find the plane region between

Question

f(x)=x² - 3x - 4 and the x-axis from x= -1 to x= 4

Answer 1

integrate x^2 -3x -4

= x^3/3 -3x^2/2 -4x

evaluate between -1 and 4

plut 4 in, then plug -1 in. Subtract them

= 4^3/3-3(4^2)/2 -4(4) - (-1/3 -3/2 -4(-1))

= 64/3 - 24 -16 +1/3 +3/2 -4

= (64-120)/3 - 13/6

= -56/3 -13/6

= (-112-13)/6

= -125/6

= -20 5/6

= total area between the upward opening parabola and the x axis

negative sign means the area is all below the x axis

(no guarantee there is no arithmetic mistake in the above somewhere, but it's the basic method. Double you money back if you find a mistake. Cheerfully refunded. But zero dollars times two is zero dollars though. That calculation has no mistakes in it guaranteed.)

It might help to graph the parabola, its x intercepts (-1,0) and (4,0), vertex (1.5,-6.25)