What is the integral from 0 to 1 of (x^2-3x+5) ?

The integral from 0 to 1 of (x^2-3x+5) is 23/6

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integral from 0 to 1 of (x^2-3x+5)

\int_{0}^{1} (x^{2} - 3 x + 5) d x

\frac{23}{6}

Decimal

3.83333 \dots

Solution steps

\int_{0}^{1} x^{2} - 3 x + 5 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} 3 x d x + \int_{0}^{1} 5 d x

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

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

\int_{0}^{1} 5 d x = 5

= \frac{1}{3} - \frac{3}{2} + 5

Simplify

\frac{1}{3} - \frac{3}{2} + 5 : \frac{23}{6}

= \frac{23}{6}

f (x) = \frac{2}{3 x ^{\frac{2}{3}}}, at x = 64

\frac{\partial}{\partial x} (64 - 8 x^{2} - y^{2})

x \to 1 + lim (\frac{1 - x}{x ( x - 1 )})

\frac{d}{d x} (cos^{2} (3 π x))

x \to - 9 - lim (\frac{x + 8}{x + 9})

What is the integral from 0 to 1 of (x^2-3x+5) ?
The integral from 0 to 1 of (x^2-3x+5) is 23/6