Write a definite integral that represents the area of the region. (do not evaluate the integral.) y1 = x2 2x 3 y2 = 2x 12


The definite integral that represents the area of the region under the given curves is:  \int\limits^3_{-3} {x^2-9} \, dx

What is the area of the region under a curve?

  • By performing a definite integral between the two locations, one can determine the area under a curve between two points.
  • Integrate y = f(x) between the limits of a and b to determine the area under the curve y = f(x) between x = a & x = b.
  • With the specified limits, integration can be used to calculate this area.

Given:

  • y_1=x^2+2x+3
  • y_2=2x+12

The points where the two intersect will be given by:

y₁ = y₂

=> x² + 2x + 3 = 2x + 12

=> x² + 3 = 12

=> x² = 9

=> x = ± 3

For x₁ = 3, y₁ = 2(3) + 12 = 18 => (x₁, y₁) = (3, 18)

For x₂ = -3, y₂ = 2(-3) + 12 = 6 => (x₂, y₂) = (-3, 6)

Now, for the area of the region under first curve y_1=x^2+2x+3:

A₁ =  \int\limits^3_{-3} {x^2+2x+12} \, dx

For the area of the region under the second curve y_2=2x+12:

A₂ = \int\limits^3_{-3} {2x+12} \, dx

For the required area of the region bounded by the two curves will be given by:

A = A₂ - A₁ = \int\limits^3_{-3} {x^2+2x+3 - (2x +12)} \, dx

A = \int\limits^3_{-3} {x^2-9} \, dx

Refer to the image for the area so bounded by the curves.

Hence, The definite integral that represents the area of the region under the given curves is:  \int\limits^3_{-3} {x^2-9} \, dx.

To learn more about area of the region under the curves, refer to the link: brainly.com/question/15122151

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