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

The integral from-1 to 5 of (3x^4+2x^2+3) is 9888/5

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

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

\frac{9888}{5}

Decimal

1977.6

Solution steps

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

\int_{- 1}^{5} 3 x^{4} d x = \frac{9378}{5}

\int_{- 1}^{5} 2 x^{2} d x = 84

\int_{- 1}^{5} 3 d x = 18

= \frac{9378}{5} + 84 + 18

Simplify

\frac{9378}{5} + 84 + 18 : \frac{9888}{5}

= \frac{9888}{5}

What is the integral from-1 to 5 of (3x^4+2x^2+3) ?
The integral from-1 to 5 of (3x^4+2x^2+3) is 9888/5