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

The integral from-2 to 2 of 4x^4-x^2+1 is 748/15

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integral from-2 to 2 of 4x^4-x^2+1

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

\frac{748}{15}

Decimal Notation

49.86666 \dots

Solution steps

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

\int_{- 2}^{2} 4 x^{4} d x = \frac{256}{5}

\int_{- 2}^{2} x^{2} d x = \frac{16}{3}

\int_{- 2}^{2} 1 d x = 4

= \frac{256}{5} - \frac{16}{3} + 4

Simplify

\frac{256}{5} - \frac{16}{3} + 4 : \frac{748}{15}

= \frac{748}{15}

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What is the integral from-2 to 2 of 4x^4-x^2+1 ?
The integral from-2 to 2 of 4x^4-x^2+1 is 748/15