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# 8.8.32-eg - q x 2 lim b →-2 Z 1-2 dx √ x 2 = lim a...

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Example Evaluate Z 1 - 4 dx q | x + 2 | . Since x + 2 0 when x ≥ - 2, it follows that | x + 2 | = x + 2 if x ≥ - 2 | x + 2 | = - ( x + 2) if x < - 2 . We remove the absolute value signs by splitting the integral into Z 1 - 4 dx q | x + 2 | = Z - 2 - 4 dx q - ( x + 2) + Z 1 - 2 dx x + 2 . Note that the function in the integrand is undefined at x = - 2 thus both integrals are improper. Therefore
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Unformatted text preview: q-( x + 2) + lim b →-2 + Z 1-2 dx √ x + 2 = lim a →-2--2 q-( x + 2) ± ± ± ± ± a-4 + lim b →-2 + 2 √ x + 2 ± ± ± ± ± 1 b = lim a →-2-²-2 q-( a + 2) + 2 √ 2 ³ + lim b →-2 + ´ 2 √ 3-2 √ b + 2 µ = 2 √ 2 + 2 √ 3 ....
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