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Question
Find the area of the region bounded by curve 4x2 = y and the line y = 8x + 12, using integration.
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Solution
Given curve is 4x2 = y and line is y = 8x + 12
On solving both equations, we get
4x2 = 8x + 12
⇒ x2 = 2x + 3
⇒ x2 – 2x – 3 = 0
⇒ x = 3, –1
Required area = `int_-1^3 {(8x + 12) - 4x^2}dx`
= `4int_-1^3 (2x + 3 - x^2)dx`
= `4[x^2 + 3x - x^3/3]_-1^3`
= `4[(9 + 9 - 9) - (1 - 3 + 1/3)]`
= `4(9 + 2 - 1/3)`
= `4(11 - 1/3)`
= `4 xx 32/2`
= `128/2` sq.units
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