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

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

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

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

- \frac{1}{x ^{3}} + \frac{1}{2}

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

- \frac{1}{x ^{3}} + \frac{1}{2}

= - \frac{1}{x ^{3}} + \frac{1}{2}

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\int_{0}^{2} t 4 t^{2} + 1 d t

\int_{0}^{2} x^{2} \cdot \frac{3}{4} x (2 - x) d x

\int_{0}^{2 π} r d r

\int_{4}^{1} - x d x

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