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Question
Find the area of the region bounded by the curve between the parabola y = 8x2 – 4x + 6 the y-axis and the ordinate at x = 2
Solution
Equation of the parabola
y = 8x2 – 4x + 6
The required region is bounded by the y-axis and the ordinate at x = 2.
∴ Required Area A = `int_0^2 y "d"x`
A = `int_0^2 (8x^2 - 4x + 6) "d"x`
= `[8(x^3/3) - 4(x^2/2) + 6x]_0^2`
= `[8/3 x^3 - 2x^2 + 6x]_0^2`
= `[(8/3) (2)^3 - 2(2)^2 + 6(2)] - [0]`
= `64/3 - 8 + 12`
= `64/3 + 4`
= `(64 + 12)/3`
A = `76/3` sq.units
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