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

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

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

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

\frac{7}{3}

Decimal Notation

2.33333 \dots

Solution steps

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

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

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

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

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

Simplify

\frac{1}{3} - 1 + 3 : \frac{7}{3}

= \frac{7}{3}

\int_{0}^{\frac{1}{2}} 7 arccos (x) d x

\frac{\partial}{\partial y} (1 - x y cos (π y))

- ln (1 - x)

x \to - \infty lim (\frac{x}{2})

\frac{d y}{d x} - tan (x) y = y^{3}, y (0) = 1

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