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

The integral from 0 to 1 of (x+1)^3 is 15/4

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

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

\frac{15}{4}

Decimal Notation

3.75

Solution steps

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

Apply u-substitution

: \int_{1}^{2} u^{3} d u

= \int_{1}^{2} u^{3} d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{15}{4}

= \frac{15}{4}

x \to - 1 lim (2 - 3 x)

x \to \infty lim (8 - 2 x)

\frac{\partial}{\partial y} (x^{2} + y^{2} - 1)

\frac{d}{d x} (- \frac{14}{x ^{6}})

4 (2 t - 5)^{2}

What is the integral from 0 to 1 of (x+1)^3 ?
The integral from 0 to 1 of (x+1)^3 is 15/4