What is area 1/4 x^2,2x^2,3-x ?

The answer to area 1/4 x^2,2x^2,3-x is 129/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area 1/4 x^2,2x^2,3-x

area

\frac{1}{4} x^{2}, 2 x^{2}, 3 - x

\frac{129}{8}

Solution steps

To find the intersection points solve

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

\frac{1}{4} x^{2} = 2 x^{2} : x = 0

\frac{1}{4} x^{2} = 3 - x : x = 2, x = - 6

2 x^{2} = 3 - x : x = 1, x = - \frac{3}{2}

= \int_{- 6}^{- \frac{3}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{3}{2}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{1} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{1}^{2} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 6}^{- \frac{3}{2}} \frac{1}{4} x^{2} - (3 - x) d x + \int_{- \frac{3}{2}}^{0} 2 x^{2} - \frac{1}{4} x^{2} d x + \int_{0}^{1} \frac{1}{4} x^{2} - 2 x^{2} d x + \int_{1}^{2} \frac{1}{4} x^{2} - (3 - x) d x

\int_{- 6}^{- \frac{3}{2}} \frac{1}{4} x^{2} - (3 - x) d x = \frac{405}{32}

\int_{- \frac{3}{2}}^{0} 2 x^{2} - \frac{1}{4} x^{2} d x = \frac{63}{32}

\int_{0}^{1} \frac{1}{4} x^{2} - 2 x^{2} d x = \frac{7}{12}

\int_{1}^{2} \frac{1}{4} x^{2} - (3 - x) d x = \frac{11}{12}

= \frac{405}{32} + \frac{63}{32} + \frac{7}{12} + \frac{11}{12}

Simplify

= \frac{129}{8}

\frac{d}{d x} (arcsin (1 - x))

\int \frac{r ^{2} - 1 0 ^{r} + 4}{r ^{3}} d r

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

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

x^{7} sin (x)