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

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

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

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

\frac{16}{3}

Solution steps

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

\int_{- 1}^{3} x^{2} d x = \frac{28}{3}

\int_{- 1}^{3} 2 x d x = 8

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

= \frac{28}{3} - 8 + 4

Simplify

\frac{28}{3} - 8 + 4 : \frac{16}{3}

= \frac{16}{3}

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