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

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

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

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

\frac{15}{4}

Decimal Notation

3.75

Solution steps

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

y = 3 x - x^{2}, y = - 4 x

\int \frac{8}{9 + 3 x} d x

t^{3} \frac{d y}{d t} + 3 t^{2} y = 3 cos (t), y (π) = 0

\int \frac{5 x ^{4} - 2}{x ^{5} - 2 x + 11} d x

y^{^{'}} + 3 y = e^{5 t}

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