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Calculus 5e_Part211 - 420 CHAPTER 5 INTEGRALS and so...

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420 ❙❙❙❙ CHAPTER 5 INTEGRALS and so Equation 8 becomes (a) If is even, then so Equation 9 gives (b) If is odd, then and so Equation 9 gives Theorem 7 is illustrated by Figure 4. For the case where is positive and even, part (a) says that the area under from to is twice the area from to because of symmetry. Recall that an integral can be expressed as the area above the -axis and below minus the area below the axis and above the curve. Thus, part (b) says the integral is because the areas cancel. EXAMPLE 10 Since satisfies , it is even and so EXAMPLE 11 Since satisfies , it is odd and so y 1 ± 1 tan x 1 ² x 2 ² x 4 dx ± 0 f ± ± x ² ± ± f ± x ² f ± x ² ± ± tan x ²³± 1 ² x 2 ² x 4 ² ± 2[ 1 7 x 7 ² x ] 0 2 ± 2( 128 7 ² 2) ± 284 7 y 2 ± 2 ± x 6 ² 1 ² ± 2 y 2 0 ± x 6 ² 1 ² f ± ± x ² ± f ± x ² f ± x ² ± x 6 ² 1 0 y ± f ± x ² x x b a f ± x ² a 0 a ± a y ± f ± x ² f y a ± a f ± x ² ± ± y a 0 f ± u ² du ² y a 0 f ± x ² ± 0 f ± ± u ² ± ± f ± u ² f y a ± a f ± x ² ± y a 0 f ± u ² du ² y a 0 f ± x ² ± 2 y a 0 f ± x ² f ± ± u ² ± f ± u ² f y a ± a f ± x ² ± y a 0 f ± ± u ² du ² y a 0 f ± x ² 9 7–44 |||| Evaluate the indefinite integral.
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