Ftc Part 1

Fundamental Theorem of Calculus, Part 1

The Fundamental Theorem of Calculus, Part 1 reveals a profound relationship: differentiation and integration are inverse operations.

Core Statement

If $f$ is continuous on $[a,b]$, then the accumulation function $F(x) = \int_a^x f(t)\,dt$ is differentiable on $(a,b)$, and $F'(x) = f(x).$

In other words, the derivative of an accumulation function simply returns the original integrand evaluated at $x$. The lower limit $a$ is constant and doesn't affect the derivative—only the rate of change matters.

Why It Works

Think of $F(x)$ as tracking total area accumulated under $f$ from $a$ to $x$. As $x$ increases, the instantaneous rate of new area being added equals the height of the curve at that point, which is $f(x)$. This intuition justifies why $F'(x) = f(x)$.

When the Upper Limit Is a Function

Often the upper limit isn't simply $x$, but a function $g(x)$, such as $\int_a^{x^2} f(t)\,dt$. Apply the chain rule: $\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x).$

Example: For $\frac{d}{dx} \int_1^{x^3} \sin(t)\,dt$, we get $\sin(x^3) \cdot 3x^2$. The factor $3x^2$ is the derivative of the upper limit.

Variable Lower Limits

If the variable appears in the lower limit, use $\int_x^b f(t)\,dt = -\int_b^x f(t)\,dt$, giving $\frac{d}{dx}\int_x^b f(t)\,dt = -f(x).$

For both limits variable, split at a constant point and apply the chain rule to each piece, combining the sign flip with differentiation.

Common Pitfall

Don't forget the chain rule multiplier when the upper limit is composite. Writing $\frac{d}{dx}\int_1^{x^3} \sin(t)\,dt = \sin(x^3)$ (without the $3x^2$) is incorrect.