From Rate Bounds to Accumulated Change
A rate describes how quickly a quantity changes. If the rate stays between two known values throughout an interval, then the accumulated change over that interval must also stay between corresponding bounds. The interval’s length determines the scale of those bounds.
The direction in which the interval is traversed matters. Accumulating from a smaller input to a larger one uses the usual order of the lower and upper bounds. Reversing the endpoints changes the sign of the integral, so it also reverses which rate bound gives the lower accumulated-change bound.
This lesson turns that observation into precise inequalities. The rate itself need not be positive: the lower and upper bounds may be negative or may straddle zero.
Rate Bounds on a Forward Interval
Let \(f\) be integrable on a closed interval \([a,b]\), where \(a<b\). Suppose there are real numbers \(m\) and \(M\) such that \(m\le f(t)\le M\) for every \(t\in[a,b]\). The accumulated change is the definite integral of the rate over the interval.
The factor \(b-a\) is the interval’s positive length. The result does not require \(m\) or \(M\) to be positive. If \(m\) is negative and \(M\) is positive, the theorem still applies and may only place the integral between a negative and a positive number.
The inequalities give bounds, not necessarily the exact integral. They become equalities at the corresponding ends if the rate is constantly equal to \(m\) or \(M\), respectively. In other cases, the actual accumulated change can lie strictly between the two bounds.
Why the Forward Bounds Hold
The key fact is that the integral of a nonnegative integrable function over a forward interval is nonnegative. The pointwise assumptions give two nonnegative functions: \(f(t)-m\) and \(M-f(t)\). Integrating each one gives the desired inequalities.
Since \(f(t)-m\ge 0\) throughout \([a,b]\), its integral is nonnegative.
\(\int_a^b(f(t)-m)\,dt=\int_a^b f(t)\,dt-m(b-a)\), so the accumulated change is at least \(m(b-a)\).
Since \(M-f(t)\ge 0\), its integral is nonnegative, which gives \(\int_a^b f(t)\,dt\le M(b-a)\).
Together, these steps prove the theorem: \(m(b-a)\le\int_a^b f(t)\,dt\le M(b-a)\). The proof relies on the order property and linearity of the definite integral, and on the interval being traversed from \(a\) to \(b\), with \(a<b\).
Worked Example: Bounding a Positive Rate
Worked Example: Bounding a Positive Rate
Let \(f(t)=2t+1\) on \([1,3]\). Find a lower and upper bound for \(\int_1^3 f(t)\,dt\) from the rate bounds, then compute the integral to check them.
For \(1\le t\le 3\), multiplying by \(2\) and adding \(1\) gives \(3\le 2t+1\le 7\). Thus \(m=3\), \(M=7\), and the interval length is \(3-1=2\). The theorem gives
To check, an antiderivative of \(2t+1\) is \(t^2+t\), since \(\frac{d}{dt}(t^2+t)=2t+1\). Therefore
The computed value satisfies \(6\le 10\le 14\), as required.
Reversing the Interval
For any integrable \(f\) and \(u<v\), reversing the endpoints changes the sign of the integral: \(\int_v^u f(t)\,dt=-\int_u^v f(t)\,dt\). This orientation rule is essential when applying rate bounds to a reversed interval.
Suppose \(m\le f(t)\le M\) on \([v,u]\), where \(v<u\). First apply the forward-interval theorem:
Negating every part of an inequality reverses its order. Using the orientation rule therefore gives
Notice that the upper rate bound \(M\) supplies the lower bound on the reversed integral, while \(m\) supplies its upper bound. This is not a new assumption about the rate; it follows from reversing the endpoints and negating the forward inequality.
Worked Example: Checking the Reversed Bounds
Worked Example: Checking the Reversed Bounds
Use \(f(t)=2t+1\) on \([1,3]\), where \(3\le f(t)\le 7\), to bound \(\int_3^1 f(t)\,dt\), and verify the result.
Here the reversed endpoints are \(u=3\) and \(v=1\), so \(u-v=2\), \(m=3\), and \(M=7\). The reversed-interval formula gives
By the orientation rule and the forward calculation, \(\int_3^1(2t+1)\,dt=-\int_1^3(2t+1)\,dt=-10\). Checking the inequalities gives \(-14\le -10\le -6\), which is true.
Rate Bounds That Are Negative
Bounds on a negative rate lead to negative accumulated-change bounds on a forward interval. For example, if \(-4\le f(t)\le -1\) across an interval of length \(3\), then multiplying the rate bounds by the positive interval length gives \(-12\le\int f\le -3\). The accumulation is negative: the quantity decreases overall. The inequalities still have their usual order.
Worked Example: A Rate Bounded Below Zero
Let \(f(t)=1-t\) on \([2,5]\). Bound its accumulated change using rate bounds, then compute and check the integral.
For \(2\le t\le5\), we have \(1-5\le1-t\le1-2\), so \(-4\le f(t)\le-1\). The interval length is \(5-2=3\). Thus
An antiderivative of \(1-t\) is \(t-\frac{t^2}{2}\), because its derivative is \(1-t\). Evaluating at both endpoints gives
Since \(-12\le-\frac{15}{2}\le-3\), the exact accumulated change satisfies the bounds.
Applying the Bounds to Accumulation Functions
An accumulation function records accumulated change from a fixed basepoint \(c\). If \(A(x)=\int_c^x f(t)\,dt\), then the interval-increment identity gives \(A(y)-A(x)=\int_x^y f(t)\,dt\). The rate bounds therefore constrain the difference between two values of \(A\).
For \(x<y\), if \(m\le f(t)\le M\) on \([x,y]\), then
If instead \(y<x\), the bounds must use the reversed-interval form. In either case, a bound on the integrand constrains the accumulated change over the interval between the inputs; it does not by itself specify the value of \(A\) at either endpoint.
A Common Sign Error
A frequent mistake is to reverse the endpoints but keep the forward-interval bounds unchanged. For example, suppose \(1\le f(t)\le4\) on \([2,5]\). The forward bounds are \(3\le\int_2^5 f(t)\,dt\le12\). A mistaken calculation might claim that \(\int_5^2 f(t)\,dt\) lies between \(1(5-2)=3\) and \(4(5-2)=12\). But the reversed integral is negative, so those positive bounds cannot be correct.
Instead, use the reversed formula, with \(u=5\), \(v=2\), \(m=1\), \(M=4\):
The sign and order are both consistent with \(\int_5^2 f(t)\,dt=-\int_2^5 f(t)\,dt\). When reversing endpoints, first write the orientation identity or directly apply the reversed-interval formula; do not simply reuse the forward bounds.
What the Bounds Tell You
Rate bounds are useful when the exact rate function or its exact integral is unavailable, but reliable lower and upper limits are known. Multiplying those limits by interval length gives an interval for the possible accumulated change. The calculation also indicates whether the change must be positive, must be negative, or could have either sign.
These conclusions concern the net change, the signed integral of the rate. When the rate can take both positive and negative values, these bounds do not directly describe the total amount accumulated without regard to sign. They are bounds on the definite integral, with cancellation included.
Before using the result, check three things: the rate is integrable on the interval, the stated bounds hold throughout it, and the endpoints are in the direction used by the formula. Keeping those conditions and the interval orientation visible makes the inequalities straightforward to apply.
Check Your Understanding
Use the assumptions stated in each question, and keep the interval direction explicit.
- If \(2\le f(t)\le6\) on \([1,4]\), what bounds follow for \(\int_1^4 f(t)\,dt\)?
- If \(-3\le f(t)\le2\) on \([0,5]\), what bounds follow for \(\int_5^0 f(t)\,dt\)?
- For \(f(t)=3-t\) on \([1,3]\), find rate bounds, use them to bound the integral, and compute the exact integral to check.
- Why does reversing the endpoints reverse the order of the accumulated-change bounds?
- If \(A(x)=\int_c^x f(t)\,dt\), what bounds constrain \(A(y)-A(x)\) when \(x<y\) and \(m\le f(t)\le M\) on \([x,y]\)?
