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

The integral from 1 to 5 of (x^2-x+1) is 100/3

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

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

\frac{100}{3}

Decimal Notation

33.33333 \dots

Solution steps

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

\int_{1}^{5} x^{2} d x = \frac{124}{3}

\int_{1}^{5} x d x = 12

\int_{1}^{5} 1 d x = 4

= \frac{124}{3} - 12 + 4

Simplify

\frac{124}{3} - 12 + 4 : \frac{100}{3}

= \frac{100}{3}

\frac{d}{d x} (- sin (4 x))

y = e^{x^{4}}

\frac{d y}{d x} = (1 + e^{2 x})

\frac{d}{d x} (\frac{x ^{2}}{x - 9})

(- 8, 9), (- 2, - 3)

What is the integral from 1 to 5 of (x^2-x+1) ?
The integral from 1 to 5 of (x^2-x+1) is 100/3