# Which definite integral would you use to compute the area enclosed by the parabola y=-x^2+4 and the line y=2x-3?

Jun 1, 2018

color(blue)[A=int_(-3.828)^(1.828)(-x^2-2x+7)*dx

#### Explanation:

the area Between two curves due to $\text{x-axis}$ given by:

color(red)[A=int_a^by_2-y_1*dx

where ${y}_{2}$ is the curve in the top and ${y}_{1}$ is the curve bottom.

in your question the area between the curves is given by:

color(blue)[A=int_(-3.828)^(1.828)(-x^2+4)-(2x-3)*dx

color(blue)[A=int_(-3.828)^(1.828)(-x^2-2x+7)*dx

show the figure below the area between the two curves:

to find the cross between the curve :

${y}_{2} = {y}_{1}$

$- {x}^{2} + 4 = 2 x - 3$

${x}^{2} + 2 x - 7 = 0$

After solving it you will get:

$x = - 3.828 \mathmr{and} x = 1.828$