What is area y=x^2,y=2,y=5 ?

The answer to area y=x^2,y=2,y=5 is -(2(5sqrt(5)-2sqrt(2)))/3-4sqrt(2)+10sqrt(5)

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area y=x^2,y=2,y=5

area

y = x^{2}, y = 2, y = 5

- \frac{2 ( 5 5 - 2 2 )}{3} - 42 + 105

Decimal Notation

11.13588 \dots

Solution steps

To find the intersection points solve

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

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

x^{2} = 5 : x = 5, x = - 5

2 = 5 :

False

= \int_{- 5}^{- 2} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- 2}^{2} ∣ f_{3} (x) - f_{2} (x) ∣ d x + \int_{2}^{5} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 5}^{- 2} x^{2} - 5 d x + \int_{- 2}^{2} ∣ 5 - 2 ∣ d x + \int_{2}^{5} x^{2} - 5 d x

\int_{- 5}^{- 2} x^{2} - 5 d x = - \frac{- 2 2 + 5 5}{3} - 52 + 55

\int_{- 2}^{2} ∣ 5 - 2 ∣ d x = 62

\int_{2}^{5} x^{2} - 5 d x = - \frac{5 5 - 2 2}{3} + 55 - 52

= - \frac{- 2 2 + 5 5}{3} - 52 + 55 + 62 - \frac{5 5 - 2 2}{3} + 55 - 52

Simplify

= - \frac{2 ( 5 5 - 2 2 )}{3} - 42 + 105

\frac{d x}{d t} = - x

y = x β + e

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

x \to 1 + lim (x - 1)

x \to 0 + lim (e^{\frac{1}{x} l n (x)})

What is area y=x^2,y=2,y=5 ?
The answer to area y=x^2,y=2,y=5 is -(2(5sqrt(5)-2sqrt(2)))/3-4sqrt(2)+10sqrt(5)