The fundamental theorem of calculus in integral equations

Differentiating an accumulated integral

If $f$ is continuous on an interval containing $a$ and $x$, the fundamental theorem of calculus gives

$$\frac{d}{dx}\int_a^x f(t)\,dt=f(x).$$

To see why, put $F(x)=\int_a^x f(t)\,dt$. For nonzero $h$,

$$\frac{F(x+h)-F(x)}h=\frac1h\int_x^{x+h}f(t)\,dt.$$

Continuity makes this average tend to $f(x)$ as $h\to0$, from either side. The corresponding evaluation rule says that if $G'=f$, then $\int_a^bf(t)\,dt=G(b)-G(a)$.

When a limit is itself a function, apply the chain rule:

$$\frac{d}{dx}\int_a^{2x}f(t)\,dt=2f(2x).$$

More generally, for differentiable $g,h$ and continuous $f$ on the relevant interval,

$$\boxed{\frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt =h'(x)f(h(x))-g'(x)f(g(x)).}$$

This formula assumes the integrand depends on $t$ alone. If it also depends explicitly on $x$, that dependence needs an additional term under suitable hypotheses. See the chain rule.

Two common applications

For an integral recurrence such as $f_{n+1}(x)=1+\int_0^xf_n(t)\,dt$, differentiation gives $f_{n+1}'=f_n$.

For an integral equation, first distinguish a fixed-limit integral—which may be an unknown constant—from an integral with a variable endpoint. In the second case, differentiating can reveal an equation for the unknown function. Keep an initial condition from the original equation: differentiation loses constant information. Finally substitute your answer back to establish sufficiency.

Example 1: a polynomial integral equation

Find the polynomial $f$ satisfying, for every real $x$,

$$x^4+2x^3-2x^2+xf(x)-\int_1^xf(t)\,dt=0.$$

At $x=1$, the integral vanishes and $f(1)=-1$. Differentiate:

$$4x^3+6x^2-4x+f(x)+xf'(x)-f(x)=0,$$

so

$$x\{f'(x)+4x^2+6x-4\}=0.$$

For $x\ne0$, the bracket is zero. Since it is a polynomial, the same identity holds at zero as well. Integrating,

$$f(x)=-\frac43x^3-3x^2+4x+C.$$

The condition $f(1)=-1$ gives $C=-2/3$, hence

$$\boxed{f(x)=-\frac43x^3-3x^2+4x-\frac23.}$$

For this polynomial, the derivative of the original left side is zero and its value at $x=1$ is zero. It therefore satisfies the original equation for all real $x$.

Example 2: two unknown functions

Let $f$ be differentiable and $g$ a cubic polynomial, with

$$\int_0^xf(t)\,dt=xf(x)+g(x)$$

for every real $x$. Suppose $g$ has a local extremum at $x=1/2$ and $g(1)=1/3$. Find $g$, and then find $f$ if $f(1)=5$.

At zero, $g(0)=0$. Differentiating gives

$$g'(x)=-xf'(x),$$

so $g'(0)=0$. The extremum also gives $g'(1/2)=0$. Since $g'$ is a genuine quadratic, it has the form

$$g'(x)=ax\left(x-\frac12\right),\qquad a\ne0.$$

Integration and $g(0)=0$ yield $g(x)=ax^3/3-ax^2/4$. Its value at one is $a/12=1/3$, so

$$\boxed{g(x)=\frac43x^3-x^2.}$$

Its derivative $2x(2x-1)$ changes sign at $1/2$, confirming the extremum.

For $x\ne0$, the differential equation now gives $f'(x)=-4x+2$. Thus on each side of zero, $f(x)=-2x^2+2x+C$. Differentiability implies continuity at zero, so the two constants coincide. The condition at one gives

$$\boxed{f(x)=-2x^2+2x+5.}$$

Direct substitution verifies

$$\int_0^xf(t)\,dt=-\frac23x^3+x^2+5x=xf(x)+g(x).$$

This argument justifies the point $x=0$ instead of dividing by zero or assuming that differentiability alone implies a continuous derivative.

Example 3: an integral recurrence

Define

$$f_0(x)=1,\qquad f_{n+1}(x)=1+\int_0^xf_n(t)\,dt\quad(n\ge0).$$

Find $f_n$. Prove that $f_n(x)>0$ for every real $x$ when $n$ is even, and that $f_n(x)=0$ has exactly one real solution when $n$ is odd.

The first few functions are $1$, $1+x$, and $1+x+x^2/2$. Induction gives

$$\boxed{f_n(x)=\sum_{k=0}^n\frac{x^k}{k!}.}$$

Indeed, integrating the sum for $f_n$ term by term and adding one produces the sum for $f_{n+1}$. The recurrence also immediately gives $f_n'=f_{n-1}$ for $n\ge1$.

For the sign and root claims, start with $f_0=1>0$. Suppose $f_{2m}>0$ everywhere. Then $f_{2m+1}'=f_{2m}>0$, so this odd-degree polynomial is strictly increasing. Its limits at the two infinities have opposite signs, and $f_{2m+1}(0)=1$, so it has exactly one real root $\alpha<0$.

Next $f_{2m+2}'=f_{2m+1}$ is negative before $\alpha$ and positive afterward. Thus $f_{2m+2}$ has a global minimum at $\alpha$, and

$$f_{2m+2}(\alpha)=f_{2m+1}(\alpha)+\frac{\alpha^{2m+2}}{(2m+2)!} =\frac{\alpha^{2m+2}}{(2m+2)!}>0.$$

This proves the even assertion for the next index and completes both claims by induction. The fundamental theorem turns the integral recurrence into a particularly simple relationship between consecutive derivatives.