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

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}

\int \frac{1}{1 - 4 x ^{2}} d x

\int \frac{1}{121 - x ^{2}} d x

d er i v a t i v e sin (x) + \frac{1}{2} cot (x)

d er i v a t i v e y = \frac{ln ( x )}{1 + ln ( x )}

d er i v a t i v e cos (3 x^{2})

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