What is area x^2+4x+3,y=-2,-2<x<0 ?

The answer to area x^2+4x+3,y=-2,-2<x<0 is 14/3

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area x^2+4x+3,y=-2,-2<x<0

area

x^{2} + 4 x + 3, y = - 2, - 2 < x < 0

\frac{14}{3}

Decimal

4.66666 \dots

Solution steps

Apply the area formula:

\int_{- 2}^{0} x^{2} + 4 x + 5 d x

Solve

\int_{- 2}^{0} x^{2} + 4 x + 5 d x : \frac{14}{3}

The area is:

= \frac{14}{3}

51

x = y + 2, x = y^{2}

\frac{1}{7 - 6 e ^{x}}

\frac{\partial}{\partial x} (- 400 x (y - x^{2}) - 2 (1 - x))

y^{^{'}} = sin (x - y)

What is area x^2+4x+3,y=-2,-2<x<0 ?
The answer to area x^2+4x+3,y=-2,-2<x<0 is 14/3