What is the integral from-1 to 1 of (4x^3+5x^2-x+1) ?

The integral from-1 to 1 of (4x^3+5x^2-x+1) is 16/3

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integral from-1 to 1 of (4x^3+5x^2-x+1)

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

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

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

\int_{- 1}^{1} 4 x^{3} d x = 0

\int_{- 1}^{1} 5 x^{2} d x = \frac{10}{3}

\int_{- 1}^{1} x d x = 0

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

= 0 + \frac{10}{3} - 0 + 2

Simplify

0 + \frac{10}{3} - 0 + 2 : \frac{16}{3}

= \frac{16}{3}

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What is the integral from-1 to 1 of (4x^3+5x^2-x+1) ?
The integral from-1 to 1 of (4x^3+5x^2-x+1) is 16/3