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Proof Strategy · Tutorial 956 of 1000

The Integration Strategy

Use positivity, comparison, and average values of integrals to turn local information about a function into rigorous global conclusions.

Advanced 9 min read

What You'll Learn

  • Rewrite function differences as integrals using the Fundamental Theorem of Calculus
  • Use the sign of an integrand to prove inequalities
  • Prove that a positive continuous function has a positive integral
  • Apply the integral mean value theorem to identify an attained average value
  • Estimate absolute values of integrals without overlooking cancellation
  • Check interval orientation and endpoint hypotheses in integral arguments

Turn a Difference Into an Integral

The Differentiation Strategy used the Mean Value Theorem to compare function values by studying a derivative. Integration offers a complementary approach: when a difference of function values can be written as an integral, the sign or size of the integrand can control that difference. The central task is to choose the interval and integrand so that the desired conclusion follows from a property of the integral.

For example, if \(F\) is continuously differentiable on \([a,b]\), the Fundamental Theorem of Calculus gives

$$ F(b)-F(a)=\int_a^b F'(t)\,dt. $$

Thus a comparison between \(F(b)\) and \(F(a)\) can be proved by examining \(F'\) throughout the interval. This is closely related to differentiation, but the proof now accumulates information across the whole interval rather than relying on one intermediate derivative value. We will use the standard properties of the Riemann integral for continuous functions: linearity, additivity over adjacent intervals, and preservation of pointwise inequalities.

The orientation of the interval matters. The expression \(\int_a^b f(t)\,dt\) is nonnegative when \(f\geq0\) and \(a<b\). If the limits are reversed, then \(\int_b^a f(t)\,dt=-\int_a^b f(t)\,dt\). In sign arguments, first check that the interval is oriented in the direction needed.

Positivity and Integral Bounds

A pointwise inequality can be integrated: if \(f(t)\leq g(t)\) on \([a,b]\), where \(a<b\), then \(\int_a^b f(t)\,dt\leq\int_a^b g(t)\,dt\). In particular, nonnegative integrands have nonnegative integrals. The next result adds a useful strictness principle: a continuous nonnegative function cannot be positive at even one point while having integral zero.

Theorem (Strict Positivity of the Integral): Let \(a<b\), and let \(f:[a,b]\to\mathbb{R}\) be continuous with \(f(t)\geq0\) for every \(t\in[a,b]\). If \(f(x_0)>0\) for some \(x_0\in[a,b]\), then \(\int_a^b f(t)\,dt>0\). Consequently, \(\int_a^b f(t)\,dt=0\) if and only if \(f\) is identically zero on \([a,b]\).

Proof. Suppose \(f(x_0)>0\), and set \(m=f(x_0)/2>0\). By continuity at \(x_0\), there is a \(\delta>0\) such that \(f(t)>m\) whenever \(t\in[a,b]\) and \(|t-x_0|<\delta\). The intersection of this neighborhood with \([a,b]\) contains an interval of positive length, including when \(x_0\) is an endpoint: since \(a<b\), there is a subinterval \(J\subseteq[a,b]\) of positive length within that neighborhood. On \(J\), \(f(t)\geq m\), and on the rest of \([a,b]\), \(f(t)\geq0\). Preservation of inequalities and additivity of the integral therefore give

$$ \int_a^b f(t)\,dt\geq m\,\operatorname{length}(J)>0. $$

This proves the first assertion. If \(f\) is identically zero, its integral is zero. Conversely, if its integral is zero, the first assertion rules out \(f(x_0)>0\) at every point; together with \(f\geq0\), this means \(f\) is identically zero. \(\square\)

This theorem depends on continuity. A function that is nonzero at a single point but zero everywhere else has integral zero in the Riemann sense, but it is not continuous at that point. Continuity ensures that a positive value persists across an interval of positive length.

Theorem (Absolute-Value Bound for an Integral): If \(f:[a,b]\to\mathbb{R}\) is continuous and \(a<b\), then $$ \left|\int_a^b f(t)\,dt\right|\leq\int_a^b |f(t)|\,dt. $$

Proof. For every \(t\), \(-|f(t)|\leq f(t)\leq |f(t)|\). Integrating both inequalities over \([a,b]\) gives

$$ -\int_a^b |f(t)|\,dt \leq \int_a^b f(t)\,dt \leq \int_a^b |f(t)|\,dt. $$

The outer quantity \(\int_a^b |f(t)|\,dt\) is nonnegative. A real number between its negative and positive is at most that quantity in absolute value, which proves the result. \(\square\)

The bound is useful when the integrand changes sign: the integral may exhibit cancellation, so an estimate based only on the sign of \(f\) may be unavailable. The integral of \(|f|\) controls the total size without that cancellation. A simpler consequence is that if \(m\leq f(t)\leq M\) on \([a,b]\), then

$$ m(b-a)\leq\int_a^b f(t)\,dt\leq M(b-a). $$

This follows by integrating the constant bounds. Dividing by \(b-a>0\) gives bounds on the average value of \(f\).

The Average Value Is Actually Attained

An integral average is more than a number lying between the minimum and maximum of a function. For continuous functions, it is a value the function takes somewhere in the interval. This fact can turn an integral identity into a statement about an intermediate point.

Definition (Integral Average): If \(a<b\) and \(f:[a,b]\to\mathbb{R}\) is continuous, the average value of \(f\) on \([a,b]\) is $$ \frac{1}{b-a}\int_a^b f(t)\,dt. $$
Theorem (Integral Mean Value Theorem): Let \(a<b\), and let \(f:[a,b]\to\mathbb{R}\) be continuous. There is a \(c\in[a,b]\) such that $$ \int_a^b f(t)\,dt=f(c)(b-a). $$

Proof. By the Extreme Value Theorem, \(f\) attains a minimum \(m\) and a maximum \(M\) on \([a,b]\). Thus \(m\leq f(t)\leq M\) throughout the interval. Integrating and dividing by the positive number \(b-a\) shows

$$ m\leq\frac{1}{b-a}\int_a^b f(t)\,dt\leq M. $$

Because \(f\) is continuous on the interval, the Intermediate Value Theorem says that it takes every value between its minimum and maximum. In particular, for some \(c\in[a,b]\), $$ f(c)=\frac{1}{b-a}\int_a^b f(t)\,dt. $$ Multiplying by \(b-a\) proves the stated identity. \(\square\)

The conclusion does not generally identify \(c\) in advance, and it need not be unique. Its value is that it replaces an integral average by an actual function value at some point in the interval.

Worked Applications

Worked Example: Prove \(\sin x\leq x\) for \(x\geq0\)

Define \(h(x)=x-\sin x\). We want to show \(h(x)\geq0\). Since \(h(0)=0\), the Fundamental Theorem of Calculus gives, for \(x\geq0\),

$$ h(x)-h(0)=\int_0^x h'(t)\,dt =\int_0^x (1-\cos t)\,dt. $$

For every real \(t\), \(\cos t\leq1\), so the integrand \(1-\cos t\) is nonnegative. The integral is therefore nonnegative, and \(h(x)\geq h(0)=0\). Hence \(\sin x\leq x\) for all \(x\geq0\). If \(x>0\), then \(1-\cos t\) is strictly positive for \(t\) in some subinterval of \((0,x)\); the strict positivity theorem shows \(h(x)>0\). Equality holds at \(x=0\).

Worked Example: Prove \(e^x\geq1+x\) for \(x\geq0\)

Set \(q(x)=e^x-1-x\). Since \(q(0)=0\), the Fundamental Theorem of Calculus yields

$$ q(x)=\int_0^x q'(t)\,dt =\int_0^x (e^t-1)\,dt. $$

For \(t\geq0\), \(e^t\geq1\), so \(e^t-1\geq0\). Thus \(q(x)\geq0\), proving \(e^x\geq1+x\) on the nonnegative axis. For \(x>0\), the integrand is positive for every \(t\in(0,x]\), so the strict positivity theorem gives \(q(x)>0\). At \(x=0\), equality holds.

Worked Example: Locate an Average Value of \(t^2\)

Find a point \(c\in[0,2]\) at which \(t^2\) equals its average value on that interval. First compute the integral:

$$ \int_0^2 t^2\,dt =\left[\frac{t^3}{3}\right]_0^2 =\frac{8}{3}-0 =\frac{8}{3}. $$

The interval length is \(2\), so the average value is

$$ \frac{1}{2}\int_0^2 t^2\,dt =\frac{1}{2}\cdot\frac{8}{3} =\frac{4}{3}. $$

The Integral Mean Value Theorem guarantees a \(c\in[0,2]\) with \(c^2=4/3\). Since \(c\geq0\), this gives \(c=2/\sqrt{3}\). This point lies in the interval because \(2/\sqrt{3}>0\) and \(2/\sqrt{3}<2\), the latter following from \(1/\sqrt{3}<1\). Substitution verifies the claim: $$ \left(\frac{2}{\sqrt{3}}\right)^2=\frac{4}{3}. $$

Choose the Integral That Matches the Goal

An integral proof is most effective when the integrand expresses exactly the information needed. To compare two values of a function, integrate its derivative between the corresponding points. To prove a nonnegative difference, look for a nonnegative integrand. To bound an integral, find constant upper and lower bounds for the integrand, or use the absolute-value bound when signs vary.

A common pitfall is to confuse the sign of an integrand with the sign of an integral when the limits are reversed. If \(f\geq0\) but \(b<a\), then $$ \int_a^b f(t)\,dt=-\int_b^a f(t)\,dt\leq0, $$ not necessarily a nonnegative number. Another pitfall is to assume that a positive integral forces the integrand to be positive everywhere. It only forces the function to be positive somewhere; the integral averages its values across the interval.

1
Express the target as a difference.
For a comparison of function values, form \(F(b)-F(a)\); for an inequality, move both sides into one difference.
2
Use an integral representation.
Apply the Fundamental Theorem of Calculus when the difference is between values of a differentiable function.
3
Check the integrand and orientation.
Verify its sign or bound on the entire interval, and ensure the limits point in the direction required.
4
Apply the matching integral principle.
Use positivity for sign conclusions, pointwise comparison for bounds, or the Integral Mean Value Theorem for an attained average.

The integration strategy converts local information distributed across an interval into a statement about total change, an inequality, or an average. Its reliability comes from making that conversion explicit and checking the sign, bounds, and orientation before drawing the conclusion.

Check Your Understanding

Use the integral principles and proof strategies developed here to answer the following questions.

  1. Why does continuity matter in the theorem that a nonnegative function with zero integral must vanish everywhere?
  2. How does the Fundamental Theorem of Calculus turn a derivative sign into a comparison of function values?
  3. What interval orientation is needed to conclude that the integral of a nonnegative function is nonnegative?
  4. What does the Integral Mean Value Theorem say about the average value of a continuous function?
  5. Why can \(\left|\int_a^b f(t)\,dt\right|\) be smaller than \(\int_a^b|f(t)|\,dt\)?

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