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March 9, 2006 Lecturer: Dr Martin Kurth Hilary Term 2006 Problem Sheet 5 – Solutions 1. First, we have to find the points where the graphs of f and g intersect. f (x) = g(x) 1 1 3 = − x+ x 2 2 leading to x2 − 3x + 2 = 0 with solutions a = 1, b = 2. Now we apply our formula for washer cross sections: Z b ³ ´ 2 2 π [g(x)] − [f (x)] dx, V = a which in our case (after a bit of rearranging) gives ¶ Z 2 µ 9 1 1 2 3 x − x + − 2 dx V = π 4 2 4 x 1 ¶¯2 µ 1 ¯¯ 1 3 3 2 9 = π x − x + x+ 12 4 4 x ¯1 π = . 12 2. As the functions f and g are the same as in question 1, we still have a = 1, b = 2. The formula for the volume generated by revolving the area between the x-axis and the graph of g about the line x = 1/2 is ¶ Z b µ 1 2π x − g(x)dx, Vg = 2 a and the corresponding formula for f ¶ Z b µ 1 f (x)dx. Vf = 2π x − 2 a To get the volume V of the solid generated by revolving the area between the graphs, we have to subtract Vf from Vg , which gives ¶ Z 2 µ 1 V = [g(x) − f (x)] dx 2π x − 2 1 1 = = = = 2 ¶· ¸ 1 1 3 1 − x+ − dx 2π x − 2 2 2 x 1 ¶ Z 2 µ 7 1 7 dx π −x2 + x − + 2 2 x 1 µ ¶¯2 ¯ 1 7 7 π − x3 + x2 − x + ln |x| ¯¯ 3 4 2 1 µ ¶ 7 π − + ln 2 . 12 Z µ 3. We start by defining a function f by f (x) = x − 1 − ln x. Taking its derivative, we get f 0 (x) = 1 − 1 , x which for x ≥ 1 means f 0 (x) ≥ 0. Now we write f (x) as f (x) = f (1) + Z x f 0 (t)dt, 1 and as the integrand is ≥ 0, so is the integral (by the domination rule), ie f (x) ≥ f (1) = 0, proving the statement. 2