiate an integral function transforms a seemingly complex operation into a straightforward application of a powerful rule, providing a direct path from the accumulation of quantities back to the original rate of change.

The Fundamental Theorem of Calculus: The Core Principle

The FTC is typically presented in two parts, each addressing a different but related idea. For the task of differentiating an integral, Part 1 is the direct and essential tool.

Part 1 of the Fundamental Theorem of Calculus states: If f is a continuous function on the closed interval [a, b] and we define a new function F by F(x) = ∫[a]^[x] f(t) dt, for x in [a, b], then F is differentiable on (a, b) and F'(x) = f(x).

In simpler terms, if you construct a function F(x) that represents the area under the curve of f(t) from a fixed starting point a to a variable endpoint x, then the instantaneous rate of change of this accumulated area—its derivative—is simply the original function f(x) evaluated at x.

This is astonishing. It means the operation of integration (finding area) and differentiation (finding slope) are inverse processes, much like addition and subtraction. The derivative of the integral "undoes" the integration, returning you to the original function, provided the lower limit is a constant and the upper limit is the variable x.

A Simple, Concrete Example

Let f(t) = t². Define F(x) = ∫[0]^[x] t² dt.

1. First, compute the integral (find the antiderivative): ∫ t² dt = (1/3)t³ + C.
2. Apply the limits: F(x) = [(1/3)x³] - [(1/3)(0)³] = (1/3)x³.
3. Now, differentiate F(x): F'(x) = d/dx [(1/3)x³] = x².

The result, x², is exactly our original function f(x). The FTC guarantees this will happen for any continuous function f.

The General Case: Variable Limits and the Chain Rule

The simple form d/dx [∫[a]^[x] f(t) dt] = f(x) is a special case where the lower limit is a constant and the upper limit is x. What happens when the limits are more complex? This is where the Leibniz Integral Rule comes into play, which is a direct application of the FTC combined with the Chain Rule.

Consider a function defined as G(x) = ∫[a(x)]^[b(x)] f(t) dt, where both the lower limit a(x) and the upper limit b(x) are differentiable functions of x. The derivative is:

G'(x) = f(b(x)) * b'(x) - f(a(x)) * a'(x)

This formula has an intuitive explanation:

• The term f(b(x)) * b'(x) accounts for how the upper limit is changing. As x changes, the endpoint b(x) moves. The rate at which area is added at the moving upper boundary is the height of the function at that boundary, f(b(x)), multiplied by the speed at which the boundary is moving, b'(x).
• The term - f(a(x)) * a'(x) does the same for the lower limit. If the lower limit moves to the right (increasing), it subtracts area from the total accumulation. Hence the negative sign.
• If a limit is a constant (e.g., a(x) = c), then a'(x) = 0, and that term vanishes, recovering the simpler FTC Part 1.

Example with Two Variable Limits

Let f(t) = sin(t). Find d/dx [ ∫[x²]^[e^x] sin(t) dt ]. Here, a(x) = x² and b(x) = e^x.

• f(b(x)) = sin(e^x)
• b'(x) = e^x
• f(a(x)) = sin(x²)
• a'(x) = 2x

Applying the formula: G'(x) = sin(e^x) * e^x - sin(x²) * (2x) G'(x) = e^x sin(e^x) - 2x sin(x²)

The Critical Role of the Dummy Variable

A common and crucial point of confusion must be addressed. In the expression ∫[a]^[x] f(t) dt, the variable t is called a dummy variable or variable of integration. It is a placeholder. The value of the definite integral depends only on the limits a and x, not on the letter t. We could just as easily write ∫[a]^[x] f(u) du and get