Evaluating Integrals with FTC Part 2

Computing integrals as limits of sums is slow. Part 1 of the theorem (Accumulation Functions and FTC Part 1) leads to a shortcut that replaces all of that with one subtraction, using Antiderivatives.

Key idea

Fundamental Theorem of Calculus, Part 2. If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$, then $$\int_a^b f(x)\,dx = F(b) - F(a) = \Big[F(x)\Big]_a^b.$$

Why it works

Let $A(x) = \int_a^x f(t)\,dt$. By Part 1, $A$ is an antiderivative of $f$. So is $F$, and two antiderivatives differ by a constant: $A(x) = F(x) + C$. At $x = a$, $A(a) = 0$, so $C = -F(a)$. At $x = b$: $\int_a^b f = A(b) = F(b) - F(a)$. Any constant added to $F$ cancels in the subtraction, so no "$+ C$" is needed.

For $f(x) = x^2$, the accumulated area $A(x)$ is $\frac{x^3}{3}$, an antiderivative of $x^2$. At $x = 2$ the area is $\frac{8}{3}$.

Worked example: A polynomial

Evaluate $\displaystyle\int_1^3 (x^2 + 1)\,dx$.

Antiderivative: $F(x) = \frac{x^3}{3} + x$.
$F(3) = 9 + 3 = 12$; $F(1) = \frac{1}{3} + 1 = \frac{4}{3}$.
$12 - \frac{4}{3} = \frac{32}{3}$.

Answer: $\frac{32}{3}$

Worked example: Sine

Evaluate $\displaystyle\int_0^{\pi}\sin x\,dx$.

Antiderivative: $-\cos x$.
$\big[-\cos x\big]_0^{\pi} = -\cos\pi - (-\cos 0) = 1 + 1$.

Answer: $2$: one arch of the sine curve has area exactly 2.

Worked example: Careful with signs

Evaluate $\displaystyle\int_{-1}^{2}(x^2 - 2x)\,dx$.

$F(x) = \frac{x^3}{3} - x^2$.
$F(2) = \frac{8}{3} - 4 = -\frac{4}{3}$.
$F(-1) = -\frac{1}{3} - 1 = -\frac{4}{3}$.
$F(2) - F(-1) = -\frac{4}{3} - \left(-\frac{4}{3}\right) = 0$.

Answer: $0$: the area below the axis on $(0, 2)$ exactly balances the area above it on $(-1, 0)$.

Together, the two parts of the theorem say that differentiation and integration undo each other. Part 1: integrate, then differentiate, and you get the function back. Part 2: differentiate, then integrate, and you get the total change. A problem about areas (adding up infinitely many slivers) becomes a problem about slopes (finding an antiderivative).

The theorem needs $f$ continuous on the whole interval. Ignoring that leads to nonsense: "$\int_{-1}^{1}\frac{1}{x^2}\,dx = \Big[-\frac{1}{x}\Big]_{-1}^{1} = -2$" is wrong, since a positive function cannot have a negative integral. The integrand blows up at 0; such improper integrals are handled in Calculus 2.

Common mistake

Computing $F(a) - F(b)$ (bottom minus top), or dropping a minus sign when $F(a)$ is negative, as in $F(2) - F(-1)$ above. Write the subtraction with parentheses: $F(b) - \big(F(a)\big)$. Check the sign of your answer against a sketch: mostly above the axis should give a positive number.

Evaluating Integrals with FTC Part 2

Practice

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