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

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

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

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

\frac{15}{4}

Decimal Notation

3.75

Solution steps

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

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}

s \cdot + 5

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