What is the integral from 0 to 1 of t^7(1+t^8)^7 ?

The integral from 0 to 1 of t^7(1+t^8)^7 is 255/64

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integral from 0 to 1 of t^7(1+t^8)^7

\int_{0}^{1} t^{7} (1 + t^{8})^{7} d t

\frac{255}{64}

Decimal Notation

3.984375

Solution steps

\int_{0}^{1} t^{7} (1 + t^{8})^{7} d t

Apply u-substitution

: \int_{1}^{2} \frac{u ^{7}}{8} d u

= \int_{1}^{2} \frac{u ^{7}}{8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int_{1}^{2} u^{7} d u

Apply the Power Rule

: [\frac{u ^{8}}{8}]_{1}^{2}

= \frac{1}{8} [\frac{u ^{8}}{8}]_{1}^{2}

Compute the boundaries:

\frac{255}{8}

= \frac{1}{8} \cdot \frac{255}{8}

Simplify

= \frac{255}{64}

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What is the integral from 0 to 1 of t^7(1+t^8)^7 ?
The integral from 0 to 1 of t^7(1+t^8)^7 is 255/64