Fundamental theorem of calculus
The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a topic in Integration in the California Calculus Standards. It is aligned to Standard 15.0, which requires students to demonstrate knowledge and proof of the Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus links derivatives and integrals: it evaluates a definite integral as \(\int_a^b f\,dx=F(b)-F(a)\) using an antiderivative, and shows that differentiation and integration are inverse operations.
Theory
The Fundamental Theorem of Calculus links derivatives and integrals. Part 2 evaluates a definite integral with an antiderivative: \(\int_a^b f=F(b)-F(a)\). Part 1 says differentiating an area-accumulation function gives back the original function.
The theorem has two halves that are two sides of the same coin.
Part 2 (evaluation): if \(F\) is any antiderivative of \(f\), thenThe two parts:
How to evaluate a definite integral
- Find an antiderivative \(F\) of the integrand.
- Evaluate \(F(b)\) and \(F(a)\) separately.
- Subtract: \(F(b)-F(a)\). No \(+C\) is needed — it cancels.
Antiderivative \(\dfrac{x^3}{3}\); apply \(F(3)-F(0)\).
| \(\int_0^3 x^2\,dx\) | \(=\) | \(\left[\dfrac{x^3}{3}\right]_0^3\) |
| \(=\) | \(\dfrac{27}{3}-0=9\) |
Antiderivative \(x^2\); evaluate at both limits.
| \(\int_1^2 2x\,dx\) | \(=\) | \(\left[x^2\right]_1^2\) |
| \(=\) | \(4-1=3\) |
Part 1 returns the integrand with \(t\) replaced by \(x\).
| \(\dfrac{d}{dx}\int_0^x t^2\,dt\) | \(=\) | \(x^2\) |
Antiderivative \(-\cos x\); substitute the limits, using \(\cos\pi=-1\) and \(\cos 0=1\).
| \(\int_0^{\pi}\sin x\,dx\) | \(=\) | \(\big[-\cos x\big]_0^{\pi}\) |
| \(=\) | \(-\cos\pi-(-\cos 0)\) | |
| \(=\) | \(1+1=2\) |
Common pitfalls
Frequently asked questions
What is the Fundamental Theorem of Calculus?
It connects integrals and derivatives. Part 2 evaluates \(\int_a^b f=F(b)-F(a)\) using an antiderivative; Part 1 says the derivative of \(\int_a^x f(t)\,dt\) is \(f(x)\).
How do you evaluate a definite integral?
Find an antiderivative \(F\), compute \(F(b)\) and \(F(a)\), and subtract: \(F(b)-F(a)\).
Why is there no + C in a definite integral?
Because the constant cancels: \((F(b)+C)-(F(a)+C)=F(b)-F(a)\). The \(+C\) only matters for indefinite integrals.
What does Part 1 of the theorem say?
Differentiating the area accumulated up to \(x\), \(\int_a^x f(t)\,dt\), gives back the integrand \(f(x)\). Integration and differentiation undo each other.