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

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

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

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

- 3

Solution steps

\int_{1}^{4} x^{2} - 4 x + 2 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

= \int_{1}^{4} x^{2} d x - \int_{1}^{4} 4 x d x + \int_{1}^{4} 2 d x

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} 4 x d x = 30

\int_{1}^{4} 2 d x = 6

= 21 - 30 + 6

Simplify

= - 3

y = 12 x sin (x)

9 x^{2} \frac{d y}{d x} = \frac{d y}{d x} + 2 x e^{- y}

t^{4}

\frac{\partial}{\partial x} (ln (\frac{x}{y} + \frac{y}{x}))

\frac{d}{d x} (ln (sin (7 x)))

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