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Unformatted text preview: MATH 138 Assignment 1solution Due by: 11 a.m. , Friday, January 15. 1. This exercise is intended to review some aspects of the fundamental theorem of calculus, the increasing function theorem, as well as inverse functions. The calculations are not hard, but you need to keep the concepts straight in your head. (a) Prove that the integral function g ( x ) = Z x 1 1 t 2 dt is increasing on the interval [ 1 , 1] . This takes very little writing. (b) Sketch the graph of g over [ 1 , 1] . Hint. This can be done by thinking of g ( x ) as the area from 1 to x under the semi circle y = 1 t 2 . There is no need to explicitly calculate g ( x ) . (c) What is the range of [ 1 , 1] under the function g ? Again, think of area under y = 1 t 2 . (d) Explain brie f y why g (0) = / 4 . If h is the inverse function of g , F nd h ( / 4) . Solution: (a) By the fundamental theorem of calculus, we have g ( x ) = 1 x 2 , which is for x [ 1 , 1] . Thus g ( x ) is increasing. (b) Since g ( x ) = 1 x 2 , we have g ( x ) = 1 2 (1 x 2 ) 1 / 2 ( 2 x ) = x 1 x 2 . Thus g ( x ) is increasing, and is concave upward on [ 1 , 0] and downward on [0 , 1] . x 6 y y = g ( x ) 1 1 (c) Notice that g ( 1) = 0 . Also, since g (1) represents the area of a semicircle of radius 1, we have g (1) = / 2 . Since g ( x ) is continuous and increasing, the range of g is [0 , / 2] . (d) Since g (0) represents 1 / 4 of the the area of a circle of radius 1, we have g (0) = / 4 . Also, if h ( g ( x )) = x , then h ( g ( x )) g ( x ) = 1 . By putting x = 0 and recalling g ( x ) = 1 x 2 , we have h ( g (0)) g (0) = 1 = h ( / 4) = 1 g (0) = 1 1 2 = 1 . 1 2. Find the following inde f nite integrals. A suitable substitution should work in each case. (a) Z 1 x ln x dx, (b) Z arctan x 1 + x 2 dx, (c) Z sin 5 x cos x dx, (d) Z dx 1 + sin x ....
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This note was uploaded on 01/06/2011 for the course MATH 138 taught by Professor Anoymous during the Winter '07 term at Waterloo.
 Winter '07
 Anoymous
 Fundamental Theorem Of Calculus, Inverse Functions

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