Question

Se consideră funcția $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=(x+2) \sin x$.

$5 \mathbf{a}$ a) Arătați că $\int_{0}^{\frac{\pi}{2}} \frac{f(x)}{x+2} d x=1$

$5 \mathbf{p} \mid$ b) Calculați $\int_{0}^{\frac{\pi}{2}} f(x) d x .$

Answer 1

Explicație pas cu pas:

$f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=(x+2) \sin(x)$

a)

$\int_{0}^{\frac{\pi}{2}} \frac{f(x)}{x+2} d x = \int_{0}^{\frac{\pi}{2}} \frac{(x+2) \sin(x) }{x+2} d x \\ = \int_{0}^{\frac{\pi}{2}} \sin(x) d x = - \cos(x) |_{0}^{\frac{\pi}{2}} \\ = - \cos\left( \frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1$

b)

$\int fg' = fg - \int f'g \\ f = x + 2 => f' = 1 \\ g' = \sin(x) => g = - \cos(x)$

→

$\int_{0}^{\frac{\pi}{2}} f(x) d x = \int_{0}^{\frac{\pi}{2}} (x+2) \sin(x) d x \\ = \left( - (x + 2) \cos(x)\right) |_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} - \cos(x) d x \\ = \left( - (x + 2) \cos(x)\right) |_{0}^{\frac{\pi}{2}} + \sin(x)|_{0}^{\frac{\pi}{2}} \\ = (0 + 2) + (1 - 0) = 3$