The fundamental theorem of calculus and integration by parts

Assume first $a<b$ and let $H(x)={\int }_a^x{f}^{\mathrm{\prime }}(t)dt$. By [L1], $H^{\mathrm{\prime }}(x)=f^{\mathrm{\prime }}(x)$ on an open interval containing $[a,b]$, and $H(a)=0$. By [L3], $H$ is continuous there. Thus $q=f-H$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $q^{\mathrm{\prime }}=0$. Applying [L4] gives $q(b)-q(a)=0$, so $f(b)-H(b)=f(a)$. Rearranging proves the first identity.

For $a=b$ the first identity is $0=0$; reversing endpoints proves it for $a>b$.

Assume $a<b$. By [L3], $(fg{)}^{\mathrm{\prime }}=f^{\mathrm{\prime }}g+f{g}^{\mathrm{\prime }}$. These terms are continuous: for continuous $u,v$, $\mathrm{\mid }u(t)v(t)-u(t_0)v(t_0)\mathrm{\mid }\le \mathrm{\mid }u(t)\mathrm{\mid }\mathrm{\mid }v(t)-v(t_0)\mathrm{\mid }+\mathrm{\mid }v(t_0)\mathrm{\mid }\mathrm{\mid }u(t)-u(t_0)\mathrm{\mid },$ and continuity of $u$ bounds $\mathrm{\mid }u(t)\mathrm{\mid }$ near $t_0$. Thus $fg$ is continuously differentiable. Apply step 1.1 to $fg$ and then [L2]: $f(b)g(b)-f(a)g(a)={\int }_a^b(f^{\mathrm{\prime }}g+f{g}^{\mathrm{\prime }})={\int }_a^b{f}^{\mathrm{\prime }}g+{\int }_a^{b}f{g}^{\mathrm{\prime }}\mathrm{.}$ Rearranging gives the second identity. For $a=b$ both sides are zero, and for $a>b$ reverse endpoints using [L1].