4Admitere Matematică 2023 — iulie — Varianta ESă se calculeze l=limα→∞∫0α2x+1x4+2x3+3x2+2x+2 dxl=\lim_{\alpha\to\infty}\displaystyle\int_0^{\alpha}\dfrac{2x+1}{x^4+2x^3+3x^2+2x+2}\,dxl=limα→∞∫0αx4+2x3+3x2+2x+22x+1dx.a) l=π3l=\dfrac{\pi}{3}l=3π; b) l=arctg2l=\operatorname{arctg} 2l=arctg2; c) l=π2l=\dfrac{\pi}{2}l=2π; d) l=arctg13l=\operatorname{arctg}\dfrac{1}{3}l=arctg31; e) l=arctg3l=\operatorname{arctg} 3l=arctg3; f) l=π4l=\dfrac{\pi}{4}l=4π.nerezolvatăAnaliză — integrale
10Pre-admitere Matematică 2023 — Varianta F — admitere anticipatăFie f:(1,∞)→Rf:(1,\infty)\to\mathbb{R}f:(1,∞)→R, f(y)=∫0y1x2−2x+y dxf(y)=\displaystyle\int_0^y \dfrac{1}{x^2-2x+y}\,dxf(y)=∫0yx2−2x+y1dx. Calculați ∫210f(y) dy\displaystyle\int_2^{10} f(y)\,dy∫210f(y)dy.a) 3π3\pi3π; b) 2π2\pi2π; c) 5π3\dfrac{5\pi}{3}35π; d) π\piπ; e) π2\dfrac{\pi}{2}2π; f) 3π2\dfrac{3\pi}{2}23π.nerezolvatăAnaliză — integrale