What is the integral from 1 to 3 of-(x^2-4x+3) ?

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

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integral from 1 to 3 of-(x^2-4x+3)

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

\int_{1}^{3} 4 x d x = 16

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

= - (\frac{26}{3} - 16 + 6)

Simplify

- (\frac{26}{3} - 16 + 6) : \frac{4}{3}

= \frac{4}{3}

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