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

The integral from 1 to 3 of (x^2-6x+5) is -16/3

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integral from 1 to 3 of (x^2-6x+5)

\int_{1}^{3} (x^{2} - 6 x + 5) d x

- \frac{16}{3}

Decimal

- 5.33333 \dots

Solution steps

\int_{1}^{3} x^{2} - 6 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_{1}^{3} x^{2} d x - \int_{1}^{3} 6 x d x + \int_{1}^{3} 5 d x

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} 6 x d x = 24

\int_{1}^{3} 5 d x = 10

= \frac{26}{3} - 24 + 10

Simplify

\frac{26}{3} - 24 + 10 : - \frac{16}{3}

= - \frac{16}{3}

\int_{2}^{4} 3 x - 1 + 2 x^{2} d x

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\int_{0}^{y} x e^{- y} d x

\int_{1}^{e} t ln (t) d t

What is the integral from 1 to 3 of (x^2-6x+5) ?
The integral from 1 to 3 of (x^2-6x+5) is -16/3