Combining Integrands
A definite integral can be known even when the integrand is written in a new way. If the new integrand is a sum or difference of functions whose integrals are already available, linearity lets us combine those values directly. This is useful in calculations with many terms: instead of evaluating a new integral from scratch, we keep track of each function’s coefficient and integral.
The key requirement is that the functions being combined are integrable on the same interval. Under that condition, integration respects addition and subtraction, just as it respects scalar multiplication.
The same rules hold with any real coefficients. For example, \(\int_a^b (2f-3g)\,dx=2\int_a^b f\,dx-3\int_a^b g\,dx\). The coefficient is applied to the integral of its function, with its sign unchanged.
Using the Rules with Known Values
When the integral values are given, the calculation has two stages. First, translate each term of the integrand into the corresponding coefficient times a known integral. Then simplify the resulting numerical expression, taking particular care with negative values.
Worked Example: A Sum with Positive and Negative Integral Values
Suppose \(f\), \(g\), and \(h\) are Riemann integrable on \([a,b]\), and \(\int_a^b f(x)\,dx=\frac{7}{3}\), \(\int_a^b g(x)\,dx=-\frac{2}{3}\), and \(\int_a^b h(x)\,dx=5\). Find \(\int_a^b (2f(x)-3g(x)+h(x))\,dx\).
All three functions are integrable on the same interval, so linearity applies term by term:
Check: The \(g\)-term contributes \((-3)(-2/3)=2\), which is positive. Adding the three contributions gives \(14/3+2+5=35/3\), as calculated.
This example illustrates an important sign principle: subtracting a function whose integral is negative adds a positive amount. The minus sign in front of the function and the negative sign in its integral both matter.
Expressions with Several Terms
Linearity extends to any finite number of terms. If \(f_1,\ldots,f_n\) are Riemann integrable on \([a,b]\) and \(c_1,\ldots,c_n\) are real numbers, then their linear combination is integrable and its integral is the corresponding linear combination of the individual integrals.
Here, a linear combination means a finite sum of functions multiplied by constant real coefficients. A difference is included because subtracting a function is the same as multiplying it by \(-1\) and adding it.
Why the Finite-Term Rule Follows
The finite-term formula follows by repeatedly applying the two-function sum rule and the scalar-multiple rule. The key is that after combining some terms, their sum is still integrable, so the two-function rule can be applied again.
For \(n=1\), the formula is \(\int_a^b c_1f_1(x)\,dx=c_1\int_a^b f_1(x)\,dx\), which is the scalar-multiple rule.
Write \(F_n(x)=\sum_{k=1}^{n}c_kf_k(x)\). By the assumption, \(F_n\) is integrable and \(\int_a^b F_n(x)\,dx=\sum_{k=1}^{n}c_k\int_a^b f_k(x)\,dx\).
Since \(F_n\) and \(f_{n+1}\) are integrable on \([a,b]\), their sum \(F_n+c_{n+1}f_{n+1}\) is integrable. Apply the sum rule and then the scalar-multiple rule.
The integral is \(\int_a^b F_n\,dx+c_{n+1}\int_a^b f_{n+1}\,dx\). Substituting the assumed formula gives the sum of all \(n+1\) coefficient-weighted integrals.
In equations, the last step is \(\int_a^b(F_n+c_{n+1}f_{n+1})\,dx =\sum_{k=1}^{n}c_k\int_a^b f_k\,dx+c_{n+1}\int_a^b f_{n+1}\,dx =\sum_{k=1}^{n+1}c_k\int_a^b f_k\,dx\). Thus the formula holds for one term, and whenever it holds for \(n\) terms it holds for \(n+1\) terms. By induction, it holds for every positive integer \(n\).
This argument explains how many-term calculations work: each term contributes independently, scaled by its coefficient. The underlying two-function sum rule and scalar-multiple rule are the essential ingredients.
Worked Example: Simplifying Before Substituting Values
Suppose \(f\), \(g\), and \(h\) are integrable on \([a,b]\), with \(\int_a^b f(x)\,dx=4\), \(\int_a^b g(x)\,dx=-1\), and \(\int_a^b h(x)\,dx=2\). Find \(\int_a^b ((f(x)+g(x))-(2g(x)-h(x)))\,dx\).
First remove the parentheses in the integrand. Subtracting the entire expression \(2g-h\) changes the signs of both its terms:
Now apply linearity and substitute the known values:
Check: The original expression contributes \(4+(-1)-2(-1)+2=4-1+2+2=7\). This agrees with the result after simplifying the integrand.
Solving for an Unknown Integral
Linearity can also be used in reverse. If an integral of a combination is known, and all but one of the component integrals are known, rearrange the linear equation to find the remaining value. This is ordinary algebra applied to the integral values.
Worked Example: Finding a Missing Integral
Suppose \(p\), \(q\), and \(r\) are integrable on \([a,b]\), and \(\int_a^b p(x)\,dx=2\), \(\int_a^b q(x)\,dx=-1\), and \(\int_a^b (3p(x)-2q(x)+r(x))\,dx=11\). Find \(\int_a^b r(x)\,dx\).
Set \(R=\int_a^b r(x)\,dx\). By linearity, the given combined integral satisfies
Subtract \(8\) from both sides to obtain \(R=3\). Therefore \(\int_a^b r(x)\,dx=3\).
Check: Substituting this value into the original linear combination gives \(3(2)-2(-1)+3=6+2+3=11\), which matches the given integral.
Common Sign and Scope Errors
A frequent mistake is to treat the integral of a difference as though the second integral were always positive. For example, if \(\int_a^b f(x)\,dx=5\) and \(\int_a^b g(x)\,dx=-2\), the incorrect calculation \(\int_a^b(f-g)\,dx=5-2=3\) has replaced the value of \(\int g\) by its magnitude.
The correct calculation keeps the signed integral value:
Correction: Subtract the integral value, including its sign. Do not subtract the absolute value unless the problem explicitly asks for one.
Another limitation is that linearity cannot be applied without the required integrability and common interval. The rule above concerns functions that are all Riemann integrable on the same interval \([a,b]\). If a function is not integrable there, or if the given values refer to different intervals, this theorem alone does not justify combining those values. Check the conditions before using the formula.
Why This Property Matters
Many integrands are easier to understand as combinations of simpler functions. Linearity allows a calculation to be organized around the functions whose integrals are known, rather than around the expanded expression as a whole. It also makes algebraic rearrangement useful: terms can be grouped, coefficients collected, and unknown component integrals isolated.
For a reliable calculation, use this sequence:
- Confirm that every function is integrable on the same interval.
- Expand or simplify the integrand carefully, distributing any minus signs.
- Attach each coefficient to the integral of its corresponding function.
- Substitute the known signed values and simplify the arithmetic.
- Check the signs by comparing the contribution of each term with the original expression.
These steps keep the meaning of the integral visible throughout the computation. In particular, a definite integral is a signed value, so a negative integral remains negative when substituted into a linearity formula.
Check Your Understanding
Use the same-interval condition and track all coefficients and signs in each problem.
- If \(\int_a^b f(x)\,dx=3\) and \(\int_a^b g(x)\,dx=-4\), what is \(\int_a^b(2f(x)+g(x))\,dx\)?
- Write the integral of \(u(x)-3v(x)+2w(x)\) as a combination of the three individual integrals.
- If \(\int_a^b u(x)\,dx=1\), \(\int_a^b v(x)\,dx=2\), and \(\int_a^b w(x)\,dx=-3\), evaluate \(\int_a^b((u(x)-v(x))-(v(x)-w(x)))\,dx\).
- What integrability and interval conditions are needed to apply the finite-term linearity formula?
- If \(\int_a^b(2p(x)+q(x))\,dx=9\) and \(\int_a^b p(x)\,dx=3\), what is \(\int_a^b q(x)\,dx\)?
