What is area y= 1/5 x^2-6/5 x+7/5 ,(-5,5) ?

The answer to area y= 1/5 x^2-6/5 x+7/5 ,(-5,5) is (460+16sqrt(2))/(15)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area y= 1/5 x^2-6/5 x+7/5 ,(-5,5)

area

y = \frac{1}{5} x^{2} - \frac{6}{5} x + \frac{7}{5}, (- 5, 5)

\frac{460 + 16 2}{15}

Decimal Notation

32.17516 \dots

Solution steps

Apply the area formula:

\int_{- 5}^{5} \frac{1}{5} x^{2} - \frac{6}{5} x + \frac{7}{5} d x

Solve

\int_{- 5}^{5} \frac{1}{5} x^{2} - \frac{6}{5} x + \frac{7}{5} d x : \frac{460 + 16 2}{15}

The area is:

= \frac{460 + 16 2}{15}

x \to - 1 lim (e^{x - 2})

\int x sin (4 - x) d x

\int ln^{2} (x) d x

y^{^{''}} - 4 y^{^{'}} + 4 y = 12 t e^{2 t} + 6 e^{2 t} + 24 - 12 t

\int \frac{1}{( D \cdot E ) - ( D \cdot \frac{F}{G} x )} d x

What is area y= 1/5 x^2-6/5 x+7/5 ,(-5,5) ?
The answer to area y= 1/5 x^2-6/5 x+7/5 ,(-5,5) is (460+16sqrt(2))/(15)