19fa-1910-homework02-solutions.pdf - H OMEWORK 2 S OLUTIONS...

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H OMEWORK 2 S OLUTIONS Math 1910, Fall 2019 5.4.24 Evaluate the integral using FTC I: 8 / 27 t 2 dt Z 1 10t 4 / 3 - 8t 1 / 3
5.4.30 Evaluate the integral using FTC I: 0 sec θ tan θdθ
5.4.50 Show that the area of the shaded parabolic arch in Figure 7 is equal to four thirds of the area of the triangle shown.
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The indicated triangle has a base of length b - a and a height of ( a + b 2 - a )( b - a + b 2 ) = ( b - a 2 ) 2 . Then, the area of the triangle is 1 2 ( b - a )( b - a 2 ) 2 = 1 8 ( b - a ) 3 . Finally, we note that 1 6 ( b - a ) 3 = 4 3 · 1 8 ( b - a ) 3 , as required. 5.4.52 (a) Apply the Comparison theorem to the inequality sin x x (valid for x 0 ) to prove that 1 - x 2 2 cos x 1 . (b) Apply it again to prove that x - x 3 6 sin x x ( for x 0

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