What is the area of the region bounded by the curve y=2x^2+2x, x axis and lines x=1 and x=2?
To find the area bound by y = 2x^2 + 2x, the x axis and the lines x=1 and x=2, we need to find the definite integral of y = 2x^2 + 2x between x = 1 and x = 2
Int [2x^2 + 2x dy]
=> 2*x^3/3 + 2*x^2/2 + C
=> (2/3)x^3 + x^2 + C
At x = 2, (2/3)x^3 + x^2 + C = 16/3 + 4 + C
At x = 1, (2/3)x^3 + x^2 + C = (2/3) + 1 + C
The definite integral is the difference = 14/3 + 3
The required area enclosed is equal to 23/3 square units.
We'll have to determine the definite integral of the given function. We'll use Leibniz-Newton formula to calculate the definite integral:
Int f(x)dx = F(b)-F(a)
We'll calculate the indefinite integral of f(x):
Int f(x)dx = Int (2x^2 + 2x)dx
We'll use the property of integral to be additive:
Int (2x^2 + 2x)dx = Int 2x^2dx + Int 2xdx
Int 2x^2dx = 2*x^3/3 + C
Int 2xdx = 2*x^2/2 + C
We'll reduce and we'll get:
Int 2xdx = x^2 + C
Int (2x^2 + 2x)dx = 2x^3/3 + x^2 + C
F(2) - F(1) = 2^4/3 + 2^2 - 2/3 - 1^2
F(2) - F(1) = 14/3 + 3
F(2) - F(1) = 23/3
The area bounded by the curve of f(x), x axis and the lines x=1, x=2 is A=23/3 square units.