Barrons AP Calculus - David Bock
(C) Note that dx = sec^2 θ dθ and that Be sure to express the limits as values of θ: 1 = tan θ yields ...
(B) If then u^2 = x + 1, and 2u du = dx. When you substitute for the limits, you get Since u ≠ 0 on its interval of integration ...
(D) f ′(x) dx = f (0) − f (8) = 11 − 7 = 4 ...
(D) On [0,6] with n = 3, Δx = 2. Heights of rectangles at x = 1, 3, and 5 are 5, 9, and 5, respectively; M(3) = (5 + 9 + 5)(2). ...
28. (D) ...
(E) For L(2) use the circumscribed rectangles: for R(2) use the inscribed rectangles: ...
(D) On [0, 1] f (x) = cos x is decreasing, so R < L. Furthermore, f is concave downward, so T < A. ...
(D) On the interval [−1,3] the area under the graph of y = |x| is the sum of the areas of two triangles: ...
(E) Note that the graph y = |x + 1| is the graph of y = |x| translated one unit to the left. The area under the graph y = |x + ...
(C) Because is a semicircle of radius 8, its area is 32π. The domain is [−8,8], or 16 units wide. Hence the average height of t ...
(C) The average value is equal to ...
(C) The average value is equal to ...
(E) The average value is The integral represents the area of a trapezoid: (5 + 3) · 10 = 40. The average value is (40). ...
(B) Since x^2 + y^2 = 16 is a circle, the given integral equals the area of a semicircle of radius 4. ...
(B) Use a graphing calculator. ...
(D) A vertical line at x = 2 divides the area under f into a trapezoid and a triangle; hence, Thus, the average value of f on [ ...
40. (B) ...
(D) g ′(x) = f (2x) · 2; thus g ′(1) = f (2) · 2 ...
(C) (Why 14? See the solution for question 39.) ...
(C) This is the Mean Value Theorem for Integrals. ...
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