can you explain the workings (1 point) Find the area under the curve y = 1/(7x)...
1 point) Find the area under the curve y = 1/(6x3) from x = 1 to x = t and evaluate it for t = 10,t = 100. Then find the total area under this curve for x > 1. a) t = 10 b) t = 100 c) Total area
(1 point) Find the area under the curve y = 1/(4x) from x = 1 to x = t and evaluate it for t = 10, t = 100. Then find the total area under this curve for x > 1. (a) t = 10 99/800 (b) t = 100 9999/80000 (c) Total area 1/8
Find the area under the curve y = 25/x3 from x = 1 to x = t. Evaluate the area under the curve for t = 10, t = 100, and t = 1000. t = 10 t = 100 t = 1000 Find the total area under this curve for x > 1.
(1 point) Find the Z-score such that: (a) The area under the standard normal curve to its left is 0.8369 Z= (b) The area under the standard normal curve to its left is 0.78 Z= (C) The area under the standard normal curve to its right is 0.1217 Z= (d) The area under the standard normal curve to its right is 0.1206 Z=
296. Area under a curve. The area of the region bounded by the curve y = (-2<x< 2), the x-axis, V4 - x4 V4- and the lines x = a and x = b(a < b) is given by sin - €) - sin-"). a. Find the exact area if a 1 and 1 b. Find the exact area if a = -V3 and 5 = vā.
Peer Leading Exercise 7 Spring 2019: Area Under the Given a function (x), the area under the curve is the area of the region bordered by the x -sxis and the graph of y(x). Area under the curve is somehow related to anti-derivatives. We wish to Example: Let f(x) -10-2x. Find the area under the curve between x 0 and x graph to help you visualize what is going on. Do you recognize the shape? 5. We include a 2...
10 Given the area under the curve y = x3 on the interval 1 < x < b is 600. Use the Fundamental Theorem of Calculus to find b.
Find the area of the region y that lies under the given curve y = f(x) over the indicated interval a <x<b. 2 Under y = 8x e over 0 < x < 2 2 over 0 < x < 2 is Round your answer to six decimal 2 The area under y = 8x e * places.
(1 point) Find the area of the surface obtained by rotating the curve y = yæ about y-axis for 1 < y < 2. Area:
Find the maximum possible area of a rectangle in quadrant 1 under the curve y = (x − 6)^2. (Include a test showing that your rectangle’s area is the maximum possible.)