Next ph, the horizontal axis, and the vertical lines at t = 0 and Let A(I) represent the area bounded b t = T for the graph below. Evaluate A(I) for a - 1, 2, 3, and 4. A(1) = A(2) - A(3) 4(4) > Next Question hp . . . .... D.I 6 f2 4101 ? # $ % & 3 4 5 6 W E R T Y U S D F\fSolve the separable differential equation 8 dt and find the particular solution satisfying the initial condition (0) - 9. z(t) > Next Question hp . . ..... . . 18 12 10I 10 # % & 2 3 5y(v)do = 18 and y(v)du = 4.9 what does the following integral equal? y(v) du - > Next Question . hp 13 15 16 10 18 14The shaded region between the graphs of y - - 2x- + 20 and y = 3x - 15 is displayed below in -6 + Ax - -40 orange. The shaded region lies between,I - and z = A typical rectangle is shown in the diagram located at position x. What is the height of this rectangle, in terms of ? Now find the total area of the orange shaded region. A = > Next Question hp . . . . . . . . 18 fio 6 $7 144 D.1 13 # % 3 4 5 6 8 N E R T Y U OEvaluate the definite integral . 8 (18x2 - 81 + 4) dr > Next Question op . . . ..... ..... ..... ..... fg 144 6 12 13 f4 " 15 # % 2 6 8 3 4 5\fTo find the blue shaded area above, we would calculate: f(z)di - area Where: a = . 6 6 - f(z) - Area = > Next Question hp . . . . 6 .... ... .. .... . .... .... . . . . 8 12 10 144 $ % - 2 3 4 5 6 8 WEvaluate the integral ( 72 2 + 9)4 de by making the appropriate substitution: u = + 0 (7:2 + 9)4 > Next Question hp . . . . . .. ... . . .... . .. ..... . . . . . . ... fg 10 18 144 6 $7 # $ % & - 2 5 6 8 3 4 W E R TSolve y' - 12xy - 2z, y(0) - 13 y(z) = hp . . . . . ... ..... . . ..... .... .. . . 18 fg F10 16 10 4- 12 $3 % & 9 - 2 # 4 5 6 8 3 W E R T Y U