Functions are from R to N or R × R R \times R to N, where R denotes the real numbers and N denotes a normed complete ring. If S, T and G are functions from R × R R \times R to N, each of S ( p − , p ) , S ( p − , p − ) , T ( p − , p ) S({p^ - },p),S({p^ - },{p^ - }),T({p^ - },p) and T ( p − , p − ) T({p^ - },{p^ - }) exists for a > p ⩽ b a > p \leqslant b , each of S ( p , p + ) , S ( p + , p + ) , T ( p , p + ) S(p,{p^ + }),S({p^ + },{p^ + }),T(p,{p^ + }) and T ( p + , p + ) T({p^ + },{p^ + }) exists for a ⩽ p > b a \leqslant p > b , G has bounded variation on [a, b] and ∫ a b G \smallint _a^bG exists, then each of \[ ∫ a b S [ G − ∫ G ] T and ∫ a b S [ 1 + G − ∏ ( 1 + G ) ] T \int _a^b S \left [ {G - \int G } \right ]T\quad {\text {and}}\quad \int _a^b {S\left [ {1 + G - \prod {(1 + G)} } \right ]} \;T \] exists and is zero. These results can be used to solve integral equations without the existence of integrals of the form \[ ∫ a b | G − ∫ G | = 0 and ∫ a b | 1 + G − ∏ ( 1 + G ) | = 0. \int _a^b {\left | {G - \int G } \right | = 0} \quad {\text {and}}\quad \int _a^b {\left | {1 + G - \prod {(1 + G)} } \right |} = 0. \] This is demonstrated by solving the linear integral equation \[ f ( x ) = h ( x ) + ( L R ) ∫ a x ( f G + f H ) f(x) = h(x) + (LR)\int _a^x {(fG + fH)} \] and the Riccati integral equations \[ f ( x ) = w ( x ) + ( L R L R ) ∫ a x ( f H + G f + f K f ) f(x) = w(x) + (LRLR)\int _a^x {(fH + Gf + fKf)} \] without the existence of the previously mentioned integrals.