Chapter 1.1, Problem 23E

Find the area of the shaded region. (a) 5- y 3 2- (1,4) (5,0) 1 3 4 5 6 (b) 3 y 2 Decide whether the problem can be solved using precalculus, or whether calculus is required. If the problem can be solved using precalculus, solve it. If the problem seems to require calculus, use a graphical or numerical approach to estimate the solution. STEP 1: Consider the figure in part (a). Since this region is simply a triangle, you may use precalculus methods to solve this part of the problem. First determine the height of the triangle and the length of the triangle's base. height 4 units units base 5 STEP 2: Compute the area of the triangle by employing a formula from precalculus, thus finding the area of the shaded region in part (a). 10 square units STEP 3: Consider the figure in part (b). Since this region is defined by a complicated curve, the problem seems to require calculus. Find an approximation of the shaded region by using a graphical approach. (Hint: Treat the shaded regi as…

Solve this differential equation: dy 0.05y(900 - y) dt y(0) = 2 y(t) =

0 5 -1 2 1 N = 1 to x = 3 Based on the graph above, estimate to one decimal place the average rate of change from x =

Complete the description of the piecewise function graphed below. Use interval notation to indicate the intervals. -7 -6 -5 -4 30 6 5 4 3 0 2 1 -1 5 6 + -2 -3 -5 456 -6 - { 1 if x Є f(x) = { 1 if x Є { 3 if x Є

Complete the description of the piecewise function graphed below. 6 5 -7-6-5-4-3-2-1 2 3 5 6 -1 -2 -3 -4 -5 { f(x) = { { -6 if -6x-2 if -2< x <1 if 1 < x <6

Let F = V where (x, y, z) x2 1 + sin² 2 +z2 and let A be the line integral of F along the curve x = tcost, y = t sint, z=t, starting on the plane z = 6.14 and ending on the plane z = 4.30. Then sin(3A) is -0.598 -0.649 0.767 0.278 0.502 0.010 -0.548 0.960

Let C be the intersection of the cylinder x² + y² = 2.95 with the plane z = 1.13x, with the clockwise orientation, as viewed from above. Then the value of cos (₤23 COS 2 y dx xdy+3 z dzis 3 z dz) is 0.131 -0.108 -0.891 -0.663 -0.428 0.561 -0.332 -0.387

2 x² + 47 The partial fraction decomposition of f(x) g(x) can be written in the form of + x3 + 4x2 2 C I where f(x) = g(x) h(x) = h(x) + x +4