${\displaystyle \int _{0}^{2}(1-|x-1|)^{3}\,\mathrm {d} x=\int _{-1}^{1}(1-|t|)^{3}\,\mathrm {d} t=2\int _{0}^{1}(1-t)^{3}\,\mathrm {d} t={\frac {1}{2}}}$.

${\displaystyle {\begin{aligned}&\left(\int _{-6}^{-1}+\int _{3}^{4}-\int _{-1}^{3}\right)\left(x^{3}-2x^{2}-3x\right)\,\mathrm {d} x\\&=\left[{\frac {x^{4}}{4}}-{\frac {2x^{3}}{3}}-{\frac {3x^{2}}{2}}\right]_{-6}^{-1}+]_{3}^{4}-]_{-1}^{3}\\&={\frac {1}{4}}\left(1-6^{4}+4^{4}-2.3^{4}+1\right)-{\frac {2}{3}}\left(-1+6^{3}+4^{3}-2.3^{3}+1\right)-{\frac {3}{2}}\left(1-6^{2}+4^{2}-2.3^{2}+1\right)\\&=-300-{\frac {452}{3}}+54={\frac {-1190}{3}}.\end{aligned}}}$

${\displaystyle \int _{-1}^{-2/3}\left(x+1+3x+2\right)\,\mathrm {d} x+\int _{-2/3}^{2}\left(x+1-3x-2\right)\,\mathrm {d} x=}$

${\displaystyle \left[2x^{2}+3x\right]_{-1}^{-2/3}+\left[-x^{2}-x\right]_{-2/3}^{2}=-{\frac {19}{3}}}$.

${\displaystyle \int _{0}^{1/2}\left(x-x^{2}+1-2x-x\right)\,\mathrm {d} x+\int _{1/2}^{1}\left(x-x^{2}+2x-1-x\right)\,\mathrm {d} x+\int _{1}^{2}\left(x^{2}-x+2x-1-x\right)\,\mathrm {d} x=}$

${\displaystyle \int _{0}^{1/2}\left(-x^{2}+1-2x\right)\,\mathrm {d} x+\int _{1/2}^{1}\left(-x^{2}+2x-1\right)\,\mathrm {d} x+\int _{1}^{2}\left(x^{2}-1\right)\,\mathrm {d} x=}$

${\displaystyle \left[{\frac {-x^{3}}{3}}\right]_{0}^{1}+\left[{\frac {x^{3}}{3}}\right]_{1}^{2}+\left[x\right]_{0}^{1/2}-\left[x\right]_{1/2}^{2}-\left[x^{2}\right]_{0}^{1/2}+\left[x^{2}\right]_{1/2}^{1}={\frac {3}{2}}}$.

${\displaystyle \int _{|x|<3,\,x>k}{\frac {3-x+k}{x-k+3}}\,\mathrm {d} x+\int _{|x|<3,\,x<k}{\frac {3-x+k}{k-x+3}}\,\mathrm {d} x=}$

${\displaystyle \int _{|x|<3,\,x>k}\left(-1+{\frac {6}{x-k+3}}\right)\,\mathrm {d} x+\int _{|x|<3,\,x<k}\mathrm {d} x=}$

${\displaystyle \left[-x+6\ln(x-k+3)\right]_{|x|<3,\,x>k}+\left[x\right]_{|x|<3,\,x<k}=}$

${\displaystyle {\begin{cases}{\text{si}}&k\geq 3&:&\left[x\right]_{-3}^{3}=6\\{\text{si}}&k\leq -3&:&\left[-x+6\ln(x-k+3)\right]_{-3}^{3}=-6+6\ln(1-6/k)\\{\text{si}}&-3\leq k\leq 3&:&\left[x\right]_{-3}^{k}+\left[-x+6\ln(x-k+3)\right]_{k}^{3}=2k+6\ln(2-k/3)\end{cases}}}$

(dans chacun des deux cas ${\displaystyle k=3}$ ou ${\displaystyle k=-3}$, communs à deux expressions, les deux fonctions de ${\displaystyle k}$ coïncident).

${\displaystyle \int _{0}^{1}{\frac {t+1}{3-t}}\,\mathrm {d} t=\int _{0}^{1}\left(-1+{\frac {4}{3-t}}\right)\,\mathrm {d} t=\left[-t-4\ln(3-t)\right]_{0}^{1}=-1+4\ln(3/2)}$.

${\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\left(\cos -\sin x\right)-\int _{\frac {\pi }{4}}^{\frac {5\pi }{4}}\left(\cos -\sin x\right)=\left[\sin +\cos \right]_{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}-\left[\sin +\cos \right]_{\frac {\pi }{4}}^{\frac {5\pi }{4}}=2{\sqrt {2}}}$.

${\displaystyle 8\int _{0}^{\pi /2}\cos =8}$.

${\displaystyle \left[{\frac {x^{3}}{3}}\right]_{0}^{1}+\left[{\frac {x^{2}}{2}}\right]_{1}^{2}+\left[-x^{2}+5x\right]_{2}^{3}={\frac {1}{3}}+{\frac {3}{2}}+0={\frac {11}{6}}}$. (Le fait que la troisième partie de l'intégrale est nulle pouvait se prévoir graphiquement : le milieu du segment ${\displaystyle 2\leq x\leq 3,\;y=-2x+5}$ a pour ordonnée ${\displaystyle -2{\frac {2+3}{2}}+5=0}$.)