The Fundamental Theorem of Calculus, Part 2

FTC Part 2: Evaluating Definite Integrals

By the end of this lesson, you will be able to use the Fundamental Theorem of Calculus Part 2 to evaluate definite integrals efficiently.

The Evaluation Theorem (FTC Part 2)

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ (meaning $F' = f$), then:

$$\int_a^b f(x), dx = F(b) - F(a)$$

Notation: $F(b) - F(a) = \Big[F(x)\Big]_a^b$

This converts the limit of Riemann sums into a simple subtraction — the connection that makes calculus computationally powerful.

Using FTC Part 2

Step 1: Find any antiderivative $F$ of $f$ (no $+C$ needed — it cancels)

Step 2: Evaluate $F(b) - F(a)$

Example: $\displaystyle\int_1^4 3x^2, dx$

$F(x) = x^3$

$= [x^3]_1^4 = 64 - 1 = 63$

Example: $\displaystyle\int_0^{\pi} \sin x, dx$

$F(x) = -\cos x$

$= [-\cos x]_0^{\pi} = -\cos\pi - (-\cos 0) = -(-1) - (-1) = 1 + 1 = 2$

Example: $\displaystyle\int_1^e \frac{1}{x}, dx$

$F(x) = \ln x$

$= [\ln x]_1^e = \ln e - \ln 1 = 1 - 0 = 1$

When the Antiderivative Requires Work

Sometimes you need to rewrite the integrand before finding the antiderivative:

$\displaystyle\int_1^4 \frac{x+1}{\sqrt{x}}, dx = \int_1^4 \left(x^{1/2} + x^{-1/2}\right), dx = \left[\frac{2x^{3/2}}{3} + 2x^{1/2}\right]_1^4$

$= \left(\frac{16}{3} + 4\right) - \left(\frac{2}{3} + 2\right) = \frac{14}{3} + 2 = \frac{20}{3}$

What FTC Part 2 Tells Us

FTC links the two major operations of calculus:

• Differentiation and integration are inverse operations
• Every continuous function has an antiderivative (via the accumulation function from FTC Part 1)
• Computing definite integrals doesn't require Riemann sums — just find the antiderivative

Summary

FTC Part 2: $\int_a^b f(x), dx = F(b) - F(a)$ where $F' = f$.

Steps: find $F$, compute $F(b) - F(a)$, no $+C$ needed.

Common integrals: $\int x^n = \frac{x^{n+1}}{n+1}$; $\int \sin x = -\cos x$; $\int \cos x = \sin x$; $\int \frac{1}{x} = \ln|x|$; $\int e^x = e^x$.