Properties of Definite Integrals

Definite integrals follow a handful of simple rules. They let you break a hard integral into easier pieces, combine information you’re given, and evaluate integrals of straight-line and circular graphs using nothing but area formulas. AP questions use these rules constantly, especially with a graph or a list of given integral values.

Key ideas

The rules

For functions $f$ and $g$ that are continuous on the intervals involved, and any constant $k$:

Property	Rule
Zero-width interval	$\displaystyle\int_a^a f(x)\,dx = 0$
Reversing the limits	$\displaystyle\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$
Adding intervals	$\displaystyle\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx$
Constant multiple	$\displaystyle\int_a^b k\,f(x)\,dx = k\int_a^b f(x)\,dx$
Sum and difference	$\displaystyle\int_a^b \big(f(x) \pm g(x)\big)\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$
Constant function	$\displaystyle\int_a^b k\,dx = k(b - a)$

Each one makes sense if you think about signed area. Reversing the limits means going “backwards” across the strips, so every $\Delta x$ is negative. Adding intervals just glues two regions side by side. The “adding intervals” rule works even when $b$ isn’t between $a$ and $c$.

There is no product rule: $\int_a^b f(x)\,g(x)\,dx$ is generally not $\int_a^b f(x)\,dx \cdot \int_a^b g(x)\,dx$.

Using geometry

If the graph is made of lines and circle pieces, find the area of each piece and attach a sign:

• positive for regions above the $x$-axis,
• negative for regions below it.

Useful formulas: rectangle $bh$; triangle $\tfrac{1}{2}bh$; trapezoid $\tfrac{a + b}{2}h$; circle of radius $r$: $\pi r^2$ (half of it for a semicircle, a quarter for a quarter circle). The graph of $y = \sqrt{r^2 - x^2}$ is the top half of a circle of radius $r$ centred at the origin.

Net area versus total area

$\int_a^b f(x)\,dx$ gives the net (signed) area. To get the total area between the graph and the axis, add the areas as positive numbers. That’s the same as $\int_a^b |f(x)|\,dx$.

Worked examples

Example 1: Combining given values

Suppose $\displaystyle\int_0^5 f(x)\,dx = 8$, $\displaystyle\int_0^2 f(x)\,dx = 3$, and $\displaystyle\int_0^5 g(x)\,dx = -2$. Find:

(a) $\displaystyle\int_2^5 f(x)\,dx$ (b) $\displaystyle\int_5^0 f(x)\,dx$ (c) $\displaystyle\int_0^5 \big(3f(x) - 2g(x)\big)\,dx$ (d) $\displaystyle\int_0^5 \big(f(x) + 4\big)\,dx$

Solution.

(a) $\displaystyle\int_0^2 f + \int_2^5 f = \int_0^5 f$, so $\displaystyle\int_2^5 f(x)\,dx = 8 - 3 = 5$.

(b) Reversing the limits: $\displaystyle\int_5^0 f(x)\,dx = -8$.

(c) $3(8) - 2(-2) = 24 + 4 = 28$.

(d) $\displaystyle\int_0^5 f(x)\,dx + \int_0^5 4\,dx = 8 + 4(5) = 28$.

Example 2: Integrals from a graph

The graph of $f$ on $[-4, 4]$ is made of a semicircle and two line segments, as shown.

Areas above the axis count as positive; areas below count as negative.

Find (a) $\displaystyle\int_{-4}^{0} f(x)\,dx$ (b) $\displaystyle\int_0^4 f(x)\,dx$ (c) $\displaystyle\int_{-4}^{4} f(x)\,dx$ (d) $\displaystyle\int_4^{-2} f(x)\,dx$

Solution.

(a) A semicircle of radius $2$, above the axis: $\tfrac{1}{2}\pi(2)^2 = 2\pi$.

(b) The triangle has base $2$ and height $2$, area $2$; the rectangle is $2 \times 2 = 4$. Both are below the axis: $-2 - 4 = -6$.

(c) Add the intervals: $2\pi + (-6) = 2\pi - 6 \approx 0.283$.

(d) First find $\displaystyle\int_{-2}^{4} f(x)\,dx$. From $-2$ to $0$ is a quarter circle, area $\pi$, then $-6$ from $0$ to $4$: total $\pi - 6$. Reversing the limits:

$$\int_4^{-2} f(x)\,dx = -(\pi - 6) = 6 - \pi \approx 2.858$$

Example 3: An absolute value graph

Evaluate $\displaystyle\int_{-2}^{3} \big(2 - |x|\big)\,dx$.

Solution. The graph of $y = 2 - |x|$ is a “tent” with its peak at $(0, 2)$, crossing the axis at $x = -2$ and $x = 2$.

• From $-2$ to $2$: a triangle above the axis with base $4$ and height $2$, area $4$.
• From $2$ to $3$: a triangle below the axis with base $1$ and height $1$ (since $y(3) = -1$), area $\tfrac{1}{2}$.

$$\int_{-2}^{3} \big(2 - |x|\big)\,dx = 4 - \frac{1}{2} = \frac{7}{2}$$

Example 4: Splitting into a circle and a rectangle

Evaluate $\displaystyle\int_0^3 \left(\sqrt{9 - x^2} + 2\right)dx$.

Solution. Use the sum rule:

$$\int_0^3 \sqrt{9 - x^2}\,dx + \int_0^3 2\,dx$$

The first is a quarter of a circle of radius $3$: $\tfrac{1}{4}\pi(3)^2 = \tfrac{9\pi}{4}$. The second is a rectangle: $2 \times 3 = 6$.

$$\int_0^3 \left(\sqrt{9 - x^2} + 2\right)dx = \frac{9\pi}{4} + 6 \approx 13.069$$

Common mistakes

Forgetting to make areas below the axis negative. In Example 2(b), the answer is $-6$, not $6$. Ask “is this region above or below the axis?” for every piece.

Ignoring the order of the limits. $\int_4^{-2}$ runs backwards. Find the integral the normal way, then change the sign.

Using a product rule that doesn’t exist. You can pull out constants and split sums, but you can’t split a product or a quotient into separate integrals.

Forgetting the width when integrating a constant. $\int_0^5 4\,dx = 20$, not $4$. A constant integrates to a rectangle.

Mixing up net area and total area. “Find $\int_{-4}^4 f(x)\,dx$” wants the net (signed) value. “Find the total area” wants everything counted as positive.

Practice

1. (Warm-up) Evaluate $\displaystyle\int_3^3 (x^5 + e^x)\,dx$.

Solution

The interval has zero width, so the integral is $0$.

2. (Warm-up) Given $\displaystyle\int_1^4 f(x)\,dx = 7$, find $\displaystyle\int_4^1 f(x)\,dx$ and $\displaystyle\int_1^4 5f(x)\,dx$.

Solution

$\displaystyle\int_4^1 f(x)\,dx = -7$ and $\displaystyle\int_1^4 5f(x)\,dx = 5(7) = 35$.

3. (Warm-up) Evaluate $\displaystyle\int_{-1}^{3} 4\,dx$.

Solution

A rectangle of height $4$ and width $3 - (-1) = 4$: the integral is $16$.

4. (Core) Given $\displaystyle\int_1^6 f(x)\,dx = 10$ and $\displaystyle\int_4^6 f(x)\,dx = 3$, find $\displaystyle\int_1^4 f(x)\,dx$ and $\displaystyle\int_4^1 2f(x)\,dx$.

Solution

$\displaystyle\int_1^4 f(x)\,dx = 10 - 3 = 7$.

$\displaystyle\int_4^1 2f(x)\,dx = -2\int_1^4 f(x)\,dx = -14$.

5. (Core) Use the graph in Example 2 to find $\displaystyle\int_{-4}^{-2} f(x)\,dx$, $\displaystyle\int_2^4 f(x)\,dx$, and $\displaystyle\int_{-2}^{2} f(x)\,dx$.

Solution

$\displaystyle\int_{-4}^{-2} f(x)\,dx = \pi$ (a quarter circle of radius $2$, above the axis).

$\displaystyle\int_2^4 f(x)\,dx = -4$ (the rectangle below the axis).

$\displaystyle\int_{-2}^{2} f(x)\,dx = \pi - 2$ (a quarter circle, then the triangle below the axis).

6. (Core) Use the graph in Example 2 to evaluate $\displaystyle\int_0^4 \big(f(x) + 2\big)\,dx$.

Solution

$$\int_0^4 f(x)\,dx + \int_0^4 2\,dx = -6 + 8 = 2$$

7. (Core) Evaluate $\displaystyle\int_{-2}^{2} \left(3 + \sqrt{4 - x^2}\right)dx$.

Solution

$\displaystyle\int_{-2}^{2} 3\,dx = 3 \times 4 = 12$, and $\displaystyle\int_{-2}^{2} \sqrt{4 - x^2}\,dx$ is a semicircle of radius $2$: $\tfrac{1}{2}\pi(2)^2 = 2\pi$.

$$\int_{-2}^{2} \left(3 + \sqrt{4 - x^2}\right)dx = 12 + 2\pi \approx 18.283$$

8. (Challenge) For the graph in Example 2, find $\displaystyle\int_{-4}^{4} |f(x)|\,dx$. Explain why it is different from $\displaystyle\int_{-4}^{4} f(x)\,dx$.

Solution

$|f(x)|$ flips the parts below the axis up, so every area counts as positive:

$$\int_{-4}^{4} |f(x)|\,dx = 2\pi + 2 + 4 = 2\pi + 6 \approx 12.283$$

$\displaystyle\int_{-4}^{4} f(x)\,dx = 2\pi - 6 \approx 0.283$ is the net area, where the regions below the axis subtract. The first is the total area.

9. (Challenge) Find all values of $k \gt 0$ such that $\displaystyle\int_0^k (4 - x)\,dx = 6$. Use geometry.

Solution

For $0 \lt k \le 4$, the region is a trapezoid with parallel sides $4$ and $4 - k$ and width $k$:

$$\frac{4 + (4 - k)}{2} \cdot k = 4k - \frac{k^2}{2}$$

The same formula works for $k \gt 4$: the triangle from $0$ to $4$ has area $8$, and the triangle below the axis from $4$ to $k$ has signed area $-\tfrac{1}{2}(k - 4)^2$, and $8 - \tfrac{1}{2}(k - 4)^2 = 4k - \tfrac{k^2}{2}$.

$$4k - \frac{k^2}{2} = 6 \quad\Rightarrow\quad k^2 - 8k + 12 = 0 \quad\Rightarrow\quad (k - 2)(k - 6) = 0$$

So $k = 2$ or $k = 6$. Check $k = 6$: $8 - \tfrac{1}{2}(2)^2 = 6$. ✓