What is the integral from 0 to 1 of (x^2-4x+7) ?

The integral from 0 to 1 of (x^2-4x+7) is 16/3

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integral from 0 to 1 of (x^2-4x+7)

\int_{0}^{1} (x^{2} - 4 x + 7) d x

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

\int_{0}^{1} x^{2} - 4 x + 7 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} 7 d x

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

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

\int_{0}^{1} 7 d x = 7

= \frac{1}{3} - 2 + 7

Simplify

\frac{1}{3} - 2 + 7 : \frac{16}{3}

= \frac{16}{3}

\frac{d}{d x} (- 3 sin (x))

x \to 4 lim (\frac{x + 1}{x + 4})

\int x^{2} cos (\frac{1}{6} x) d x

\frac{d y}{d x} = 6 e^{2 x - y}

\int_{0}^{π} x cos (x) d x

What is the integral from 0 to 1 of (x^2-4x+7) ?
The integral from 0 to 1 of (x^2-4x+7) is 16/3