Calculer une intégrale d'une fonction usuelle en passant par la primitive directement Exercice

\int_{0}^{1} x \, \mathrm{d}x = \dfrac{1}{2}

\int_{0}^{1} x \, \mathrm{d}x = -\dfrac{1}{2}

\int_{0}^{1} x \, \mathrm{d}x = 2

\int_{0}^{1} x \, \mathrm{d}x = 1

\int_{1}^{5} \dfrac{2}{x^2} \, \mathrm{d}x = \dfrac{8}{5}

\int_{1}^{5} \dfrac{2}{x^2} \, \mathrm{d}x = \dfrac{8}{3}

\int_{1}^{5} \dfrac{2}{x^2} \, \mathrm{d}x = \dfrac{1}{5}

\int_{1}^{5} \dfrac{2}{x^2} \, \mathrm{d}x = \dfrac{1}{2}

\int_{0}^{\pi} \cos(x) = 0 

\int_{0}^{\pi} \cos(x) = 1 

\int_{0}^{\pi} \cos(x) = \dfrac{1}{2} 

\int_{0}^{\pi} \cos(x) = −1 

\int_{0}^{1} \dfrac{x^2+1}{3} \, \mathrm{d}x = \dfrac{4}{9} 

\int_{0}^{1} \dfrac{x^2+1}{3} \, \mathrm{d}x = \dfrac{1}{9} 

\int_{0}^{1} \dfrac{x^2+1}{3} \, \mathrm{d}x = \dfrac{4}{3} 

\int_{0}^{1} \dfrac{x^2+1}{3} \, \mathrm{d}x = \dfrac{4}{5} 

\int_{1}^{2} \dfrac{1}{\sqrt{x}} \, \mathrm{d}x = 2

\int_{1}^{2} \dfrac{1}{\sqrt{x}} \, \mathrm{d}x = \sqrt{3} 

\int_{1}^{2} \dfrac{1}{\sqrt{x}} \, \mathrm{d}x = 2\sqrt{2}-2 

\int_{1}^{2} \dfrac{1}{\sqrt{x}} \, \mathrm{d}x = 1