What is the integral from 1 to x^2 of 1/(t^4) ?

The integral from 1 to x^2 of 1/(t^4) is -1/(3x^6)+1/3

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integral from 1 to x^2 of 1/(t^4)

\int_{1}^{x^{2}} \frac{1}{t ^{4}} d t

- \frac{1}{3 x ^{6}} + \frac{1}{3}

Solution steps

\int_{1}^{x^{2}} \frac{1}{t ^{4}} d t

Apply the Power Rule

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

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

Compute the boundaries:

- \frac{1}{3 x ^{6}} + \frac{1}{3}

= - \frac{1}{3 x ^{6}} + \frac{1}{3}

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\int_{\frac{π}{2}}^{π} - 12 cos (x) d x

What is the integral from 1 to x^2 of 1/(t^4) ?
The integral from 1 to x^2 of 1/(t^4) is -1/(3x^6)+1/3