hw15solutions - Z π 1 2 (cos x + | cos x | ) dx. 1 Answer:...

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Math 2250 Written HW #15 Solutions 1. Graph the function f ( x ) = 1 + 1 - x 2 and use some geometry (and not the Fundamental Theorem of Calculus) to compute the definite integral Z 1 - 1 ± 1 + p 1 - x 2 ² dx. Answer: Recall that y = 1 - x 2 is the equation of a semicircle of radius 1 centered at the origin and lying above the x -axis. Adding 1 shifts the graph up by 1, so the graph of y = 1 + 1 - x 2 looks like this: -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -0.4 0.4 0.8 1.2 1.6 2 2.4 Hence, the area under the curve is equal to the area of a rectangle of width 2 and height 1 plus the area of a semicircle of radius 1, which is 1 2 π (1) 2 = π/ 2. Therefore, Z 1 - 1 ± 1 + p 1 - x 2 ² dx = 2 + π 2 . 2. Using properties of the definite integral, show that if f ( x ) 0 for all x in the interval [ a,b ], then Z b a f ( x ) dx 0 . In other words, the definite integral of a non-negative function is non-negative. Answer: Using property 6 from the list I wrote on the board in class, since f ( x ) 0 for all x in [ a,b ], we know that Z b a f ( x ) 0 · ( b - a ) = 0 , as desired. 3. Compute the definite integral
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Unformatted text preview: Z π 1 2 (cos x + | cos x | ) dx. 1 Answer: For x between 0 and π/ 2, cos x is nonnegative, so | cos x | = cos x . On the other hand, cos x ≤ 0 for x between π/ 2 and π , so | cos x | =-cos x on [ π/ 2 ,π ]. Therefore, using property 5 of the definite integral, Z π 1 2 (cos x + | cos x | ) dx = Z π/ 2 1 2 (cos x + | cos x | ) dx + Z π π/ 2 1 2 (cos x + | cos x | ) dx = Z π/ 2 1 2 (cos x + cos x ) dx + Z π π/ 2 1 2 (cos x-cos x ) dx = Z π/ 2 cos xdx + Z π π/ 2 dx. The second term is zero, so we just need to compute R π/ 2 cos xdx . Using the Fundamental Theorem of Calculus, since sin x is an antiderivative of cos x , we know that Z π/ 2 cos xdx = [sin x ] π/ 2 = sin( π/ 2)-sin(0) = 1-0 = 1 . Putting this all together, then, we see that R π 1 2 (cos x + | cos x | ) dx = 1. 2...
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This note was uploaded on 01/18/2012 for the course MATH 2250 taught by Professor Chestkofsky during the Fall '08 term at University of Georgia Athens.

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hw15solutions - Z π 1 2 (cos x + | cos x | ) dx. 1 Answer:...

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