6.6 Applying Properties of Definite Integrals

Definite integral properties let you find the value of an integral by using area and algebraic rules, often without finding an antiderivative. You use rules like reversing limits, splitting an interval into adjacent pieces, and pulling out constants to combine known integral values into new ones. For AP Calculus, track signs carefully when intervals reverse or areas fall below the x-axis.

Why This Matters for the AP Calculus Exam

Definite integral properties show up constantly on the AP Calculus exam, especially when a problem gives you a few integral values and asks you to build a new one. You often see these in multiple-choice questions where a table or graph gives areas, and in free-response questions where you need to break an integral into adjacent intervals or combine pieces. Getting fluent with these properties means you can solve quickly without grinding through antiderivatives, which saves time for harder problems.

This topic also connects to reading integrals as signed area. When part of a region sits below the x-axis, that area counts as negative, and these properties help you keep the signs straight.

Key Takeaways

• A definite integral $\int_{a}^{b} f(x)\, dx$ gives a single number, which you can read as the signed area between $f(x)$ and the x-axis on $[a, b]$.
• When the upper and lower limits match, the integral is 0: $\int_{a}^{a} f(x)\, dx = 0$.
• Swapping the limits flips the sign: $\int_{b}^{a} f(x)\, dx = -\int_{a}^{b} f(x)\, dx$.
• Constants pull out front, and sums or differences of functions split into separate integrals.
• Adjacent intervals add: $\int_{a}^{b} f(x)\, dx + \int_{b}^{c} f(x)\, dx = \int_{a}^{c} f(x)\, dx$, and you can rearrange this to solve for any piece.
• You can evaluate some integrals using geometry (triangles, rectangles, semicircles), and the definition still works for functions with removable or jump discontinuities.

What Is a Definite Integral?

A definite integral integrates a function over an interval $[a, b]$ and gives you an actual number as the answer, not an expression with $+ C$. You can also read it as the signed area between $f(x)$ and the x-axis on $[a, b]$.

$$\int_{a}^{b}f(x)\, dx$$

Each part of the notation matters:

• $b$ is the upper limit of integration.
• $a$ is the lower limit of integration.
• $f(x)$ is the function you are integrating with respect to $x$.

"Signed area" is the key idea. Area above the x-axis counts as positive, and area below the x-axis counts as negative. So an integral can come out negative even though area itself is never negative.

Properties of Definite Integrals

Learn and get comfortable with these properties. They let you combine known integrals into new ones.

1. The Zero Rule. When the upper and lower limits are the same, the integral equals zero. As an area, both limits at the same spot give you no width, so there is no area.

$$\int_{a}^{a}f(x)\, dx = 0$$

2. Reversing the limits. If the limits are swapped (the larger number on the bottom), you can rewrite the integral by negating it.

$$\int_{b}^{a}f(x)\, dx = -\int_{a}^{b}f(x)\, dx$$

3. Constant multiple. A constant factor $k$ can move to the front of the integral.

$$\int_{a}^{b}k\cdot f(x)\, dx = k\int_{a}^{b}f(x)\, dx$$

4. Sum and difference. If two functions are added or subtracted in the integrand, you can integrate them separately.

$$\int_{a}^{b}[f(x)+g(x)]\, dx = \int_{a}^{b}f(x)\, dx + \int_{a}^{b}g(x)\, dx$$ $$\int_{a}^{b}[f(x)-g(x)]\, dx = \int_{a}^{b}f(x)\, dx - \int_{a}^{b}g(x)\, dx$$

5. Adjacent intervals. An integral from $a$ to $c$ can be split at any point $b$ in between by adding the two pieces. You can rearrange this to solve for either piece.

$$\int_{a}^{b}f(x)\, dx +\int_{b}^{c}f(x)\, dx = \int_{a}^{c}f(x)\, dx$$

For example, you can rearrange it like this:

$$\int_{a}^{c}f(x)\, dx -\int_{a}^{b}f(x)\, dx = \int_{b}^{c}f(x)\, dx$$

Evaluating With Geometry

Some definite integrals are easiest to evaluate by recognizing the shape under the curve. If $f(x)$ is linear or made of straight pieces, the region is often a triangle, rectangle, or trapezoid, and you can use familiar area formulas. A semicircle works the same way when the curve is the top half of a circle. Remember to count any area below the x-axis as negative.

The definition of the definite integral still works for functions that have removable or jump discontinuities, so you can integrate piecewise-defined functions as long as the breaks are isolated points.

How to Use This on the AP Calculus Exam

Problem Solving

When a problem gives you several integral values and asks for a new one, line up the intervals and decide which property connects them. Most of these reduce to the adjacent-interval rule plus the constant multiple and sum/difference rules.

Walkthrough. If $\int_{1}^{5}f(x) \, dx = 3$, $\int_{5}^{10}f(x) \, dx = 5$, and $\int_{10}^{13}f(x) \, dx = -7$, find $\int_{1}^{13}f(x) \, dx$.

The target interval runs from 1 to 13, and the given pieces line up end to end, so add them.

$$\int_{1}^{13}f(x) \, dx = \int_{1}^{5}f(x) \, dx + \int_{5}^{10}f(x) \, dx + \int_{10}^{13}f(x) \, dx$$ $$\int_{1}^{13}f(x) \, dx = 3+5-7$$ $$\int_{1}^{13}f(x) \, dx = 1$$

Example 2. If $\int_{1}^{10}g(x) \, dx = 12$ and $\int_{6}^{10}g(x) \, dx = -7$, find $\int_{1}^{6}g(x) \, dx$.

Here you are given the larger interval and one piece of it, so subtract.

$$\int_{1}^{6}g(x)\, dx = \int_{1}^{10}g(x)\, dx - \int_{6}^{10}g(x)\, dx$$ $$\int_{1}^{6}g(x)\, dx = 12-(-7)$$ $$\int_{1}^{6}g(x)\, dx = 19$$

Common Trap

Example 3. If $\int_{1}^{10}f(x)\, dx = 15$ and $\int_{10}^{6}f(x)\, dx = 12$, find $\int_{1}^{6}f(x)\, dx$.

Watch the limits. One integral runs from 10 to 6, so rewrite it with the smaller number on the bottom by flipping the sign (the reversing-limits rule).

$$\int_{6}^{10}f(x)\, dx = -12$$

Now combine the pieces.

$$\int_{1}^{6}f(x)\, dx = \int_{1}^{10}f(x)\, dx \, - \int_{6}^{10}f(x)\, dx$$ $$\int_{1}^{6}f(x)\, dx = 15 - (-12)$$ $$\int_{1}^{6}f(x)\, dx = 27$$

Practice Problems

Try each one before checking the solution.

1. If $\int_{1}^{19}h(x)\, dx = 17$, $\int_{6}^{19}h(x)\, dx = 2$, and $\int_{4}^{6}h(x)\, dx = -3$, find $\int_{1}^{4}h(x)\, dx$.
2. If $\int_{1}^{8}f(x)\, dx = -8$ and $\int_{8}^{30}f(x)\, dx = 200$, find $\int_{1}^{30}f(x)\, dx$.
3. If $\int_{1}^{4}g(x)\, dx = -8$ and $\int_{4}^{2}g(x)\, dx = 3$, find $\int_{1}^{2}g(x)\, dx$.

Question 1 Solution

$$\int_{1}^{4}h(x)\, dx = \int_{1}^{19}h(x)\, dx - \int_{6}^{19}h(x)\, dx - \int_{4}^{6}h(x)\, dx$$ $$\int_{1}^{4}h(x)\, dx = 17-2-(-3)$$ $$\int_{1}^{4}h(x)\, dx = 18$$

Question 2 Solution

$$\int_{1}^{30}f(x) \, dx = \int_{1}^{8}f(x) \, dx + \int_{8}^{30}f(x) \, dx$$ $$\int_{1}^{30}f(x)\, dx = -8+200$$ $$\int_{1}^{30}f(x)\, dx = 192$$

Question 3 Solution

Since $\int_{4}^{2}g(x)\, dx = 3$ runs from 4 to 2, flip it: $\int_{2}^{4}g(x)\, dx = -3$.

$$\int_{1}^{2}g(x)\, dx = \int_{1}^{4}g(x)\, dx - \int_{2}^{4}g(x)\, dx$$ $$\int_{1}^{2}g(x)\, dx = -8-(-3)$$ $$\int_{1}^{2}g(x)\, dx= -5$$

Common Misconceptions

• A definite integral is not always positive. Area below the x-axis counts as negative, so the result can be negative or zero even when the curve covers real area.
• Reversing limits is not the same as leaving them alone. Swapping the upper and lower limits flips the sign of the whole integral.
• The constant multiple rule only lets you pull out constants, not variables. You cannot move a factor like $x$ outside the integral.
• The sum and difference rule applies to functions being added or subtracted, not multiplied or divided. There is no property that splits the integral of a product into a product of integrals.
• Splitting at a point only works when the point lies in the interval and you keep the order consistent. Mixing up which interval is the whole and which is the piece leads to sign errors.
• Equal limits give 0, not the value of $f$. The integral $\int_{a}^{a} f(x)\, dx$ is always 0 because there is no width.

Related AP Calculus Guides

• Unit 6 Overview: Integration and Accumulation of Change
• 6.11 Integrating Using Integration by Parts
• 6.1 Integration and Accumulation of Change
• 6.12 Integrating Using Linear Partial Fractions
• 6.3 Riemann Sums, Summation Notation, and Definite Integral Notation
• 6.4 The Fundamental Theorem of Calculus and Accumulation Functions