What is the integral from 0 to 1 of (\sqrt[3]{1+7x}) ?

The integral from 0 to 1 of (\sqrt[3]{1+7x}) is 45/28

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integral from 0 to 1 of (\sqrt[3]{1+7x})

\frac{45}{28}

Solution steps

Apply u-substitution

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} u^{\frac{4}{3}}]_{1}^{8}

= \frac{1}{7} [\frac{3}{4} u^{\frac{4}{3}}]_{1}^{8}

Compute the boundaries:

\frac{45}{4}

= \frac{1}{7} \cdot \frac{45}{4}

Simplify

= \frac{45}{28}

\int x^{2} - 9 x + 3 d x

f (x) = x x, (9, 27)

n = 1 \sum \infty \frac{9}{6 + 9 ^{n}}

x \to - 3 lim (\frac{x ^{2} - 5}{3 - x})

\int \frac{1}{1 - 9 x ^{2}} d x

What is the integral from 0 to 1 of (\sqrt[3]{1+7x}) ?
The integral from 0 to 1 of (\sqrt[3]{1+7x}) is 45/28