What is the integral from 0 to 1 of (7x^2-4x+6) ?

The integral from 0 to 1 of (7x^2-4x+6) is 19/3

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integral from 0 to 1 of (7x^2-4x+6)

\int_{0}^{1} (7 x^{2} - 4 x + 6) d x

\frac{19}{3}

Decimal

6.33333 \dots

Solution steps

\int_{0}^{1} 7 x^{2} - 4 x + 6 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} 7 x^{2} d x - \int_{0}^{1} 4 x d x + \int_{0}^{1} 6 d x

\int_{0}^{1} 7 x^{2} d x = \frac{7}{3}

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

\int_{0}^{1} 6 d x = 6

= \frac{7}{3} - 2 + 6

Simplify

\frac{7}{3} - 2 + 6 : \frac{19}{3}

= \frac{19}{3}

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What is the integral from 0 to 1 of (7x^2-4x+6) ?
The integral from 0 to 1 of (7x^2-4x+6) is 19/3