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Question
The distribution of a continuous random variable X in range (– 3, 3) is given by p.d.f.
f(x) = `{{:(1/16(3 + x)^2",", - 3 ≤ x ≤ - 1),(1/16(6 - 2x^2)",", - 1 ≤ x ≤ 1),(1/16(3 - x)^2",", 1 ≤ x ≤ 3):}`
Verify that the area under the curve is unity.
Solution
Since(– 3, 3) in the range of given p.d.f
Area under the curve A = `int_oo^oo "f"(x) "d"x = int_3^3 "f"(x) "d"x`
A = `int_(-3)^(-1) "f"(x) "d"x + int_(-1)^1 "f"(x) "d"x + int_1^3 "f"(x) "d"x`
A = `int_(-3)^(-1) 1/16 (3 + x)^2 "d"x + int_(-1)^1 1/16 (6 - 2x^2) "d"x + int_1^3 1/16 (3 - x)^2 "d"x`
= `1/16{[(3 + x)^3/3]_(-3)^(-1) + [6x - (3x^3)/3]_(-1)^1 + [(3 - x)^3/(3(-1))]_1^3}`
= `1/16 {(3 - 1)^3/3 - (3 - 3)^3/3 + [6 - 2/3] - [-6 + 2/3] + [((3 - 3)/(-3))^3] - ((3 - 1)/(-3))^3}`
= `1/16 {(2)^3/3 - 0/3 + 6 - 2/3 + 6 - 2/3 + 0 + (2)^3/3}`
= `1/16{8/3 + 6 - 2/3 + 6 - 2/3 + 8/3}`
= `1/16 {12 + 12/3}`
= `1/16 {12 + 4}`
= `1/16 {16}`
= 1
∴ A = 1 sq.units
Hence the Area under the curve is unity.
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