What is the integral from 1 to 3 of 2pix(-x^2+4x-3) ?

The integral from 1 to 3 of 2pix(-x^2+4x-3) is (16pi)/3

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integral from 1 to 3 of 2pix(-x^2+4x-3)

\int_{1}^{3} 2 π x (- x^{2} + 4 x - 3) d x

\frac{16 π}{3}

Decimal

16.75516 \dots

Solution steps

\int_{1}^{3} 2 π x (- x^{2} + 4 x - 3) d x

Take the constant out:

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

= 2 π \cdot \int_{1}^{3} x (- x^{2} + 4 x - 3) d x

Expand

x (- x^{2} + 4 x - 3) : - x^{3} + 4 x^{2} - 3 x

= 2 π \cdot \int_{1}^{3} - x^{3} + 4 x^{2} - 3 x 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_{1}^{3} x^{3} d x + \int_{1}^{3} 4 x^{2} d x - \int_{1}^{3} 3 x d x)

\int_{1}^{3} x^{3} d x = 20

\int_{1}^{3} 4 x^{2} d x = \frac{104}{3}

\int_{1}^{3} 3 x d x = 12

= 2 π (- 20 + \frac{104}{3} - 12)

Simplify

2 π (- 20 + \frac{104}{3} - 12) : \frac{16 π}{3}

= \frac{16 π}{3}

d er i v a t i v e y = x^{\frac{3}{2}} (x + c e^{x})

\int 10 - (14 - 2 (16 - 2 x)^{\frac{1}{2}}) d x

\int x^{4} - 8 x^{2} + 16 d x

x \to - 3 lim (\frac{x}{( x + 3 ) ^{4}})

y^{^{''}} - k y = k x e^{- x}

What is the integral from 1 to 3 of 2pix(-x^2+4x-3) ?
The integral from 1 to 3 of 2pix(-x^2+4x-3) is (16pi)/3