What is the area of the region bounded by the curve y=2x^2+2x, x axis and lines x=1 and x=2?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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