What is area \sqrt[4]{x},sqrt(2-x),0 ?

The answer to area \sqrt[4]{x},sqrt(2-x),0 is 22/15

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area \sqrt[4]{x},sqrt(2-x),0

area

\frac{22}{15}

Solution steps

To find the intersection points solve

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

2 - x = 0 : x = 2

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

\int_{1}^{2} 0 - 2 - x d x = \frac{2}{3}

= \frac{4}{5} + \frac{2}{3}

Simplify

= \frac{22}{15}

(3 x^{2} + 1)^{2}

\frac{d}{d x} (- 4 sin (x) - 9 ln (x) + 11)

\frac{d ^{5}}{d x ^{5}} (\frac{1}{x - 1})

\int (- 2 sec^{2} (x)) d x

\int_{1}^{2} \frac{2}{x ^{3}} d x

What is area \sqrt[4]{x},sqrt(2-x),0 ?
The answer to area \sqrt[4]{x},sqrt(2-x),0 is 22/15