The Fundamental Theorem of Calculus, Part 1

FTC Part 1: Derivatives of Integrals

By the end of this lesson, you will be able to apply FTC Part 1 to differentiate accumulation functions and handle chain rule complications with variable upper limits.

The Accumulation Function

Define the accumulation function (or area function):

$$F(x) = \int_a^x f(t), dt$$

$F(x)$ gives the net signed area accumulated from $a$ to $x$.

FTC Part 1

If $f$ is continuous on an interval containing $a$, then $F(x) = \int_a^x f(t), dt$ is differentiable and:

$$F'(x) = \frac{d}{dx}\int_a^x f(t), dt = f(x)$$

The derivative of an integral (with variable upper limit) is the integrand evaluated at the upper limit.

This says: differentiation and integration are inverse processes.

Direct Applications

Example 1: $\frac{d}{dx}\int_1^x \cos(t^2), dt = \cos(x^2)$

No need to actually evaluate the integral — just substitute $x$ for $t$ in the integrand.

Example 2: $\frac{d}{dx}\int_x^5 \sqrt{1+t^3}, dt$

Flip the limits first: $= \frac{d}{dx}\left[-\int_5^x \sqrt{1+t^3}, dt\right] = -\sqrt{1+x^3}$

Chain Rule Complication

When the upper limit is a function of $x$ (not just $x$ itself), apply the chain rule:

$$\frac{d}{dx}\int_a^{g(x)} f(t), dt = f(g(x)) \cdot g'(x)$$

Example: $\frac{d}{dx}\int_0^{x^3} \sin t, dt = \sin(x^3) \cdot 3x^2 = 3x^2\sin(x^3)$

Example: $\frac{d}{dx}\int_2^{e^x} \ln t, dt = \ln(e^x) \cdot e^x = x \cdot e^x$

Both Limits Variable

When both limits are functions of $x$, split the integral at a constant:

$$\frac{d}{dx}\int_{h(x)}^{g(x)} f(t), dt = \frac{d}{dx}\left[\int_c^{g(x)} f(t), dt - \int_c^{h(x)} f(t), dt\right]$$

$= f(g(x)) \cdot g'(x) - f(h(x)) \cdot h'(x)$

Example: $\frac{d}{dx}\int_{x^2}^{x^3} \sqrt{t}, dt = \sqrt{x^3}(3x^2) - \sqrt{x^2}(2x) = 3x^2 \cdot x^{3/2} - 2x \cdot |x| = 3x^{7/2} - 2x^2$ (for $x > 0$)

What FTC Part 1 Tells Us

FTC Part 1 shows that every continuous function has an antiderivative — namely the accumulation function $F(x) = \int_a^x f(t), dt$.

It also explains why the notation $\int f(x), dx = F(x) + C$ makes sense: $F(x)$ is a function whose derivative is $f(x)$.