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

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

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

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

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

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

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

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

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

= - \frac{26}{3} + 16 - 2

Simplify

- \frac{26}{3} + 16 - 2 : \frac{16}{3}

= \frac{16}{3}

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\int_{1}^{3} 5 (2 x - 4)^{4} d x

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