What is the integral from 0 to 7 of 4x\sqrt[3]{x+1} ?

The integral from 0 to 7 of 4x\sqrt[3]{x+1} is 1209/7

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integral from 0 to 7 of 4x\sqrt[3]{x+1}

\frac{1209}{7}

Decimal Notation

172.71428 \dots

Solution steps

Take the constant out:

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

Apply u-substitution

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int_{1}^{8} u^{\frac{4}{3}} d u = \frac{381}{7}

= 4 (\frac{381}{7} - \frac{45}{4})

Simplify

4 (\frac{381}{7} - \frac{45}{4}) : \frac{1209}{7}

= \frac{1209}{7}

\int_{0}^{1} (5 x^{e} + 4 e^{x}) d x

\int_{0}^{3} (x^{2} - 5 x + 9) d x

\int_{2}^{4} x^{2} + 5 x + 7 + 1 d x

\int_{- 7}^{4} \frac{1}{2} x d x

\int_{3}^{6} (2 x + 9) d x

What is the integral from 0 to 7 of 4x\sqrt[3]{x+1} ?
The integral from 0 to 7 of 4x\sqrt[3]{x+1} is 1209/7