Calculer une intégrale de sommes de fonctions usuelles en passant par la primitive directement Exercice

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

\int_{1}^{2} \dfrac{1}{\sqrt{x}} + x^2 \, \mathrm{d}x = \sqrt{2} + 6

\int_{1}^{2} \dfrac{1}{\sqrt{x}} + x^2 \, \mathrm{d}x = \sqrt{3} + 6

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

\int_{1}^{2} \dfrac{1}{\sqrt{3x}} + \dfrac{1}{\sqrt{2x}} \, \mathrm{d}x = 2 \dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{6}} (\sqrt{2}-1)

\int_{1}^{2} \dfrac{1}{\sqrt{3x}} + \dfrac{1}{\sqrt{2x}} \, \mathrm{d}x = \dfrac{2 + \sqrt{3}}{\sqrt{3}} ( \sqrt{2} - 1)

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

\int_{1}^{2} \dfrac{1}{\sqrt{3x}} + \dfrac{1}{\sqrt{2x}} \, \mathrm{d}x = \dfrac{\sqrt{6}}{3} + 1 - \dfrac{ \sqrt{2}}{3} - \dfrac{ \sqrt{2}}{2}

\int_{0}^{\dfrac{\pi}{2}} \cos(x) + \sin(x) \, \mathrm{d}x = 2

\int_{0}^{\dfrac{\pi}{2}} \cos(x) + \sin(x) \, \mathrm{d}x = 1

\int_{0}^{\dfrac{\pi}{2}} \cos(x) + \sin(x) \, \mathrm{d}x = -1

\int_{0}^{\dfrac{\pi}{2}} \cos(x) + \sin(x) \, \mathrm{d}x = -2

\int_{0}^{1} x^4 + \exp(2x) \, \mathrm{d}x = \dfrac{\exp(2)}{2} - \dfrac{1}{4}

\int_{0}^{1} x^4 + \exp(2x) \, \mathrm{d}x = 1 + \exp(2)

\int_{0}^{1} x^4 + \exp(2x) \, \mathrm{d}x = 1 - \exp(1)

\int_{0}^{1} x^4 + \exp(2x) \, \mathrm{d}x = 1 + \exp(4) - \exp(1)

\int_{1}^{2} \dfrac{1}{x^2} + 5 \, \mathrm{d}x = \dfrac{11}{2}

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

\int_{1}^{2} \dfrac{1}{x^2} + 5 \, \mathrm{d}x = -\dfrac{11}{2}

\int_{1}^{2} \dfrac{1}{x^2} + 5 \, \mathrm{d}x = -\dfrac{9}{2}