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asst1soln

asst1soln - MATH 138 Assignment 1-solution Due by 11 a.m...

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MATH 138 Assignment 1-solution 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 ) = 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 fl y why g (0) = π/ 4 . If h is the inverse function of g , fi nd h ( π/ 4) . Solution: (a) By the fundamental theorem of calculus, we have g ( x ) = 1 x 2 , which is 0 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 y y = g ( x ) 1 1 (c) Notice that g ( 1) = 0 . Also, since g (1) represents the area of a semi-circle of radius 1, we have g (1) = π/ 2 . Since g ( x ) is continuous and increasing, the range of g is [0 , π/ 2] . (d) Since g (0) represents 1 / 4 of the the area of a circle of radius 1, we have g (0) = π/ 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 ( π/ 4) = 1 g (0) = 1 1 0 2 = 1 . 1

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2. Find the following inde fi nite integrals. A suitable substitution should work in each case. (a) 1 x ln x dx, (b) arctan x 1 + x 2 dx, (c) sin 5 x cos x dx, (d) dx 1 + sin x . Hint for (d). Multiply the top and bottom by 1 sin x , use a trigonometric identity, and then make a substitution. Solution: (a) Let u = ln x and du = 1 x dx . Then 1 x ln x dx = 1 d du = ln | u | + C = ln | ln x | + C.
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