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

The integral from 0 to 3 of (3x^2-4x+1) is 12

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

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

12

Solution steps

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

\int_{0}^{3} 3 x^{2} d x = 27

\int_{0}^{3} 4 x d x = 18

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

= 27 - 18 + 3

Simplify

= 12

\frac{d}{d x} (x sin (\frac{y}{x}))

\frac{\partial}{\partial x} (cos (x) cos (y) e^{z})

\int \frac{1}{( x - 7 ) ^{2}} d x

y^{^{''}} - 5 y^{^{'}} - 6 y = 12 e^{2 t}

x \to 0 lim (\frac{1}{( x - 1 ) ^{2}} + 2)

What is the integral from 0 to 3 of (3x^2-4x+1) ?
The integral from 0 to 3 of (3x^2-4x+1) is 12