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

The integral from 1 to 3 of (-x^3+4x^2-3) is 26/3

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

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

\frac{26}{3}

Decimal Notation

8.66666 \dots

Solution steps

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

\int_{1}^{3} x^{3} d x = 20

\int_{1}^{3} 4 x^{2} d x = \frac{104}{3}

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

= - 20 + \frac{104}{3} - 6

Simplify

- 20 + \frac{104}{3} - 6 : \frac{26}{3}

= \frac{26}{3}

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What is the integral from 1 to 3 of (-x^3+4x^2-3) ?
The integral from 1 to 3 of (-x^3+4x^2-3) is 26/3