MCQ
$\int \limits_{-\pi}^{\pi}|\pi-| x || d x$ is equal to :
- ✓$\pi^{2}$
- B$2 \pi^{2}$
- C$\sqrt{2} \pi^{2}$
- D$\frac{\pi^{2}}{2}$
$=2 \int_{0}^{\pi}(\pi- x ) d x$
$=2\left[\pi x -\frac{ x ^{2}}{2}\right]_{0}^{\pi}=\pi^{2}$
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$\frac{d y}{d x}=1+x e^{y-x},-\sqrt{2}\,<\,x\,<\,\sqrt{2}, y (0)=0$ then, the minimum value of $y(x)$ , $\mathrm{x} \in(-\sqrt{2}, \sqrt{2})$ is equal to: