Two ideas have been built entirely separately so far: the antiderivative (reverse a derivative) and the definite integral (a limit of Riemann sums, measuring exact area). Nothing so far suggests they should be related at all. One is about slopes, the other about area. The Fundamental Theorem of Calculus is the proof that they are, secretly, the same computation. That's the reason a definite integral almost never gets evaluated by taking a Riemann-sum limit directly.
Imagine an odometer that only reports your car's SPEED at every instant, never the distance itself. The Fundamental Theorem says: find any function whose rate of change matches that speed reading. The total distance traveled is just that function's value at the end minus its value at the start. You never needed to add up infinitely many tiny speed readings directly. A single subtraction does the same job exactly.
The Fundamental Theorem of Calculus is the single result connecting the two ideas above, antiderivatives and definite integrals, by proving they're secretly the same computation. Part 2 is the practical engine: to evaluate \int_a^b f(x)\,dx, find ANY antiderivative F of f and compute F(b)-F(a). The +C from the previous section cancels automatically in the subtraction. That's why a definite integral's answer is a single number, never a family. Part 1, covered in depth next chapter, is the deeper reason this works at all. Differentiating that running total, A(x)=\int_a^x f(t)\,dt, gives back f(x) itself. That's what makes "any antiderivative" the right thing to plug in. The theorem earns the word "Fundamental" precisely because it turns two ideas that look unrelated on the surface, a rate of change and an accumulated total, into two sides of the exact same computation.
Theorem
Fundamental Theorem of Calculus, Part 2
also known as: FTC2, the Evaluation Theorem
Recalled from Calculus I — Differential & Integral Calculus · FTC Part 2: evaluating definite integrals — first taught there.
Suppose f is continuous on [a, b], and let F be ANY antiderivative of f on [a, b] (that is, F'(x) = f(x) for every x in [a,b]). Then
\int_a^b f(x)\,dx = F(b) - F(a).FTC Part 1 already exhibits ONE particular antiderivative of f: the accumulation function A(x) = \int_a^x f(t)\,dt satisfies A'(x) = f(x) on (a,b). Since F is also given to be an antiderivative of the same f, the General Antiderivative Theorem (a Mean Value Theorem corollary, Chapter 7 Section 1) says F and A differ by a constant: F(x) = A(x) + C \quad\text{for some constant } C. Evaluate this identity at x=a: A(a) = \int_a^a f(t)\,dt = 0 (a degenerate interval of zero width bounds zero signed area), so F(a) = A(a) + C = 0 + C = C -- the constant C is exactly F(a). Evaluate at x=b: F(b) = A(b) + C = A(b) + F(a), so A(b) = F(b) - F(a). But A(b) = \int_a^b f(t)\,dt by A's own definition, which is exactly \int_a^b f(x)\,dx with the dummy variable renamed -- giving \int_a^b f(x)\,dx = F(b)-F(a), the claim.
This is why FTC Part 2 makes definite integrals COMPUTABLE without ever building a Riemann sum: Chapter 7 could only ever approach \int_a^b f as a limit of finite sums, refining n forever without literally reaching it by hand. FTC Part 2 replaces that limit with a single subtraction, F(b)-F(a), PROVIDED an antiderivative F can be written down in closed form -- which every basic antiderivative rule from Chapter 7 Sections 1-4 (power, constant multiple, sum/difference, trig, exponential) now feeds directly into. The theorem does not care WHICH antiderivative F is used: any two antiderivatives of f differ by a constant C, and that same C cancels exactly in F(b)-F(a), so the computation is independent of which member of the antiderivative family gets chosen -- picking C=0 is always a valid, and always the simplest, choice.
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Watch A(x) build as the shaded area grows
Drag x — the shaded signed area above and the value of A(x) below move together, since A is defined by exactly that area.
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Worked example
Evaluating a definite integral with FTC Part 2
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1
Evaluate \displaystyle\int_1^3 (2x+1)\,dx.
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2
Find an antiderivative of 2x+1: F(x) = x^2+x (choosing C=0, since it will cancel regardless).
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3
Apply Part 2: F(3) - F(1) = (9+3) - (1+1) = 12 - 2.
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4
\boxed{\int_1^3(2x+1)\,dx = 10}. The +C never had to be carried through at all.
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Where this goes next
Every application chapter ahead
Area between curves, volumes, arc length, work: every application two chapters from now sets up a definite integral and then leans on this theorem to compute it. Nothing downstream re-derives integration from Riemann sums directly. FTC Part 2 is the computational bridge every one of them crosses.
Check your understanding (counts toward your progress)
Evaluate \displaystyle\int_0^{\pi} \sin x\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
F(x)=-\cos x; F(\pi)-F(0) = 1-(-1) = 2.
Correct answer
Also accepted:
Why
The antiderivative of \sin x is -\cos x -- evaluate it at \pi and at 0 and subtract.
What is the value of \displaystyle\int_0^2 (4x^3-2x)\,dx?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
F(x)=x^4-x^2; F(2)-F(0) = (16-4)-0 = 12.
Correct answer
Also accepted:
Why
Find an antiderivative of 4x^3-2x term by term, then evaluate it at 2 and at 0 and subtract.
g(x)=\displaystyle\int_2^x\left(-5t^2+1t-1\right)dt. Using FTC Part 1, find g'(-2).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
g'(x)=-5x^2+1x-1 (the integrand itself), so g'(-2)=-23.
Correct answer
Also accepted:
Why
FTC Part 1: g'(x) is exactly the integrand evaluated at x.
g(x)=\displaystyle\int_2^x\left(-6t^2-4t+4\right)dt. Using FTC Part 1, find g'(-3).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
g'(x)=-6x^2-4x+4 (the integrand itself), so g'(-3)=-38.
Correct answer
Also accepted:
Why
FTC Part 1: g'(x) is exactly the integrand evaluated at x.