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

The integral from 0 to 1 of (x^2-3x+6) is 29/6

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integral from 0 to 1 of (x^2-3x+6)

\int_{0}^{1} (x^{2} - 3 x + 6) d x

\frac{29}{6}

Decimal

4.83333 \dots

Solution steps

\int_{0}^{1} x^{2} - 3 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} x^{2} d x - \int_{0}^{1} 3 x d x + \int_{0}^{1} 6 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} 6 d x = 6

= \frac{1}{3} - \frac{3}{2} + 6

Simplify

\frac{1}{3} - \frac{3}{2} + 6 : \frac{29}{6}

= \frac{29}{6}

x \to 0 lim (+ (x sin (\frac{1}{x})))

y^{^{'''}} + y^{^{''}} = 3 e^{x} + 4 x^{2}

\frac{d y}{d t} - 2 t y = 9 t^{2} e^{t^{2}}

\frac{d y}{d x} = y sin (x), y (0) = 1

\int e^{x y} d x

What is the integral from 0 to 1 of (x^2-3x+6) ?
The integral from 0 to 1 of (x^2-3x+6) is 29/6