The Fundamental Theorem of Calculus and Definite Integrals
Introduced| English |
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| antiderivative/ˌæntɪdɪˈrɪvətɪv/ |
Compute an area without adding rectangles
- Riemann sums give the idea of a definite integral, but summing infinitely many rectangles is impractical.
- The second part of the Fundamental Theorem of Calculus gives a shortcut.
- Find an antiderivative 原函数 $F$ (a function whose derivative is $f$), then subtract its values at the limits.
- Integration becomes "undo the derivative, then plug in two numbers."
A function $F$ with $F'=f$ is called an ____ of $f$.
FTC Part 2 evaluates an antiderivative at the limits.
The evaluation theorem
- If $F'=f$ (so $F$ is an antiderivative of $f$), then:
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$$\int_a^b f(x)\,dx = F(b)-F(a)$$
- Compute $F$, evaluate it at the top limit and the bottom limit, and subtract.
- The exact signed area drops out — no rectangles, no limits of sums.
The exact area, computed
y = x²
FTC Part 2 turns this shaded area into $F(2)-F(0)$ — an antiderivative evaluated at the limits.
FTC Part 2 says $\int_a^b f\,dx=$
Evaluate an antiderivative at upper minus lower limit.
The evaluation bar notation
- A tidy way to show the two-step evaluation: $\Big[F(x)\Big]_a^b = F(b)-F(a)$.
- Write the antiderivative in brackets with the limits, then subtract "top minus bottom."
- Keep the order: upper limit value minus lower limit value.
- Reversing it flips the sign (matching the swap-limits property).
Evaluate $\int_0^2 x^2\,dx$ using $F(x)=\tfrac{x^3}{3}$. (Enter a decimal.)
$\tfrac{8}{3}-0=\tfrac83\approx2.667$.
Computing $F(a)-F(b)$ instead of $F(b)-F(a)$ gives the correct value.
That flips the sign — use upper minus lower.
Evaluate $\int_1^3 2x\,dx$ using $F(x)=x^2$.
$F(3)-F(1)=9-1=8$.
Any antiderivative works
- If $F$ is one antiderivative, so is $F+C$ for any constant — but the $+C$ cancels in $F(b)-F(a)$.
- So for a definite integral you can pick the simplest antiderivative and ignore $+C$.
- (The constant only matters for indefinite integrals, lesson 6.8.)
- Double-check by differentiating your $F$: you should get $f$ back.
For a definite integral, the constant $+C$ of the antiderivative...
$(F(b)+C)-(F(a)+C)=F(b)-F(a)$.
You must use a genuine antiderivative ($F'=f$), then compute $F(b)-F(a)$ in that order. Swapping to $F(a)-F(b)$ flips the sign. The $+C$ cancels in a definite integral, so don't fret over it here — but do verify $F'=f$, or the whole answer is wrong.
Evaluate $\displaystyle\int_0^2 x^2\,dx$.
- An antiderivative of $x^2$ is $F(x)=\dfrac{x^3}{3}$ (since $F'=x^2$).
- $\displaystyle\int_0^2 x^2\,dx = \Big[\tfrac{x^3}{3}\Big]_0^2 = \tfrac{2^3}{3}-\tfrac{0^3}{3}=\tfrac{8}{3}$.
- That confirms the earlier Riemann-sum estimate of $\approx2.67$.
FTC Part 2: $\int_a^b f(x)\,dx = F(b)-F(a)$, where $F$ is any antiderivative of $f$ ($F'=f$). Find $F$, evaluate at the upper minus the lower limit. The $+C$ cancels, so any antiderivative works — this turns area into simple arithmetic.