What is area y=sqrt(3-x),y=-sqrt(3-x),y=x-1 ?

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

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area y=sqrt(3-x),y=-sqrt(3-x),y=x-1

area

y = 3 - x, y = - 3 - x, y = x - 1

\frac{9}{2}

Decimal Notation

4.5

Solution steps

To find the intersection points solve

f_{i} (x) = f_{j} (x)

3 - x = - 3 - x : x = 3

3 - x = x - 1 : x = 2

- 3 - x = x - 1 : x = - 1

= \int_{- 1}^{2} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{2}^{3} ∣ f_{1} (x) - f_{2} (x) ∣ d x

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

\int_{- 1}^{2} - 3 - x - (x - 1) d x = \frac{19}{6}

\int_{2}^{3} 3 - x - (- 3 - x) d x = \frac{4}{3}

= \frac{19}{6} + \frac{4}{3}

Simplify

= \frac{9}{2}

x \to 1 lim (4 x^{2} - 7 x + 5)

\int e^{x} cos (5 x) d x

- 60 = \frac{d ^{2} y}{d x ^{2}}

f (x) = 4 e^{0.5 x}

\frac{\partial}{\partial x} (x^{5} + x^{2} y + x + 5)

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