Determine the area of the region bounded by y = \(2x^2\) + 10 and Y = \(4x…

Explanation

To find the point of intersection, equate the two equations

⇒ 2x\(^2\) + 10 = 4x + 16

⇒ 2x\(^2\) + 10 - 4x - 16 = 0

⇒ 2x\(^2\) - 4x - 6 = 0

Factorize

⇒ 2(x + 1)(x - 3) = 0

∴ x = -1 or 3

The curves will intersect at -1 and 3

Area = \(\int^b_a\) [upper function] - [lower function] dx

= \(\int^3_1 4x + 16 - (2x^2 + 10) dx\)

= \(\int^3_1 -2x^2 + 4x + 6dx\)

= \((\frac {-2x^3}{3} + 2x^2 + 6x)^3_1\)

= \((\frac {-2(3)^3}{3} + 2(3)^2 + 6(3)) - (\frac {-2(-1)^3}{3} + 2(-1)^2 + 6(-1))\)

= 18 - \((\frac {10}{3})\)

= 18 + \(\frac {10}{3}\)

= \(\frac {64}{3}\)
 

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