Question

Write a definite integral that represents the area of the region. (Do not evaluate the integral.) y1 = x2 + 2x + 3 y2 = 2x + 12Figure:A parabola was given in the figure

Answer 1

Answer:

$A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36$

Step-by-step explanation:

The equations are:

$y = x^{2} + 2x + 3$

$y = 2x + 12$

The two graphs intersect when:

$x^{2} + 2x + 3 = 2x + 12$

$x^{2} = 0$

$x_{1} = 3\\x_{2} = -3$

To find the area under the curve for the first equation:

$A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

To find the area under the curve for the second equation:

$A_{2} = \int\limits^3__-3}{2x + 12} \, dx$

To find the total area:

$A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

Simplifying the equation:

$A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx$

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).