What is the integral from 1 to 8 of 9\sqrt[3]{x} ?

The integral from 1 to 8 of 9\sqrt[3]{x} is 405/4

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integral from 1 to 8 of 9\sqrt[3]{x}

\frac{405}{4}

Decimal Notation

101.25

Solution steps

Take the constant out:

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

Apply the Power Rule

: [\frac{3}{4} x^{\frac{4}{3}}]_{1}^{8}

= 9 [\frac{3}{4} x^{\frac{4}{3}}]_{1}^{8}

Compute the boundaries:

\frac{45}{4}

= 9 \cdot \frac{45}{4}

Simplify

= \frac{405}{4}

\int - 10 ln (x^{2} - 1) d x

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\frac{s}{s ^{2} + 1}

\int \frac{1}{36 - 6 t ^{2}} d t

What is the integral from 1 to 8 of 9\sqrt[3]{x} ?
The integral from 1 to 8 of 9\sqrt[3]{x} is 405/4