31&35都是不存在 不存在要如何判斷

adrant adrant Evaluate the following improper integrals: x dx 28. dx 广 1 29. = -2 (x² – 4) 3x vertical 2 X 30. dx 31. 1.0 • dx (9 - x2) ? x - 1 x² dx 32. si 33. (x – 8) 3 dx (x - 3)2/5 - | r 5 X 1 34. dx 35. dx (x² – 4)² -3 (x + 2)4

第52.自己算了一個跟解答很接近的答案，但還是不知道哪裡不對，求解過程🙏感謝（附解答）

Evaluating a Definite Integral In Exercise KO 47-52, evaluate the definite integral. 63. C! the and tial x(x2 + 1) dx 47. 2x2x3 + 1 dx 48. 1 1 49. 1 • dx 、+ 1 (2x + 1 so. S tant contra dx x(1 + [atta de 51. [ are de 、 。 64.0 es 03/ 52. 1 dx 13 72

9）想問中間值那個什麼的是怎麼算？ 網路上的定義搜不到 謝謝

9-12 Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places. *10 1/2 9. 8,5°/x3 + 1dx, n=4 , 10. ("* cos*x dx, n= 4 4 12 JO

第十題求解題過程，感謝🙏

where ci n n Writing a Limit as a Definite Integral In Exercises 7–10, write the limit as a definite integral on the given interval, is any point in the ith subinterval. 7. lim § (30; + 10) Ax; 3c 8. lim Ë Vc? + 4 Ax; [-1,5] [O, 3] 9. lim Š 10. lim § (2-4 sin c;) Ax; [1, 5] [0, 1] A»0 n n 3 1 + Ax; 141-0 ||A||→0 i=1

求微積分解答!

2. Evaluate the integral SSSE (xz – ys) dV, where E = {(x, y, z) | -1 <x< 1,0 < y = 2,0 < z=1} using three different orders of integration.

作業寫不出來😭微積分

if (x, y) = (0,0) ( 20 points ) Let f(x, y) = x²y² x4 + y + 0, if (x,y) = (0,0) (a) Does lim f(x,y) exist ? Give reasons. (x,y) → (0,0) (b ) Find the partial derivatives fx(0,0) and fy(0,0).

大一微積分 求求大家幫忙🙏🏻😭

(b ) Find the radius of convergence and interval of convergence of the power series Enzo (-1)n (2x + 3)n 2n

求微積分解答

Evaluate the iterated integral sole ex dxdy. (Hint: by reversing the order of integration)

有人可以解釋一下箭頭畫的地方是怎麼來的嗎？我知道目標是要把1-sin^2θ換成cos^2θ

985 Double Integrals and Volume EXAMPLE 5 14.2 Volume of a Region Bounded by Two Surfaces Find the volume of the solid region bounded above by the paraboloid Paraboloid z = 1 - x2 - y2 and below by the plane z = 1- y as shown in Figure 14.20. Plane Paraboloid: z=1 - x2 - y2 Plane: z=1-y Figure 14.20 Solution Equating z-values, you can determine that the intersection of the two surfaces occurs on the right circular cylinder given by 1- y = 1 - x2 - y2 x² = y - y². So, the region R in the xy-plane is a circle, as shown in Figure 14.21. Because the volume of the solid region is the difference between the volume under the paraboloid and the volume under the plane, you have Volume = (volume under paraboloid) – (volume under plane) Wyz? (1 – x2 - y2) dx dy – (1 – y) dx dy y-y2 Wy-y² 3 y-y2 R: 0 Sys1 -dy-y2 sxs dy-y2 Figure 14.21 - TT 4 SL -SI*6 = y2 + x) dx dy 6 – ymla - Kdy – year2 dy = 990 1.0 – (23 – 1,276? dy 4 2. 2y - 1 = sin e (7/2 cos4 e de 2 -1/2 1/2 cos4 do 1 d." 311 Wallis's Formula 6 16 TT 32

請問這題怎麼算

Find the integral Sx(x - 1) 1 dx