Find the area or the region under the graph of f(x)=3x^2+6x and above the x-axis from x=0 to x=3

Answer 1

F(x) = x^3 +3x^2 = (3)^3 + 3(3)^2 = 27+27 = 54

Answer 2

F(x) = x^3 +3x^2 = (3)^3 + 3(3)^2 = 27+27 = 54

Answer 3

To solve a problem like this, we can set up the following integral

3

∫ 3x^2 + 6x dx

0

Evaluating the integral, we get

x^3 + 3x^2 [evaluated from 0 to 3]

(3)^3 + 3(3)^2 - (0)^3 + 3(0)^2

= 27 + 27 - 0 - 0

= 54