Solving an Integral Equation by Differentiating with the FTC

An integral equation defines an unknown function $F$ implicitly through an integral whose upper limit is the variable $x$, for example

$G\bigl(F(x)\bigr)=\int_{a}^{x} K\bigl(t,F(t)\bigr)\,dt.$

The right-hand side is an accumulation function, so the Fundamental Theorem of Calculus applies: differentiating it in $x$ simply evaluates the integrand at the top limit. That single move removes the integral and leaves an ordinary equation (usually a first-order ODE) for $F$.

The algorithm

1. Differentiate both sides in $x$. By the FTC, $\dfrac{d}{dx}\displaystyle\int_{a}^{x} g(t)\,dt = g(x)$, so the integral becomes $K\bigl(x,F(x)\bigr)$. Differentiate the left side with the chain rule (e.g. $\tfrac{d}{dx}F^{2}=2F\,F'$).
2. Simplify to an ODE. Cancel common factors of $F(x)$ — but only after noting why you may (see step 4). You are usually left with $F'(x)=h(x)$ or a separable equation.
3. Get the initial value. Plug $x=a$ into the original equation. The integral from $a$ to $a$ is $0$, which pins down $F(a)$.
4. Solve and integrate. Solve the ODE, fixing the constant with the initial value: $F(x)=F(a)+\int_{a}^{x}h(t)\,dt$. Evaluate at the requested point.

Cancelling a factor like $F(x)$ is only legal where it is nonzero. A hypothesis such as "$F$ is non-constant" (hence not identically $0$) is exactly what licenses the division — never drop that justification.

Mini-example

Suppose $F$ is non-constant with

$F^{2}(x)=\int_{0}^{x}F(t)\,e^{\sqrt{t}}\,dt,\qquad x\ge 0.$

Differentiate. The FTC gives the right side derivative $F(x)\,e^{\sqrt{x}}$; the chain rule gives $2F(x)F'(x)$ on the left:

$2F(x)F'(x)=F(x)\,e^{\sqrt{x}}.$

Cancel $F(x)$ (allowed since $F\not\equiv 0$): $\;F'(x)=\tfrac{1}{2}e^{\sqrt{x}}.$

Initial value. At $x=0$ the original equation reads $F^{2}(0)=\int_{0}^{0}=0$, so $F(0)=0$.

Integrate. With $u=\sqrt{t}$ one finds $\int_{0}^{x}\tfrac{1}{2}e^{\sqrt{t}}\,dt=\bigl(\sqrt{t}-1\bigr)e^{\sqrt{t}}\big|_{0}^{x}=(\sqrt{x}-1)e^{\sqrt{x}}+1$, so

$F(x)=(\sqrt{x}-1)\,e^{\sqrt{x}}+1.$

Hence $F(4)=(2-1)e^{2}+1=e^{2}+1$.

Exam trick: the instant you see the unknown function under an integral sign with a variable upper limit, differentiate the whole equation — the FTC is doing all the work. And always read off the initial condition by setting $x$ equal to the lower limit, where the integral vanishes.