Evaluating Definite Integrals with the FTC

Riemann sums and limits get the exact value of a definite integral, but they are slow. The second part of the Fundamental Theorem of Calculus gives a shortcut that is almost magical: to find the area under $f$ from $a$ to $b$, find any antiderivative $F$ and compute $F(b) - F(a)$. This is how you’ll evaluate most definite integrals from now on.

Key ideas

The FTC, part 2

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

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

The shorthand $\Big[F(x)\Big]_a^b$ means $F(b) - F(a)$.

Why it works: by the first part of the FTC, $g(x) = \displaystyle\int_a^x f(t)\,dt$ is an antiderivative of $f$. Any other antiderivative is $F(x) = g(x) + C$. So

$$F(b) - F(a) = \big(g(b) + C\big) - \big(g(a) + C\big) = g(b) - 0 = \int_a^b f(x)\,dx .$$

The $+ C$ always cancels, so you can leave it out of definite integrals.

The steps

1. Rewrite the integrand if needed (powers, split fractions, expand).
2. Find an antiderivative using the rules from indefinite integrals.
3. Substitute the upper limit, then the lower limit, and subtract. Use brackets so the subtraction applies to everything.

The net change theorem

Applying the FTC to a function’s own derivative gives

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

The integral of a rate of change is the net change. Rearranged, this lets you find a later value from an earlier one:

$$F(b) = F(a) + \int_a^b F'(x)\,dx .$$

This is a favourite on the AP exam. When the antiderivative is hard or impossible to find, the question is usually calculator-active, and you evaluate the integral numerically. Give decimal answers correct to three decimal places.

Worked examples

Example 1: A power function

Evaluate $\displaystyle\int_1^3 x^2\,dx$.

Solution. An antiderivative of $x^2$ is $\tfrac{x^3}{3}$:

$$\int_1^3 x^2\,dx = \left[\frac{x^3}{3}\right]_1^3 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$$

The shaded area under $y = x^2$ from $1$ to $3$ is exactly $\tfrac{26}{3} \approx 8.667$.

Example 2: Trig functions

Evaluate (a) $\displaystyle\int_0^{\pi/2} \cos x\,dx$ and (b) $\displaystyle\int_0^{\pi} \sin x\,dx$. (Angles are in radians.)

Solution.

(a) $\displaystyle\int_0^{\pi/2} \cos x\,dx = \Big[\sin x\Big]_0^{\pi/2} = \sin\frac{\pi}{2} - \sin 0 = 1 - 0 = 1$

(b) An antiderivative of $\sin x$ is $-\cos x$:

$$\int_0^{\pi} \sin x\,dx = \Big[-\cos x\Big]_0^{\pi} = -\cos\pi - (-\cos 0) = -(-1) + 1 = 2$$

So one “hump” of the sine curve has area exactly $2$.

Example 3: Rewrite first

Evaluate $\displaystyle\int_1^4 \left(3\sqrt{x} - \frac{2}{x^2}\right)dx$.

Solution. Write the terms as powers: $3x^{1/2} - 2x^{-2}$. Antidifferentiate each term with the power rule:

$$3 \cdot \frac{x^{3/2}}{3/2} - 2 \cdot \frac{x^{-1}}{-1} = 2x^{3/2} + \frac{2}{x}$$ $$\begin{aligned} \int_1^4 \left(3\sqrt{x} - \frac{2}{x^2}\right)dx &= \left[2x^{3/2} + \frac{2}{x}\right]_1^4 \\ &= \left(2(8) + \frac{2}{4}\right) - \left(2(1) + \frac{2}{1}\right) \\ &= 16.5 - 4 = \frac{25}{2} \end{aligned}$$

Example 4: Net change (calculator active)

A function $f$ satisfies $f(2) = 5$ and $f'(x) = \sqrt{1 + x^3}$. Find $f(4)$.

Solution. By the net change theorem,

$$f(4) = f(2) + \int_2^4 f'(x)\,dx = 5 + \int_2^4 \sqrt{1 + x^3}\,dx .$$

There’s no simple antiderivative of $\sqrt{1 + x^3}$, so use a calculator’s numerical integration: $\displaystyle\int_2^4 \sqrt{1 + x^3}\,dx \approx 10.742$.

$$f(4) \approx 5 + 10.742 = 15.742$$

On the AP exam, write the integral expression (with the $5 +$) before giving the number. That setup is worth points on its own.

Common mistakes

Subtracting in the wrong order. It’s $F(\text{upper}) - F(\text{lower})$, always.

Losing a sign when subtracting. Put the whole $F(a)$ in brackets. In Example 3, $-\big(2 + 2\big)$, not $-2 + 2$.

Antidifferentiating sine with the wrong sign. $\dfrac{d}{dx}(-\cos x) = \sin x$, so $\int \sin x\,dx = -\cos x + C$. Check by differentiating.

Using degrees. $\cos\tfrac{\pi}{2}$ means radians. Make sure your calculator is in radian mode for calculus.

Thinking a zero integral means zero area. $\int_{-1}^{2} (x^2 - 2x)\,dx = 0$ (see Practice 6), but the graph has regions above and below the axis whose signed areas cancel.

Forgetting the starting value. In net change problems, the integral is the change. Add it to the initial value.

Practice

1. (Warm-up) Evaluate $\displaystyle\int_0^2 (3x^2 + 1)\,dx$.

Solution

$$\Big[x^3 + x\Big]_0^2 = (8 + 2) - 0 = 10$$

2. (Warm-up) Evaluate $\displaystyle\int_1^2 4x^3\,dx$.

Solution

$$\Big[x^4\Big]_1^2 = 16 - 1 = 15$$

3. (Warm-up) Evaluate $\displaystyle\int_0^1 e^x\,dx$.

Solution

$$\Big[e^x\Big]_0^1 = e - 1 \approx 1.718$$

4. (Core) Evaluate $\displaystyle\int_0^{\pi/4} \sec^2 x\,dx$.

Solution

$\dfrac{d}{dx}\tan x = \sec^2 x$, so

$$\Big[\tan x\Big]_0^{\pi/4} = \tan\frac{\pi}{4} - \tan 0 = 1$$

5. (Core) Evaluate $\displaystyle\int_1^9 \frac{x - 1}{\sqrt{x}}\,dx$.

Solution

Split the fraction: $\dfrac{x - 1}{\sqrt{x}} = x^{1/2} - x^{-1/2}$.

$$\begin{aligned} \int_1^9 \left(x^{1/2} - x^{-1/2}\right)dx &= \left[\frac{2}{3}x^{3/2} - 2x^{1/2}\right]_1^9 \\ &= \left(\frac{2}{3}(27) - 2(3)\right) - \left(\frac{2}{3} - 2\right) \\ &= 12 - \left(-\frac{4}{3}\right) = \frac{40}{3} \end{aligned}$$

6. (Core) Evaluate $\displaystyle\int_{-1}^{2} (x^2 - 2x)\,dx$. What does the answer tell you about the graph?

Solution

$$\left[\frac{x^3}{3} - x^2\right]_{-1}^{2} = \left(\frac{8}{3} - 4\right) - \left(-\frac{1}{3} - 1\right) = -\frac{4}{3} - \left(-\frac{4}{3}\right) = 0$$

The net signed area is $0$: the region above the axis (from $-1$ to $0$) has exactly the same area as the region below the axis (from $0$ to $2$).

7. (Core) Evaluate $\displaystyle\int_1^e \left(\frac{2}{x} + x\right)dx$.

Solution

$$\left[2\ln|x| + \frac{x^2}{2}\right]_1^e = \left(2 + \frac{e^2}{2}\right) - \left(0 + \frac{1}{2}\right) = \frac{3}{2} + \frac{e^2}{2} \approx 5.195$$

8. (Challenge) A tank holds $200$ L of water at $t = 0$. Water drains out at a rate of $r(t) = 10e^{-0.1t}$ litres per minute. How much water is in the tank after $5$ minutes? (Calculator active.)

Solution

The water is leaving, so subtract the amount drained:

$$200 - \int_0^5 10e^{-0.1t}\,dt \approx 200 - 39.347 = 160.653$$

There are about $160.653$ L left. (Exactly, the integral is $100(1 - e^{-0.5})$.)

9. (Challenge) Evaluate $\displaystyle\int_0^2 |x^2 - 1|\,dx$.

Solution

$x^2 - 1$ is negative on $[0, 1)$ and positive on $(1, 2]$, so split at $x = 1$:

$$\begin{aligned} \int_0^2 |x^2 - 1|\,dx &= \int_0^1 (1 - x^2)\,dx + \int_1^2 (x^2 - 1)\,dx \\ &= \left[x - \frac{x^3}{3}\right]_0^1 + \left[\frac{x^3}{3} - x\right]_1^2 \\ &= \frac{2}{3} + \left(\frac{2}{3} - \left(-\frac{2}{3}\right)\right) = \frac{2}{3} + \frac{4}{3} = 2 \end{aligned}$$