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Question
If `int_0^"a" (2x + 1) "d"x` = 2, find a
Solution
Given, `int_0^"a" (2x + 1) "d"x` = 2
∴ `[(2x^2)/2 + x]_0^"a"` = 2
∴ `[x^2 + x]_0^"a"` = 2
∴ [(a2 + a) – (0)] = 2
∴ a2 + a = 2
∴ a2 + a – 2 = 0
∴ a2 + 2a – a – 2 = 0
∴ a(a + 2) – 1(a + 2) = 0
∴ (a + 2)(a – 1) = 0
∴ a + 2 = 0 or a – 1 = 0
∴ a = – 2 or a = 1
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