What is the integral from 2 to x^3 of 1/(t^3) ?

The integral from 2 to x^3 of 1/(t^3) is -1/(2x^6)+1/8

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integral from 2 to x^3 of 1/(t^3)

\int_{2}^{x^{3}} \frac{1}{t ^{3}} d t

- \frac{1}{2 x ^{6}} + \frac{1}{8}

Solution steps

\int_{2}^{x^{3}} \frac{1}{t ^{3}} d t

Apply the Power Rule

: [- \frac{1}{2 t ^{2}}]_{2}^{x^{3}}

= [- \frac{1}{2 t ^{2}}]_{2}^{x^{3}}

Compute the boundaries:

- \frac{1}{2 x ^{6}} + \frac{1}{8}

= - \frac{1}{2 x ^{6}} + \frac{1}{8}

\int_{0}^{5} 6

\int_{1}^{2} (4 x^{3} - 6) d x

\int_{\frac{1}{3}}^{3} 5 - \frac{5}{x ^{2}} d x

\int_{2}^{8} (68 + 47) d x

\int_{0}^{3} \frac{x}{x ^{2} + 9} d x

What is the integral from 2 to x^3 of 1/(t^3) ?
The integral from 2 to x^3 of 1/(t^3) is -1/(2x^6)+1/8