Introduction
A definite integral records accumulated signed contribution across an interval. If every function value is multiplied by the same constant, each contribution to that accumulation is scaled by that constant. The resulting rule is simple, but using it correctly requires attention to the constant, the interval, and the direction of integration.
This tutorial develops the scalar multiple rule for definite integrals. It explains why the rule holds from the Riemann-sum definition and shows how to evaluate an integral when the integral of the unscaled function is already known.
The Scalar Multiple Rule
Let \(f\) be a real-valued Riemann integrable function on \([a,b]\), where \(a<b\), and let \(c\) be a real constant. The function \(cf\) is defined by \((cf)(x)=c f(x)\) for every \(x\) in the interval. The scalar multiple rule states that its definite integral is the same constant times the integral of \(f\).
The word constant matters: \(c\) does not vary with \(x\) over the interval. A parameter can serve as \(c\) if it is held fixed while integrating with respect to \(x\). A factor that changes with \(x\) is part of the integrand, not a scalar factor covered by this rule.
The rule applies to positive, negative, and zero values of \(c\). A negative factor reverses the sign of the integral, while a zero factor gives a zero integral. In each case, the factor multiplies the value of the original integral; it does not change the integration bounds.
Why the Rule Works
The definite Riemann integral is defined as a limit of tagged Riemann sums. For a partition \(P\) of \([a,b]\), write its subinterval widths as \(\Delta x_i\), and choose a tag \(\xi_i\) in each subinterval. The tagged sum for \(f\) is \(\sum_{i=1}^{n} f(\xi_i)\Delta x_i\). The integral exists when these sums approach a single value as the partition mesh shrinks to zero.
For the scaled function \(cf\), use the same partition and the same tags. Each term of the new sum is multiplied by \(c\). Therefore the whole sum is multiplied by \(c\), not merely one of its terms.
Since the sums for \(f\) approach \(\int_a^b f(x)\,dx\), the sums for \(cf\) approach \(c\int_a^b f(x)\,dx\). Multiplication by a fixed real number preserves convergence, including when that number is negative or zero. Thus \(cf\) is Riemann integrable, and its integral has the claimed value.
Using a Known Integral
In many applications, the main work is not finding the original integral. Instead, a value for \(\int_a^b f(x)\,dx\) is supplied, and the goal is to find the integral of a scaled version of \(f\). The procedure is direct: identify the scalar, identify the known integral on the same interval and in the same orientation, and multiply.
Rewrite the integrand as \(c f(x)\), with \(c\) independent of \(x\).
Check that the known value uses the same function and bounds, in the same order.
Multiply the known integral by \(c\), then substitute the known value into the rule to verify the result.
Worked Example: A Negative Scalar
Suppose \(f\) is integrable on \([-1,2]\) and \(\int_{-1}^{2} f(x)\,dx=\frac{5}{2}\). Evaluate \(\int_{-1}^{2} -4f(x)\,dx\).
The scalar is \(c=-4\), and the known integral is over exactly the same interval. Applying the rule gives
Check by substituting the supplied integral value into the rule: the scaled integral must equal \(-4\) times \(\frac{5}{2}\), which is \(-10\). The negative sign is required because the scalar is negative.
Reversed Bounds and Scalar Factors
The scalar multiple rule is compatible with signed integral orientation. For \(a>b\), the oriented integral is defined by \(\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx\). For equal bounds, \(\int_a^a f(x)\,dx=0\). Consequently, the same scalar rule holds for any order of bounds, provided the integrals are defined.
For reversed bounds, either apply the scalar rule directly to the integral as written, or first reverse the bounds and change the sign. Do not do both changes without accounting for their signs. The following example uses a known value with reversed bounds.
Worked Example: A Scaled Integral with Reversed Bounds
Suppose \(g\) is integrable on \([-1,2]\) and \(\int_{2}^{-1} g(x)\,dx=-\frac{3}{4}\). Evaluate \(\int_{2}^{-1} 6g(x)\,dx\).
The known value and the required integral both have bounds from \(2\) to \(-1\), so no bound reversal is needed. The scalar is \(6\):
Check the orientation explicitly. Reversing the bounds in the supplied value gives \(\int_{-1}^{2}g(x)\,dx=\frac{3}{4}\). The scalar rule then gives \(\int_{-1}^{2}6g(x)\,dx=6(\frac{3}{4})=\frac{9}{2}\). Reversing those bounds changes the sign, so \(\int_{2}^{-1}6g(x)\,dx=-\frac{9}{2}\), agreeing with the result.
Finding an Unknown Scalar
The rule can also be used in reverse. If an integral of a scaled function is known, and the integral of the original function is known, the scalar can be found by solving the resulting equation. Division is valid only when the known integral of the original function is nonzero. If that integral is zero, the rule gives a scaled integral of zero for every value of the scalar, so this information alone cannot determine the scalar.
Worked Example: Determine the Scalar
Suppose \(h\) is integrable on \([0,3]\), \(\int_0^3 h(x)\,dx=-2\), and \(\int_0^3 k h(x)\,dx=7\), where \(k\) is a real constant. Find \(k\).
By the scalar multiple rule, the second integral equals \(k\) times the first. Substitute both given values:
Divide both sides by \(-2\), which is nonzero:
Check by substitution: \((-\frac{7}{2})(-2)=\frac{14}{2}=7\), which matches the given scaled integral. Thus \(k=-\frac{7}{2}\).
A Common Factor Mistake
A frequent error is to divide by the scalar when the task is to evaluate the integral of the scaled function. For example, if \(\int_0^1 p(x)\,dx=-2\), a mistaken calculation might say
This is incorrect. The scalar multiple rule says to multiply the original integral by \(-3\), not divide by \(-3\). The corrected calculation is
The division step belongs to a different situation: if the value of the scaled integral is given and the original integral is known, one may solve for the scalar by dividing by the original integral, provided that integral is nonzero.
Scope and Conditions
Before using the rule, confirm that the original function is integrable on the interval. The theorem guarantees that a constant multiple of a Riemann integrable function is also Riemann integrable. It does not require the function to be positive: integrals are signed, and negative function values contribute negatively to the integral.
Also check that the factor really is constant with respect to the variable of integration. For instance, when integrating with respect to \(x\), a fixed parameter \(r\) can be taken outside the integral. A factor that depends on \(x\) cannot be pulled out by this rule, because it changes from one point of the interval to another.
The rule is useful because it separates scaling from accumulation. Once the integral of a function is known, multiplying the function by a constant requires no new partition, geometric calculation, or evaluation of the original integral. The constant acts on the accumulated value in exactly the same way it acts on each Riemann-sum contribution.
Check Your Understanding
Use the scalar multiple rule and show the key arithmetic in each answer.
- If \(\int_1^4 q(x)\,dx=3\), what is \(\int_1^4 -5q(x)\,dx\)?
- If \(\int_{-2}^{1} r(x)\,dx=-\frac{2}{3}\), what is \(\int_{1}^{-2} 3r(x)\,dx\)?
- If \(\int_0^2 s(x)\,dx=4\) and \(\int_0^2 c s(x)\,dx=-10\), find \(c\).
- Why must a factor be constant with respect to the integration variable to use this rule?
- If \(\int_a^b f(x)\,dx=0\), can the value of \(\int_a^b c f(x)\,dx\) determine \(c\)? Explain.
