If [itex]x\ge 0[/itex], |x|= x so this is 6x= x^2- 7 which is the same as x^2- 6x- 7= 0.apiwowar said:i know that the way to solve this is by evaluating the integral from a to b of the first function minus the second one but how would i solve for x to find out what the limits of integration should be?
if you set them equal to each other you get 6|x| = x^2 - 7
but I am not exactly sure what to do with the |x|
The region enclosed by these two equations is a combination of a parabola and two line segments. It resembles a "V" shape, with the vertex at the origin and the two arms extending upwards and downwards.
The x-intercepts can be found by setting y=0 and solving for x. This gives us two points: (0,0) and (6,0). The y-intercept can be found by setting x=0, giving us a point at (0,-7).
To find the area of this region, we need to split it into two parts: the area under the parabola and the area between the two line segments. We can use integration to find the area under the parabola, and then find the area of the two triangles formed by the line segments using the formula A = 1/2 * base * height. Finally, we add these two areas together to get the total area of the region.
The domain of this region is all real numbers, as both equations are defined for any value of x. The range, however, is limited. The lower bound of the range is -7, which is the y-intercept of the parabola. The upper bound is infinity, as the parabola and line segments extend infinitely upwards and downwards.
Changing the coefficient of x in y= 6|x| affects the steepness of the two line segments that form the region. A larger coefficient will result in steeper lines, while a smaller coefficient will result in more gradual lines. The overall shape of the region will remain the same, but the size of the V will change.