Determine the area of the region bounded by y = $2x^2$ + 10 and Y = $4x + 16$ .

$\frac {-10}{3}$

$\frac {44}{3}$

$\frac {64}{3}$

Correct answer is D

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

Open photo
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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