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

The integral from 1 to 2 of (x^3) is 15/4

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integral from 1 to 2 of (x^3)

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

\frac{15}{4}

Decimal Notation

3.75

Solution steps

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

Apply the Power Rule

: [\frac{x ^{4}}{4}]_{1}^{2}

= [\frac{x ^{4}}{4}]_{1}^{2}

Compute the boundaries:

\frac{15}{4}

= \frac{15}{4}

\frac{d}{d x} (7 - (1.01)^{- x^{2}})

f (x) = x^{2} + 1, (- 5, 26)

\frac{\partial}{\partial x} (3 x^{2} - 6 x - 9)

\int \frac{1}{4 x + 1} d x

\frac{d}{d x} (\frac{ln ( sin ^{x} ( x ) )}{x})

What is the integral from 1 to 2 of (x^3) ?
The integral from 1 to 2 of (x^3) is 15/4