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Calculus BC · Unit 6 · 6.6
AP Calculus · Integration and Accumulation of Change
Applying Properties of Definite Integrals
Split, glue and factor without evaluating anything — and know exactly where the properties stop.
key idea$\int_a^b=\int_a^c+\int_c^b;\qquad \int_b^a=-\int_a^b$
The rule
These properties let you take an integral apart, glue pieces together and move constants out — all without evaluating anything. Exam questions supply a few integral values and ask you to assemble a new one, so the properties are the entire method.
| Property | Statement | Reading |
|---|
| Linearity | $\displaystyle\int_a^b\left[f\pm g\right]=\int_a^b f\pm\int_a^b g$ | split at $+$ and $-$ |
| Constant multiple | $\displaystyle\int_a^b kf=k\int_a^b f$ | a constant factor comes out |
| Additivity | $\displaystyle\int_a^b=\int_a^c+\int_c^b$ | split the interval anywhere |
| Reversal | $\displaystyle\int_b^a f=-\int_a^b f$ | swapping limits flips the sign |
| Zero width | $\displaystyle\int_a^a f=0$ | no interval, no area |
| Comparison | $f\le g\Rightarrow\displaystyle\int_a^b f\le\int_a^b g$ | a taller curve encloses more |
Additivity does not require $c$ to lie between $a$ and $b$. With reversal in hand, the splitting point can sit outside the interval.
Pattern recognition
Splitssums, differences, constant multiples, intervals
Does NOT splitproducts, quotients, compositions
Flips signswapping the two limits
what does and does not distribute
- Products and quotients do not split — $\int fg\neq\int f\int g$.
- Only constants come out front; a function factor cannot.
- Reversal is what makes an out-of-order splitting point usable.
- Assemble the target from the pieces you are given — do not evaluate.
AdditivitySplitting the interval at any point $c$ — the integrand is untouched.
LinearitySplitting the integrand at $+$ or $-$ — the interval is untouched.
ReversalSwapping the limits negates the value. It is what makes an out-of-range splitting point legal.
ComparisonIf one curve sits below another throughout, its integral is no larger.
Worked examples
EX 1
Assemble from given values
$\int_0^5$ and $\int_0^2$Given $\displaystyle\int_{0}^{5}f=12$ and $\displaystyle\int_{0}^{2}f=3$, find $\displaystyle\int_{2}^{5}f$.
1
Apply additivity
$\displaystyle\int_{0}^{5}f=\int_{0}^{2}f+\int_{2}^{5}f$.
2
Substitute the known values
$12=3+\displaystyle\int_{2}^{5}f$.
3
Solve
$\displaystyle\int_{2}^{5}f=9$.
4
Note what was not needed
No formula for $f$, and no antiderivative. The property alone answers it.
EX 2
Combine several properties
linearity plus constantsGiven $\displaystyle\int_{1}^{4}f=6$ and $\displaystyle\int_{1}^{4}g=-2$, find $\displaystyle\int_{1}^{4}\left[3f-2g\right]$.
1
Split the sum
$\displaystyle\int_{1}^{4}3f-\int_{1}^{4}2g$.
2
Pull out the constants
$3\displaystyle\int_{1}^{4}f-2\int_{1}^{4}g$.
5
Watch the double negative
$-2\times(-2)=+4$. A negative integral value meeting a negative coefficient is where this goes wrong.
EX 3
A splitting point outside the interval
reversal in actionGiven $\displaystyle\int_{0}^{7}f=10$ and $\displaystyle\int_{0}^{9}f=4$, find $\displaystyle\int_{7}^{9}f$.
1
Write the additivity relation
$\displaystyle\int_{0}^{9}f=\int_{0}^{7}f+\int_{7}^{9}f$.
2
Substitute
$4=10+\displaystyle\int_{7}^{9}f$.
3
Solve
$\displaystyle\int_{7}^{9}f=-6$.
4
Sanity-check the sign
A negative value is fine — it means $f$ sits mostly below the axis on $[7,9]$.
5
Note the structure
The total over the longer interval is smaller, which is only possible if the extra stretch contributes negatively.
EX 4
What does not distribute
the trapA student writes $\displaystyle\int_{0}^{2}x\sin x\,dx=\left(\int_{0}^{2}x\,dx\right)\left(\int_{0}^{2}\sin x\,dx\right)$. Evaluate the claim.
1
Name the assumed property
That integration distributes over a product. No such property exists.
2
Test it on something simple
$\displaystyle\int_{0}^{1}x\cdot x\,dx=\int_{0}^{1}x^{2}\,dx=\tfrac13$.
3
Compare with the claimed method
$\left(\int_{0}^{1}x\,dx\right)^{2}=\left(\tfrac12\right)^{2}=\tfrac14$.
4
Conclude
$\tfrac13\neq\tfrac14$, so the property is false.
5
What actually works
Linearity covers sums, differences and constant multiples — nothing more. A product needs a technique such as substitution or parts.
Watch out
Linearity does not extend to products
$\int fg\neq\left(\int f\right)\left(\int g\right)$, and the same for quotients. Example 4 disproves it in two lines.
Only constants factor out
$\int kf=k\int f$ requires $k$ to be a number. A function factor stays inside.
Reversing limits flips the sign
$\int_b^a=-\int_a^b$. This is what lets you use a splitting point outside the interval.
Negative values are legitimate
A definite integral is signed area. A negative result means the curve spends more of the interval below the axis.