What is the integral from 0 to 2 of (5x^2-8x+4) ?

The integral from 0 to 2 of (5x^2-8x+4) is 16/3

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integral from 0 to 2 of (5x^2-8x+4)

\int_{0}^{2} (5 x^{2} - 8 x + 4) d x

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

\int_{0}^{2} 5 x^{2} - 8 x + 4 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} 5 x^{2} d x - \int_{0}^{2} 8 x d x + \int_{0}^{2} 4 d x

\int_{0}^{2} 5 x^{2} d x = \frac{40}{3}

\int_{0}^{2} 8 x d x = 16

\int_{0}^{2} 4 d x = 8

= \frac{40}{3} - 16 + 8

Simplify

\frac{40}{3} - 16 + 8 : \frac{16}{3}

= \frac{16}{3}

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What is the integral from 0 to 2 of (5x^2-8x+4) ?
The integral from 0 to 2 of (5x^2-8x+4) is 16/3