Fundamental Theorem of Calculus

How Derivatives and Integrals Are Connected

The Fundamental Theorem of Calculus (FTC) connects derivatives and integrals. It explains why differentiating an accumulation function recovers its rate and why antiderivatives can evaluate definite integrals without recomputing a Riemann-sum limit.

Assume $f$ is continuous on the interval being considered. Part One differentiates an integral whose upper bound varies. Part Two evaluates a fixed-bound integral with any antiderivative of the integrand. Continuity is a sufficient hypothesis that guarantees both statements apply here.

Differentiating an Integral Function

Part One states that accumulating a continuous rate and then differentiating returns the original rate:

\begin{aligned} F(x) &= \int_{a}^{x} f(t) \, dt \\ &\implies F'(x) \\ &= f(x) \end{aligned}

This means that the derivative of a function defined as an integral is the function inside the integral itself.

If $f(t)$ is the inflow rate at time $t$ , then $F(x)$ is the volume added between the starting time and time $x$ . Its instantaneous rate of change, $F'(x)$ , equals the inflow rate $f(x)$ .

Example:

Find the derivative of $F(x) = \int_{1}^{x} \sqrt{2+t} \, dt$ .

Solution:

By FTC Part One, the derivative is obtained by evaluating the integrand at the variable upper bound: replace $t$ with $x$ .

\frac{d}{dx} \left( \int_{1}^{x} \sqrt{2+t} \, dt \right) = \sqrt{2+x}

The result is the original integrand evaluated at the upper limit.

Evaluating Integrals with Antiderivatives

If $F'(x)=f(x)$ , Part Two of the Fundamental Theorem of Calculus evaluates the definite integral by subtracting the values of $F$ at the two endpoints:

\int_{a}^{b} f(x) \, dx = F(b) - F(a)

Here $F(x)$ is an antiderivative of $f(x)$ . Subtracting its endpoint values gives the signed accumulation of $f(x)$ from $a$ to $b$ . For a nonnegative integrand, that value is also the geometric area.

That difference is often written with the bracket notation $[F(x)]_{a}^{b}$ , which means $F(b) - F(a)$ .

Example 1:

Find the value of $\int_{1}^{3} x^2 \, dx$ .

Solution:

Find the antiderivative: The antiderivative of $f(x) = x^2$ is $F(x) = \frac{1}{3}x^3$ .
Evaluate at the limits: Calculate $F(3) - F(1)$ .
$\int_{1}^{3} x^2 \, dx = \left[ \frac{1}{3}x^3 \right]_{1}^{3}$

Example 2:

Find the value of $\int_{1}^{5} (2x-1) \, dx$ .

Solution:

Find the antiderivative: The antiderivative of $f(x) = 2x - 1$ is $F(x) = x^2 - x$ .
Evaluate at the limits: Calculate $F(5) - F(1)$ .
$\int_{1}^{5} (2x-1) \, dx = \left[ x^2 - x \right]_{1}^{5}$

The Fundamental Theorem of Calculus calculates a definite integral from an antiderivative. Find the antiderivative, then subtract its value at the lower bound from its value at the upper bound.

How Derivatives and Integrals Are Connected

Differentiating an Integral Function

Part One states that accumulating a continuous rate and then differentiating returns the original rate:

\begin{aligned} F(x) &= \int_{a}^{x} f(t) \, dt \\ &\implies F'(x) \\ &= f(x) \end{aligned}

This means that the derivative of a function defined as an integral is the function inside the integral itself.

If $f(t)$ is the inflow rate at time $t$ , then $F(x)$ is the volume added between the starting time and time $x$ . Its instantaneous rate of change, $F'(x)$ , equals the inflow rate $f(x)$ .

Example:

Find the derivative of $F(x) = \int_{1}^{x} \sqrt{2+t} \, dt$ .

Solution:

By FTC Part One, the derivative is obtained by evaluating the integrand at the variable upper bound: replace $t$ with $x$ .

\frac{d}{dx} \left( \int_{1}^{x} \sqrt{2+t} \, dt \right) = \sqrt{2+x}

The result is the original integrand evaluated at the upper limit.

Evaluating Integrals with Antiderivatives

If $F'(x)=f(x)$ , Part Two of the Fundamental Theorem of Calculus evaluates the definite integral by subtracting the values of $F$ at the two endpoints:

\int_{a}^{b} f(x) \, dx = F(b) - F(a)

That difference is often written with the bracket notation $[F(x)]_{a}^{b}$ , which means $F(b) - F(a)$ .

Example 1:

Find the value of $\int_{1}^{3} x^2 \, dx$ .

Solution:

Find the antiderivative: The antiderivative of $f(x) = x^2$ is $F(x) = \frac{1}{3}x^3$ .
Evaluate at the limits: Calculate $F(3) - F(1)$ .
$\int_{1}^{3} x^2 \, dx = \left[ \frac{1}{3}x^3 \right]_{1}^{3}$

Example 2:

Find the value of $\int_{1}^{5} (2x-1) \, dx$ .

Solution:

Find the antiderivative: The antiderivative of $f(x) = 2x - 1$ is $F(x) = x^2 - x$ .
Evaluate at the limits: Calculate $F(5) - F(1)$ .
$\int_{1}^{5} (2x-1) \, dx = \left[ x^2 - x \right]_{1}^{5}$

The Fundamental Theorem of Calculus calculates a definite integral from an antiderivative. Find the antiderivative, then subtract its value at the lower bound from its value at the upper bound.