What is the integral from 0 to 1 of (x^2-4x+8) ?

The integral from 0 to 1 of (x^2-4x+8) is 19/3

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integral from 0 to 1 of (x^2-4x+8)

\int_{0}^{1} (x^{2} - 4 x + 8) d x

\frac{19}{3}

Decimal Notation

6.33333 \dots

Solution steps

\int_{0}^{1} x^{2} - 4 x + 8 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} x^{2} d x - \int_{0}^{1} 4 x d x + \int_{0}^{1} 8 d x

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

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

\int_{0}^{1} 8 d x = 8

= \frac{1}{3} - 2 + 8

Simplify

\frac{1}{3} - 2 + 8 : \frac{19}{3}

= \frac{19}{3}

\int_{0}^{2} 5 - x^{2} d x

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

\int_{0}^{1} \frac{1}{x ^{2} - 36} d x

\int_{1}^{5} (x^{4}) d x

\int_{- 4}^{4} x^{2} - 9 d x

What is the integral from 0 to 1 of (x^2-4x+8) ?
The integral from 0 to 1 of (x^2-4x+8) is 19/3