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

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

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

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

\frac{369}{4}

Decimal

92.25

Solution steps

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

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}^{2} x + 4 d x

\int_{0}^{a} x^{2} sin (\frac{x}{2}) d x

\int_{- 1}^{5} (x - 8)^{2} d x

\int_{1 - 2 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

\int_{2}^{5} 3 x^{2} d x

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