What is the integral from 0 to 1 of (7x^2-4x+9) ?

The integral from 0 to 1 of (7x^2-4x+9) is 28/3

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integral from 0 to 1 of (7x^2-4x+9)

\int_{0}^{1} (7 x^{2} - 4 x + 9) d x

\frac{28}{3}

Decimal

9.33333 \dots

Solution steps

\int_{0}^{1} 7 x^{2} - 4 x + 9 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} 9 d x

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

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

\int_{0}^{1} 9 d x = 9

= \frac{7}{3} - 2 + 9

Simplify

\frac{7}{3} - 2 + 9 : \frac{28}{3}

= \frac{28}{3}

What is the integral from 0 to 1 of (7x^2-4x+9) ?
The integral from 0 to 1 of (7x^2-4x+9) is 28/3