What is the integral from 1 to 8 of pi(x-1) ?

The integral from 1 to 8 of pi(x-1) is pi 49/2

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integral from 1 to 8 of pi(x-1)

\int_{1}^{8} π (x - 1) d x

π \frac{49}{2}

Decimal Notation

76.96902 \dots

Solution steps

\int_{1}^{8} π (x - 1) d x

Take the constant out:

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

= π \cdot \int_{1}^{8} x - 1 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

= π (\int_{1}^{8} x d x - \int_{1}^{8} 1 d x)

\int_{1}^{8} x d x = \frac{63}{2}

\int_{1}^{8} 1 d x = 7

= π (\frac{63}{2} - 7)

Join

\frac{63}{2} - 7 : \frac{49}{2}

= π \frac{49}{2}

f (x) = 3 x^{2} + 8 x + 4, at x = - 1

x \to 0 lim (x^{3} + 2 x^{2} - x - 2)

x \to 0 - lim (\frac{x + ∣ x ∣}{x})

\frac{d}{d x} (\frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x ^{3}})

x \to 0 lim (\frac{x - 1}{x ^{3} + x})

What is the integral from 1 to 8 of pi(x-1) ?
The integral from 1 to 8 of pi(x-1) is pi 49/2