What is the integral from 0 to 1 of (2x^4-3x^2+5) ?

The integral from 0 to 1 of (2x^4-3x^2+5) is 22/5

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integral from 0 to 1 of (2x^4-3x^2+5)

\int_{0}^{1} (2 x^{4} - 3 x^{2} + 5) d x

\frac{22}{5}

Decimal

4.4

Solution steps

\int_{0}^{1} 2 x^{4} - 3 x^{2} + 5 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_{0}^{1} 2 x^{4} d x - \int_{0}^{1} 3 x^{2} d x + \int_{0}^{1} 5 d x

\int_{0}^{1} 2 x^{4} d x = \frac{2}{5}

\int_{0}^{1} 3 x^{2} d x = 1

\int_{0}^{1} 5 d x = 5

= \frac{2}{5} - 1 + 5

Simplify

\frac{2}{5} - 1 + 5 : \frac{22}{5}

= \frac{22}{5}

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What is the integral from 0 to 1 of (2x^4-3x^2+5) ?
The integral from 0 to 1 of (2x^4-3x^2+5) is 22/5