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

The integral from 0 to 1 of 2pi(x)(7-6x-x^2) is (5pi)/2

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integral from 0 to 1 of 2pi(x)(7-6x-x^2)

\int_{0}^{1} 2 π (x) (7 - 6 x - x^{2}) d x

\frac{5 π}{2}

Decimal Notation

7.85398 \dots

Solution steps

\int_{0}^{1} 2 π x (7 - 6 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 (7 - 6 x - x^{2}) d x

Expand

x (7 - 6 x - x^{2}) : 7 x - 6 x^{2} - x^{3}

= 2 π \cdot \int_{0}^{1} 7 x - 6 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} 7 x d x - \int_{0}^{1} 6 x^{2} d x - \int_{0}^{1} x^{3} d x)

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

\int_{0}^{1} 6 x^{2} d x = 2

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

= 2 π (\frac{7}{2} - 2 - \frac{1}{4})

Simplify

2 π (\frac{7}{2} - 2 - \frac{1}{4}) : \frac{5 π}{2}

= \frac{5 π}{2}

x \to \infty lim (x - 3 x^{\frac{1}{3}})

(- 7.2) (8.2)

\frac{7 - e ^{\frac{- t}{5}}}{6}

\frac{d}{d x} (1^{\frac{1}{2}})

y^{^{''}} + 12 y^{^{'}} + 36 y = e - 7 x (x^{2} + 3 x + 6)

What is the integral from 0 to 1 of 2pi(x)(7-6x-x^2) ?
The integral from 0 to 1 of 2pi(x)(7-6x-x^2) is (5pi)/2