What is the integral from 0 to 1 of (3x^2-4x+1) ?

The integral from 0 to 1 of (3x^2-4x+1) is 0

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integral from 0 to 1 of (3x^2-4x+1)

\int_{0}^{1} (3 x^{2} - 4 x + 1) d x

0

Solution steps

\int_{0}^{1} 3 x^{2} - 4 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

= \int_{0}^{1} 3 x^{2} d x - \int_{0}^{1} 4 x d x + \int_{0}^{1} 1 d x

\int_{0}^{1} 3 x^{2} d x = 1

\int_{0}^{1} 4 x d x = 2

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

= 1 - 2 + 1

Simplify

= 0

y^{^{''}} + \frac{1}{4} y = 0

x \to - 4 lim (3)

x \to \infty lim (\frac{1}{lo g _{2} ( x )})

\int_{- 1}^{0} \frac{1}{x ^{2}} d x

\frac{4 s e ^{- 4 s}}{2 s ^{2} + 50}

What is the integral from 0 to 1 of (3x^2-4x+1) ?
The integral from 0 to 1 of (3x^2-4x+1) is 0