calculus problem please help?

Question

FIND THE AREA UNDER THE CURVE

-3X^2+X+14

Answer 1

First, find where the curve hits the x-axis.

-3x^2 + x + 14 = 0

3x^2 - x - 14 = 0

(3x - 7 )(x + 2 )= 0

x = 7/3, -2

These are the endpoints of integration.

Then integrate the function from -2 to 7/3.

-x^3 + (1/2)x^2 + 14 x from -2 to 7/3

(8 + 2 - 28) - (-243/9 + 49/18 + 98/3)

I don't have time to finish it out ... but simplify to get the answer!

Answer 2

The area under the curve is the integral of the equation.
Int(-3x^2+X+14)dx
This becomes
-3X^3/3+X^2/2+14x+C
Which then becomes
-X^3 + .5X^2+ 14X + C