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}

Decimal Notation

1.60714 \dots

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}

f (x) = \frac{r}{r ^{2} + 5}

\frac{d}{d x} (g (x) (x^{3}))

\frac{\partial}{\partial x} (y x^{2} + 6 x^{3} + x y)

\frac{\partial}{\partial x} (x^{2} + y^{2} + x^{2} y + 4)

\frac{d}{d x} (3 x^{2} + 5 x + 8)

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