What is the integral from 0 to 1 of 2pi(x)(8-7x-x^2) ?

The integral from 0 to 1 of 2pi(x)(8-7x-x^2) is (17pi)/6

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integral from 0 to 1 of 2pi(x)(8-7x-x^2)

\int_{0}^{1} 2 π (x) (8 - 7 x - x^{2}) d x

\frac{17 π}{6}

Decimal Notation

8.90117 \dots

Solution steps

\int_{0}^{1} 2 π x (8 - 7 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 (8 - 7 x - x^{2}) d x

Expand

x (8 - 7 x - x^{2}) : 8 x - 7 x^{2} - x^{3}

= 2 π \cdot \int_{0}^{1} 8 x - 7 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} 8 x d x - \int_{0}^{1} 7 x^{2} d x - \int_{0}^{1} x^{3} d x)

\int_{0}^{1} 8 x d x = 4

\int_{0}^{1} 7 x^{2} d x = \frac{7}{3}

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

= 2 π (4 - \frac{7}{3} - \frac{1}{4})

Simplify

2 π (4 - \frac{7}{3} - \frac{1}{4}) : \frac{17 π}{6}

= \frac{17 π}{6}

\int_{0}^{1} \frac{5}{x ^{5}} d x

\frac{d}{d x} (\frac{24}{x ^{2} + 12})

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

\frac{d y}{d t} = \frac{y}{2} + 4 e^{\frac{t}{2}}

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

What is the integral from 0 to 1 of 2pi(x)(8-7x-x^2) ?
The integral from 0 to 1 of 2pi(x)(8-7x-x^2) is (17pi)/6