Data la funzione \( f(x)=-\frac{x^{3}}{9}-4 x^{2}+x \), calcola \[ f(0), f(-3), f\left(\frac{1}{2}\right), f(1) \]

\( \int _ { - \infty } ^ { \infty } \frac { \sin x } { x } d x \)

solve \( \lim _{x \rightarrow 1} \frac{x-1}{\sqrt{x}-1} \)

\( \int ( x ) a x + 8 ) ^ { 2 } \cos x + 2 x \)

\( \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { x } e ^ { \frac { x } { y } } d y d x \)

put \( z=\mathrm{Re}^{i \theta} \), then \( |z|=R \) Now, \( \left|\int_{\Gamma} f(z) d z\right| \leq \frac{\pi R}{\left(R^{2}-1\right)^{2}} \rightarrow 0 \) as \( R \rightarrow \infty \) Thus, the equation (1) becomes \( \lim _{R \rightarrow \infty} \int_{-R}^{R} f(x) d x+\lim _{R \rightarrow \infty} \int_{\Gamma} f(z) d z=\frac{\pi}{2} \) \( \Rightarrow \quad \int_{-\infty}^{\infty} f(x) d x+0=\frac{\pi}{2} \Rightarrow \int_{-\infty}^{\infty} f(x) d x=\frac{\pi}{2} \)