What is area (x+2)^3,2x^2-8x+8,0 ?

The answer to area (x+2)^3,2x^2-8x+8,0 is 28/3

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area (x+2)^3,2x^2-8x+8,0

area

(x + 2)^{3}, 2 x^{2} - 8 x + 8, 0

\frac{28}{3}

Decimal Notation

9.33333 \dots

Solution steps

To find the intersection points solve

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

(x + 2)^{3} = 2 x^{2} - 8 x + 8 : x = 0

(x + 2)^{3} = 0 : x = - 2

2 x^{2} - 8 x + 8 = 0 : x = 2

= \int_{- 2}^{0} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{0}^{2} ∣ f_{3} (x) - f_{2} (x) ∣ d x

= \int_{- 2}^{0} (x + 2)^{3} - 0 d x + \int_{0}^{2} 0 - (2 x^{2} - 8 x + 8) d x

\int_{- 2}^{0} (x + 2)^{3} - 0 d x = 4

\int_{0}^{2} 0 - (2 x^{2} - 8 x + 8) d x = \frac{16}{3}

= 4 + \frac{16}{3}

Simplify

= \frac{28}{3}

x \to 1 lim (x + 4 - 12)

f (x) = x + 2, at x = 1

f (x) = lo g_{4} (3 x^{2} + 1)

x \to 14 - lim (15 x + ln (14 - x))

(1 + e^{- t})^{2}

What is area (x+2)^3,2x^2-8x+8,0 ?
The answer to area (x+2)^3,2x^2-8x+8,0 is 28/3