AM3 Soluzioni Tutorato 1 - Dipartimento di Matematica

x 2 + y 26. lim(x,y)→(0,0) x 2 + sin 2 yx 2 + y 2 ∣ ∣∣∣∣x 2 + sin 2 y − 1 =∣≤ π2 (x,y)→(0,0)y2 −→ 04quindilim(x,y)→(0,0)y 2 − sin 2 ∣y ∣∣∣x 2 + y 2 ≤x 2 + y 2x 2 + sin 2 y = 1y 4x 2 + sin 2 y ≤ π24 y2 sin 2 yx 2 + sin 2 y ≤Esercizio 2 1.2.3.4.5.∫ π20∫ π20= 3sin 2 xdx =sin 4 xdx =∫ π20= 3 4 π − 3 ∫ π2∫ +∞0= 1 2∫ π20∫ π201 − cos2x2= π 4 − [ 14 sin 2x ] π20= π 4sin 3 xsinxdx = − [ sin 3 xcos x ] π 20+ 3sin 2 x(1 − sin 2 x)dx = 30sin 4 xdx =⇒ 41(x 2 + 2) 2dx = 1 21x 2 + 2 dx − 1 2∫ ∞∫0+∞0+ 1 2∫ 0 +∞0∫ +∞= 1 4 0√2=16 π∫ 120∫ 120∫ ∞x 2(x 2 + 2) 2dx = 1 21x 2 + 2 = 1 21(x 2 + 2) 2dx = 1 21x 2 + 2 dx = 1 8∫ +∞∫ 0 +∞∫0+∞∫0+∞0∫ ∞∫0∞0∫ π20∫ π20sin 2 xdx − 3∫ π2sin 4 xdx = 3 4 π =⇒ ∫ π22(x 2 + 2) 2dx = 1 2x 2(x 2 + 2) 2dx Max ddx (− 1∫ ∞00∫ π20sin 4 xdx =0sin 2 xcos 2 xdxsin 4 xdx = 3 16 π2 + x 2 − x 2(x 2 + 2) 2 dx =x 2 + 2 )dx = 1 [−x2 x 2 + 201x 2 + 2 quindi1x 2 + 2 dx − 1 ∫ +∞x 22 0 (x 2 + 2) 2dx =√ ∫1 2 ∞( √ x 2) 2 + 1 dx = 18 y 2 + 1 dy =0] +∞x 3∫ 1x 4 − 2x 3 + 2x 2 − 2x + 1 dx = 2 x 30 (x 2 + 1)(x − 1) 2dx =1x − 1 + 1 12 (x − 1) 2 + 1 1[log(|x2 x 2 + 1 dx = − 1|) − 1 12 x − 1 ++ 1 2 arctanx ]12∫ 10= log 1 2 + 1 2 + arctan 1 2arctanxcosh xdx = 0 perchè è l’integrale di una funzione dispari−1 1 + x2 su un intervallo simmetrico rispetto all’origine+2