Math Core

You rarely compute a definite integral from scratch. Instead you break it into pieces, pull out constants, flip limits and reuse values you already know. These properties follow directly from thinking of the integral as signed area, and AP questions test them constantly, often by giving you a few integral values and asking for a new one.

Properties about the limits

Limit properties

For a function $f$ that is integrable on the intervals involved:

Zero width: $\displaystyle\int_a^a f(x)\,dx = 0$ .
Reversing the limits: $\displaystyle\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$ .
Adding adjacent intervals: $\displaystyle\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx$ .

Property 1 says a region with no width has no area. Property 2 makes sense from Riemann sums: going from $b$ to $a$ makes every width $\Delta x$ negative, which flips the sign of the whole sum.

Property 3 says that area from $a$ to $b$ plus area from $b$ to $c$ is the area from $a$ to $c$ . Thanks to Property 2, it's true no matter how $a$ , $b$ and $c$ are ordered. For example, even though 7 is outside $[2, 5]$ ,

\int_2^5 f(x)\,dx = \int_2^7 f(x)\,dx + \int_7^5 f(x)\,dx.

Worked example: Splitting and reversing

Suppose $\displaystyle\int_0^5 f(x)\,dx = 8$ and $\displaystyle\int_0^2 f(x)\,dx = 3$ . Find $\displaystyle\int_2^5 f(x)\,dx$ and $\displaystyle\int_5^2 f(x)\,dx$ .

By additivity, $\displaystyle\int_0^2 f + \int_2^5 f = \int_0^5 f$ , so

\int_2^5 f(x)\,dx = 8 - 3 = 5.

Reversing the limits changes the sign: $\displaystyle\int_5^2 f(x)\,dx = -5$ .

Properties about the integrand

Linearity

For constants $k$ and functions $f$ and $g$ :

Constant multiple: $\displaystyle\int_a^b k\,f(x)\,dx = k\int_a^b f(x)\,dx$ .
Sum and difference: $\displaystyle\int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$ .
Constant function: $\displaystyle\int_a^b k\,dx = k(b - a)$ .

Stretching a graph vertically by a factor $k$ multiplies every rectangle's height, and therefore the whole area, by $k$ . Stacking one graph on another adds the heights, so the areas add. Property 6 is just a rectangle: height $k$ , width $b - a$ .

Worked example: Combining properties

Using the values from the previous example, evaluate $\displaystyle\int_5^2 \big(3f(x) - 2\big)\,dx$ .

Flip the limits first, then split:

\begin{aligned} \int_5^2 \big(3f(x) - 2\big)\,dx &= -\int_2^5 \big(3f(x) - 2\big)\,dx \\ &= -\left(3\int_2^5 f(x)\,dx - \int_2^5 2\,dx\right) \\ &= -\big(3(5) - 2(3)\big) = -(15 - 6) = -9. \end{aligned}

Common mistake

There is no product rule for integrals: $\displaystyle\int_a^b f(x)g(x)\,dx$ is generally not $\displaystyle\int_a^b f(x)\,dx \cdot \int_a^b g(x)\,dx$ . For example, $\displaystyle\int_0^2 x \cdot x\,dx = \frac{8}{3}$ , but $\displaystyle\int_0^2 x\,dx \cdot \int_0^2 x\,dx = 2 \cdot 2 = 4$ . Linearity only lets you split sums and pull out constant factors.

Using additivity with piecewise functions

When a function changes formula, or its graph crosses the axis, split the integral at the break point and handle each piece separately.

The region under y = |x − 2| from x = 0 to x = 5 splits at x = 2 into two triangles.Open in grapher →

Worked example: An absolute value integrand

Evaluate $\displaystyle\int_0^5 |x - 2|\,dx$ .

Split at $x = 2$ , where the expression inside the absolute value changes sign:

\int_0^5 |x - 2|\,dx = \int_0^2 (2 - x)\,dx + \int_2^5 (x - 2)\,dx.

The first piece is a triangle with base 2 and height 2, area 2. The second is a triangle with base 3 and height 3, area $\dfrac{9}{2}$ . The total is $2 + \dfrac{9}{2} = \dfrac{13}{2}$ .

Comparison and symmetry

Two more facts are handy for checking answers and for quick multiple-choice decisions.

Comparison: if $f(x) \le g(x)$ for every $x$ in $[a, b]$ (with $a \lt b$ ), then $\displaystyle\int_a^b f(x)\,dx \le \int_a^b g(x)\,dx$ . In particular, if $f(x) \ge 0$ on $[a, b]$ , its integral there is nonnegative.
Symmetry: if $f$ is odd ( $f(-x) = -f(x)$ ), the areas on the two sides of the $y$ -axis cancel, so $\displaystyle\int_{-a}^{a} f(x)\,dx = 0$ . If $f$ is even ( $f(-x) = f(x)$ ), the two sides match, so $\displaystyle\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$ .

Worked example: Symmetry saves work

Evaluate $\displaystyle\int_{-3}^{3} \big(x^5 - 4x + 2\big)\,dx$ .

Split by linearity. The functions $x^5$ and $-4x$ are odd, so their integrals over $[-3, 3]$ are 0. What's left is the constant:

\int_{-3}^{3} 2\,dx = 2\big(3 - (-3)\big) = 12.

Tip

When you're given several integral values, write them on a number line with the limits marked. Seeing which intervals are adjacent makes it obvious what to add or subtract.

Practice

Practice 1

If $\displaystyle\int_1^7 f(x)\,dx = 10$ and $\displaystyle\int_1^4 f(x)\,dx = 6$ , find $\displaystyle\int_4^7 f(x)\,dx$ .