What is the integral from 0 to 9 of (2x^3+1) ?

The integral from 0 to 9 of (2x^3+1) is 6579/2

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integral from 0 to 9 of (2x^3+1)

\int_{0}^{9} (2 x^{3} + 1) d x

\frac{6579}{2}

Decimal Notation

3289.5

Solution steps

\int_{0}^{9} 2 x^{3} + 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_{0}^{9} 2 x^{3} d x + \int_{0}^{9} 1 d x

\int_{0}^{9} 2 x^{3} d x = \frac{6561}{2}

\int_{0}^{9} 1 d x = 9

= \frac{6561}{2} + 9

Simplify

\frac{6561}{2} + 9 : \frac{6579}{2}

= \frac{6579}{2}

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

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

f (x) = 4^{x} - x^{2} + 3 ln (x)

\frac{40 t ^{2}}{t ^{2} + 150}

What is the integral from 0 to 9 of (2x^3+1) ?
The integral from 0 to 9 of (2x^3+1) is 6579/2