What is area y=x^2-9,y=(2x-1)(x+3) ?

The answer to area y=x^2-9,y=(2x-1)(x+3) is 1/6

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area y=x^2-9,y=(2x-1)(x+3)

area

y = x^{2} - 9, y = (2 x - 1) (x + 3)

\frac{1}{6}

Decimal

0.16666 \dots

Solution steps

Apply the area formula:

\int_{- 3}^{- 2} x^{2} - 9 - (2 x - 1) (x + 3) d x

Solve

\int_{- 3}^{- 2} x^{2} - 9 - (2 x - 1) (x + 3) d x : \frac{1}{6}

The area is:

= \frac{1}{6}

\frac{d}{d x} (x e^{2 x y})

\frac{d}{d x} (9 (3 x + 5)^{2})

(x, y) \to (0, 3) lim (x^{3} y)

((x^{3})^{2})^{2}

\frac{\partial}{\partial x} (\frac{1}{y - 1})

What is area y=x^2-9,y=(2x-1)(x+3) ?
The answer to area y=x^2-9,y=(2x-1)(x+3) is 1/6