[proofplan] Apply the product rule to the pointwise product of $f$ and $g$, then integrate the resulting identity over $[a,b]$. The [Fundamental Theorem of Calculus (Part Two)](/theorems/191) evaluates the [integral](/page/Integral) of the [derivative](/page/Derivative) of the product as the endpoint difference $\big[f(x)g(x)\big]_a^b$. Linearity of the integral then isolates the term containing $f'g$ and gives the stated integration by parts formula. [/proofplan] [step:Differentiate the product of $f$ and $g$] Define the product [function](/page/Function) \begin{align*} h: [a,b] &\to \mathbb{R} \\ x &\mapsto f(x)g(x). \end{align*} Since $f,g: [a,b]\to\mathbb{R}$ are [continuously](/page/Continuity) differentiable, the product rule applies on $(a,b)$ and gives \begin{align*} h'(x)=f'(x)g(x)+f(x)g'(x) \end{align*} for every $x\in(a,b)$. Moreover $h'$ extends continuously to $[a,b]$ by the displayed formula, because $f'$, $g$, $f$, and $g'$ are continuous on $[a,b]$. [guided] We first package the expression whose derivative contains both terms in the desired identity. Define \begin{align*} h: [a,b] &\to \mathbb{R} \\ x &\mapsto f(x)g(x). \end{align*} The hypotheses say that both $f$ and $g$ are continuously differentiable on $[a,b]$. Therefore $f$ and $g$ are differentiable on the open interval $(a,b)$, and their derivatives $f'$ and $g'$ extend continuously to the closed interval $[a,b]$. Applying the product rule to the functions $f$ and $g$ at each point $x\in(a,b)$ gives \begin{align*} h'(x)=f'(x)g(x)+f(x)g'(x). \end{align*} This identity is the mechanism behind integration by parts: it expresses the derivative of one product as the sum of the two integrands that will later be separated. Since $f'$, $g$, $f$, and $g'$ are continuous on $[a,b]$, the functions $f'g$ and $fg'$ are continuous on $[a,b]$, so the right-hand side defines a continuous extension of $h'$ to the closed interval. [/guided] [/step] [step:Integrate the derivative of the product and evaluate the endpoints] Because $h$ is continuously differentiable on $[a,b]$, the [Fundamental Theorem of Calculus](/theorems/632) (Part Two) applies to $h$ and gives \begin{align*} \int_a^b h'(x)\,d\mathcal{L}^1(x)=h(b)-h(a). \end{align*} By the definition of $h$, \begin{align*} h(b)-h(a)=f(b)g(b)-f(a)g(a)=\big[f(x)g(x)\big]_a^b. \end{align*} Hence \begin{align*} \int_a^b h'(x)\,d\mathcal{L}^1(x)=\big[f(x)g(x)\big]_a^b. \end{align*} [/step] [step:Use linearity of the integral to isolate the $f'g$ term] Substituting $h'(x)=f'(x)g(x)+f(x)g'(x)$ into the preceding identity and using linearity of the integral for continuous functions on $[a,b]$, we obtain \begin{align*} \int_a^b f'(x)g(x)\,d\mathcal{L}^1(x) + \int_a^b f(x)g'(x)\,d\mathcal{L}^1(x) = \big[f(x)g(x)\big]_a^b. \end{align*} Subtracting the second integral from both sides gives \begin{align*} \int_a^b f'(x)g(x)\,d\mathcal{L}^1(x) = \big[f(x)g(x)\big]_a^b - \int_a^b f(x)g'(x)\,d\mathcal{L}^1(x). \end{align*} This is the stated integration by parts identity, with $d\mathcal{L}^1(x)$ writing explicitly the one-dimensional [Lebesgue measure](/page/Lebesgue Measure) represented by $dx$ in the theorem statement. [/step]