find the area between the curve y= x^2+4x and the x - axis from x=-2 to x= 0.

Answer 1

Recall the Fundamental Theorem of Calculus and Riemann sums. You just want
∫[-2,0] (0 - (x^2 + 4x)) dx = 16/3

Answer 2

Area under a curve - region bounded by the given function, vertical
lines and the x axis =
a
∫ y (x) dx
b

In this case:

∫ ( x² + 4 x ) dx = x³ / 3 + 4 x² / 2 = x³ / 3 + 2 x² + C

Area =

2
∫ ( x² + 4 x ) dx =
0

0
| x³ / 3 + 2 x² | = F(0) - F(2) =
- 2

0³ / 3 + 2 ∙ 0² - [ ( - 2 )³ / 3 + 2 ∙ ( - 2 )² ] = 0 - ( - 8 / 3 + 2 ∙ 4 ) =

0 - ( - 8 / 3 + 8 ) = 0 - ( - 8 / 3 + 24 / 3 ) = 0 - ( 16 / 3 ) = - 16 / 3

A negative area value means the area is below the x-axis

The area cannot be negative, so the area value is:

A = | - 16 / 3 |

A = 16 / 3

Answer 3

My typo.

Area is not

2
∫ ( x² + 4 x ) dx
0

Area =
0
∫ ( x² + 4 x ) dx =
- 2