What is area 4x+12,x^2,8-x^2 ?

The answer to area 4x+12,x^2,8-x^2 is 320/3

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area 4x+12,x^2,8-x^2

area

4 x + 12, x^{2}, 8 - x^{2}

\frac{320}{3}

Decimal

106.66666 \dots

Solution steps

To find the intersection points solve

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

4 x + 12 = x^{2} : x = 6, x = - 2

4 x + 12 = 8 - x^{2} : x = - 2

x^{2} = 8 - x^{2} : x = 2, x = - 2

= \int_{- 2}^{2} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- 2}^{6} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 2}^{2} x^{2} - (8 - x^{2}) d x + \int_{- 2}^{6} 4 x + 12 - x^{2} d x

\int_{- 2}^{6} 4 x + 12 - x^{2} d x = \frac{256}{3}

\int_{- 2}^{2} x^{2} - (8 - x^{2}) d x = \frac{64}{3}

= \frac{64}{3} + \frac{256}{3}

Simplify

= \frac{320}{3}

4^{2 x^{2}}

2 + 3 x

\frac{d}{d x} (cos^{2} (3 x))

\frac{\partial}{\partial x} (- sin (x^{2} y) \cdot x^{2})

\frac{\partial}{\partial y} (\frac{( 3 x + y )}{z})