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

The integral from 0 to 1 of (4+r)^3 is 369/4

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integral from 0 to 1 of (4+r)^3

\int_{0}^{1} (4 + r)^{3} d r

\frac{369}{4}

Solution steps

\int_{0}^{1} (4 + r)^{3} d r

Apply u-substitution

: \int_{4}^{5} u^{3} d u

= \int_{4}^{5} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{4}^{5}

= [\frac{u ^{4}}{4}]_{4}^{5}

Compute the boundaries:

\frac{369}{4}

= \frac{369}{4}

\int_{0}^{\frac{1}{2}} \frac{x ^{2}}{2 x + 1} d x

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

\int_{0}^{2} (5 x cos (π x)) d x

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

\int_{0}^{4} \frac{x}{x ^{2} - 3 x} d x

What is the integral from 0 to 1 of (4+r)^3 ?
The integral from 0 to 1 of (4+r)^3 is 369/4