Find the area of the region enclosed by the given curves. y=(sqrt(12-4x)) x=y So

Question

Find the area of the region enclosed by the given curves. y=(sqrt(12-4x)) x=y

Explanation / Answer

Sketch region enclosed by y = e^(3x), y = e^(7x), and x = 1 http://www.wolframalpha.com/input/?i=y+%… The two curves intersect at point (0,1) and from x = 0 to x = 1 ----> e^(7x) = e^(3x) So each rectangular cross-section perpendicular to x-axis has height e^(7x) - e^(3x) and width dx Integrating from x = 0 to x = 1, we get: A = ?0¹ (e^(7x) - e^(3x)) dx A = (1/7 e^(7x) - 1/3 e^(3x)) |0¹ A = (1/7 e7 - 1/3 e³) - (1/7 - 1/3) A = (3e7 - 7e³) / 21 - (3 - 7)/21 A = (3e7 - 7e³ + 4)/21 A ˜ 150.157176992 This seems close to your answer. Perhaps they want the exact form? ______________________________ Sketch region enclosed by 2y = 4vx, y = 3, and 2y + 3 x = 7 http://www.wolframalpha.com/input/?i=plo… From sketch it would be better to integrate with respect to y We find point of intersection of: 2y = 4vx and 2y + 3x = 7 4vx = 7 - 3x 16x = (7 - 3x)² 16x = 9x² - 42x + 49 9x² - 58x + 49 = 0 (9x - 49) (x - 1) = 0 x = 1 or x = 49/9 Only valid solution is x = 1 ---> 2y = 4v1 ----> y = 2 So we integrate from y = 2 to y = 3 Rewrite equations in terms of x as function of y: 2y = 4vx -----> x = 1/4 y² 2y + 3x = 7 ----> x = 7/3 - 2/3 y On interval 2 5 cos(6x) > 5 sin(6x) On interval (p/24, p/6) ---> 5 sin(6x) > 5 cos(6x) A1 = ?[0 to p/24] (5 cos(6x) - 5 sin(6x)) dx A1 = (5/6 sin(6x) + 5/6 cos(6x)) |[0 to p/24] A1 = 5/6 [(sin(p/4) + cos(p/4)) - (sin(0) + cos(0))] A1 = 5/6 [(v2/2 + v2/2) - (0 + 1)] A1 = 5/6 (v2 - 1) A2 = ?[p/24 to p/6] (5 sin(6x) - 5 cos(6x)) dx A2 = (-5/6 cos(6x) - 5/6 sin(6x)) |[p/24 to p/6] A2 = -5/6 [(cos(p) + sin(p)) - (cos(p/4) + sin(p/4))] A2 = -5/6 [(-1 + 0) - (v2/2 + v2/2)] A2 = -5/6 (-1 - v2) A2 = 5/6 (v2 + 1) A = A1 + A2 A = 5/6 (v2 - 1) + 5/6 (v2 + 1) A = 5/6 (v2 - 1 + v2 + 1) A = 5/6 (2v2) A = 5v2/3 A ˜ 2.35702 I think what you did wrong here was integrating over the whole interval, which causes negative area to cancel out with some of the positive area.

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.