What is the integral from 0 to 1 of 2pi(x)(9-8x-x^2) ?

The integral from 0 to 1 of 2pi(x)(9-8x-x^2) is (19pi)/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 2pi(x)(9-8x-x^2)

\int_{0}^{1} 2 π (x) (9 - 8 x - x^{2}) d x

\frac{19 π}{6}

Decimal

9.94837 \dots

Solution steps

\int_{0}^{1} 2 π x (9 - 8 x - x^{2}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 π \cdot \int_{0}^{1} x (9 - 8 x - x^{2}) d x

Expand

x (9 - 8 x - x^{2}) : 9 x - 8 x^{2} - x^{3}

= 2 π \cdot \int_{0}^{1} 9 x - 8 x^{2} - x^{3} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 π (\int_{0}^{1} 9 x d x - \int_{0}^{1} 8 x^{2} d x - \int_{0}^{1} x^{3} d x)

\int_{0}^{1} 9 x d x = \frac{9}{2}

\int_{0}^{1} 8 x^{2} d x = \frac{8}{3}

\int_{0}^{1} x^{3} d x = \frac{1}{4}

= 2 π (\frac{9}{2} - \frac{8}{3} - \frac{1}{4})

Simplify

2 π (\frac{9}{2} - \frac{8}{3} - \frac{1}{4}) : \frac{19 π}{6}

= \frac{19 π}{6}

g (x) = \frac{1 + 6 x}{9 - 4 x}

f^{^{'}} (x) = 3^{x}

\int + (8 x + 3 sin (x))^{2} d x

4 x^{2} + 1

4 y^{^{''}} - 4 y^{^{'}} + 10 y = 0

What is the integral from 0 to 1 of 2pi(x)(9-8x-x^2) ?
The integral from 0 to 1 of 2pi(x)(9-8x-x^2) is (19pi)/6