What is the integral from-6 to 6 of \sqrt[3]{x^4} ?

The integral from-6 to 6 of \sqrt[3]{x^4} is (216\sqrt[3]{6})/7

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integral from-6 to 6 of \sqrt[3]{x^4}

Decimal Notation

56.07114 \dots

Solution steps

assuming

x \geq 0

= \int_{- 6}^{6} x^{\frac{4}{3}} d x

Apply the Power Rule

: [\frac{3}{7} x^{\frac{7}{3}}]_{- 6}^{6}

= [\frac{3}{7} x^{\frac{7}{3}}]_{- 6}^{6}

Compute the boundaries:

\int_{1}^{6} x + 3 d x

\int_{1}^{\infty} \frac{1}{x e ^{x}} d x

\int_{0}^{2} \frac{7 x ^{6}}{2187} d x

\int_{- 3}^{3} (∣ y - 2 ∣)^{3} d y

\int_{0}^{2} x^{2} \cdot (x^{3} + 1) d x

What is the integral from-6 to 6 of \sqrt[3]{x^4} ?
The integral from-6 to 6 of \sqrt[3]{x^4} is (216\sqrt[3]{6})/7