Question

MAT1052(1,10-19,2,20-22,... Genel bakış Planlar Kaynaklar Durum ve izleme Katil Let I- -dx. Select all that apply x²-8 Yanıtınız: The integral is improper of Type I. The integral is improper of Type II. 8x dx = dx + dx 2 x3-8 2 X-2 2 x² + 2x+4 - 4+ () a arctar(2) + 4 = + arctanl 23 4 t-2 - limin '2t+2 arctan 1-21 \/1²+2t+4 3 23 The integral is convergent. 1 = lim t-2 ) 12 arct an 20+4 √ 1²+2t+4 38 23 12 arctan 33 18 273 The improper integral is divergent

Answer 1

And I = 0 7. x4 dx I 2x²8 f(x) 2²8 not F(x) is defined on 2 4. K x=2 we get f(x)=0 I is improper integral of Ind type. So 7 I = 4 x -da x38 7 8x talda x²8 11 7 7 X - 8 dat Sxdx x²8 7 da Il- x²8 27 S Y dx 2 B (x-2)(x + 2x+4) a² b ² = 10-b) (0²+ b²tab) A (X-2) (x²+2x+4) x²+2x+4 + X X-2 X = A (+2x+4) +B(x-2) x=2 , A = 7

And we get B= - 1 (x-2) 7 7 N-2 - I- -da 1 6 X-2 x + 2x + 2 de 2 2 1 - I da N-2 I= 1 Iz tor tor Iz =( X-2 x² + 2x +4 da x-2 = 7 7 3 I2= da S dx 9 2 (2x+2) -3 22+2 2 (n²+2x+4) 7 dlx?+2x+4) x + 2x+4 Iz = 2 [ log (x²+ 20+4)], X²+2x + 4 7. 그 x²+2x +4 7 dx dx-3 2 7 1 dx (x+1/²+3 da I3 = (x + 1)² +3 da = (x+1² + (31² 7 care ton ( x ) ] 2 ul

S3 Lt are ton, I 2 = 1/2 [In 167)- Lt In(x²+2x+4)] X-) 2+ 3. [ /s (anton () - Lt Arctan Xezt = 1 in (67) + 53 arctants Į Lt in(x²+2x+41 √3 2 - 2 + + 3 Lt are ton/x X+2+ dat = 16 [ In (x-2)] / / _ z Iz = + [Ins- lt In (x-2)] - / Iz X-2+ 11 Is In 5 - 1 1 2 167 +3 arc tan (23) I Lt In (4-2) + 1 Lt In (x²+2x+4) – √3 (+ 6 x+2+ 6 K+2+ ar tomlity + 3 arcton ( 2 /3) - 7 Lt in/x-2 Kuat = 7 in Faro 6 Vx²+2x+4 6 - 2 +

7. I= 8 14 8 II + [x2 7 =8 in E X-2 + 453 3 are ton ( 2 ) - 12/25 Lt in 6 √67 3 x 2 √x²+2x+4 4 / 3 Lt + ar ton ( 2 X 2+ + 45 २ 4/3 + ln [ 5 / 7 i 4 ancton ( 16 ) х I = 45 2 4/3 axt2 X-2 + 4 antan -ht 255 X-2+ (x²+2x+4 Statements 1 Not true 2) true 3) Not true 4) true Not true GI Not true 키 true because I is unbounded as X+ 2 +
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