Determine the area of the region bounded by y = \(2x^2\) + 10 and Y = \(4x + 16\).
A.
18
B.
\(\frac {-10}{3}\)
C.
\(\frac {44}{3}\)
D.
\(\frac {64}{3}\)
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
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 + \(\frac {10}{3}\)
= \(\frac {64}{3}\)