What is the integral from 0 to 2 of 6x^2-18x+12 ?

The integral from 0 to 2 of 6x^2-18x+12 is 4

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integral from 0 to 2 of 6x^2-18x+12

\int_{0}^{2} 6 x^{2} - 18 x + 12 d x

4

Solution steps

\int_{0}^{2} 6 x^{2} - 18 x + 12 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}^{2} 6 x^{2} d x - \int_{0}^{2} 18 x d x + \int_{0}^{2} 12 d x

\int_{0}^{2} 6 x^{2} d x = 16

\int_{0}^{2} 18 x d x = 36

\int_{0}^{2} 12 d x = 24

= 16 - 36 + 24

Simplify

= 4

f (x) = x^{2} - \frac{1}{x}, (2, 3.5)

n = 1 \sum \infty \frac{( ( - 7 ) ^{n - 1} )}{8 ^{n}}

y^{^{''}} - 14 y^{^{'}} + 49 y = t^{- 2} e^{7 t}

x \to - \infty lim (\frac{7}{x} - 4)

\int \frac{1}{w ^{3}} d w

What is the integral from 0 to 2 of 6x^2-18x+12 ?
The integral from 0 to 2 of 6x^2-18x+12 is 4