What does the fundamental theorem of calculus actually say

The first part of the Fundamental Theorem of Calculus (FTC I) tells us how to compute a definite integral using an antiderivative. If f is continuous on a closed interval [a,b] and F is any function with F'(x)=f(x) for every x in that interval, then $$\int_{a}^{b} f(x)\,dx = F(b)-F(a).$$ This formula turns the problem of summing infinitely many infinitesimal pieces into a simple subtraction, and it is the bridge that makes definite integration practical.

Part 2: The Derivative of an Integral

The second part (FTC II) describes the reverse relationship: differentiation undoes integration. Define a function F(x)=∫ₐˣ f(t)dt where the upper limit is variable. Provided f is continuous at x, the derivative of F is exactly the original integrand: $$F'(x)=f(x).$$ In other words, the process of accumulating area from a fixed start point and then differentiating brings you back to the height of the curve at each point.

Intuitively, FTC II says that the slope of the accumulated‑area function equals the height of the original curve. The continuity requirement guarantees that the area changes smoothly as the upper limit moves, preventing sudden jumps that would break the derivative. If f has a finite number of jump discontinuities, the theorem can still hold at points of continuity, but the formula for the derivative fails exactly at the jumps. Both parts of the theorem are used constantly in physics, engineering, and probability. FTC I lets you replace a hard integral with the difference of two antiderivative values, while FTC II justifies the technique of differentiating under the integral sign. Remember that the theorem only guarantees these relationships when the integrand is continuous on the interval of interest; checking that hypothesis saves you from subtle errors.

Example: evaluate $\int_{0}^{2} 3x^{2}\,dx$. First find an antiderivative: $F(x)=\int 3x^{2}dx = x^{3}+C$. Using FTC I, compute F(2)−F(0)=2^{3}−0^{3}=8. The same result follows from FTC II by defining G(x)=\int_{0}^{x}3t^{2}dt; then G'(x)=3x^{2} and G(2)=8. Thus the definite integral equals 8 square units.

The theorem requires several hypotheses:

• f is continuous on [a,b]
• The interval [a,b] is closed and bounded
• An antiderivative of f exists on the interval

To evaluate a definite integral with the FTC:

1. 1Find an antiderivative F of the integrand
2. 2Evaluate F at the upper limit
3. 3Subtract the value of F at the lower limit

Comparison of the two parts of the theorem: