Question

Sa se calculeze integrala de la 0 la 2 din
$\sqrt{ {x}^{2} + 4 }$
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Answer 1

Răspuns:

Explicație pas cu pas:

x = 2tan(u)$\int _0^2\sqrt{x^2+4}dx = \int _0^{\frac{\pi }{4}}4\sec ^3\left(u\right)du =4\cdot \int _0^{\frac{\pi }{4}}\sec ^3\left(u\right)du = 4\left(\left[\frac{\sec ^2\left(u\right)\sin \left(u\right)}{2}\right]^{\frac{\pi }{4}}_0+\frac{1}{2}\cdot \int _0^{\frac{\pi }{4}}\sec \left(u\right)du\right) = 4\left(\left[\frac{1}{2}\sec \left(u\right)\tan \left(u\right)\right]^{\frac{\pi }{4}}_0+\frac{1}{2}\ln \left(1+\sqrt{2}\right)\right) = 4\left(\frac{1}{\sqrt{2}}+\frac{1}{2}\ln \left(1+\sqrt{2}\right)\right)$