Tutorial ExerciseEvaluate the integral by interpreting it in terms of areas.(1²-14) dePart 1 of 3We are concerned with the segment of the line y == 1/2/²x-14 that begins at (0, -14) and that ends at (7, 21/2✔Part 2 of 3Part 3 of 3Therefore,y10- 10-156² ²2² × -2x - 14 dx =12/1x - 14 dx can be interpreted as the area of the triangle above the x-axis minus the area of the triangle below the x-axis. The area of the lower triangle is417546XX21/2).bh =28 and the area of the upper triangle is 63/463/4.

To evaluate the following integral by interpreting in terms of areas.

Evaluate the integral by interpreting it in terms of areas. L² (3x-14) de Part 1 of 3 We are concerned with the segment of the line y=x-14 that begins at (0, -14) and that ends at (7, 21/2✔ Part 2 of 3 Part 3 of 3 y Therefore, 10 5 -51 -10 -15 2 2-14 dx s 4 x-140 <- 14 dx can be interpreted as the area of the triangle above the x-axis minus the area of the triangle below the x-axis. The area of the lower triangle is bh = 28✔ 175 4 6 X X 21/2). 28 and the area of the upper triangle is 63

Evaluate the integral by interpreting it in terms of areas. L² (3x-14) de Part 1 of 3 We are concerned with the segment of the line y=x-14 that begins at (0, -14) and that ends at (7, 21/2✔ Part 2 of 3 Part 3 of 3 y Therefore, 10 5 -51 -10 -15 2 2-14 dx s 4 x-140 <- 14 dx can be interpreted as the area of the triangle above the x-axis minus the area of the triangle below the x-axis. The area of the lower triangle is bh = 28✔ 175 4 6 X X 21/2). 28 and the area of the upper triangle is 63

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter3: Functions And Graphs

Section3.2: Graphs Of Equations

Problem 78E

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