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

The integral from 2 to x^2 of 1/(t^3) is -1/(2x^4)+1/8

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

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

- \frac{1}{2 x ^{4}} + \frac{1}{8}

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

- \frac{1}{2 x ^{4}} + \frac{1}{8}

= - \frac{1}{2 x ^{4}} + \frac{1}{8}

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What is the integral from 2 to x^2 of 1/(t^3) ?
The integral from 2 to x^2 of 1/(t^3) is -1/(2x^4)+1/8