What is the integral from 5 to x^2 of 1/(t^5) ?

The integral from 5 to x^2 of 1/(t^5) is -1/(4x^8)+1/2500

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integral from 5 to x^2 of 1/(t^5)

\int_{5}^{x^{2}} \frac{1}{t ^{5}} d t

- \frac{1}{4 x ^{8}} + \frac{1}{2500}

Solution steps

\int_{5}^{x^{2}} \frac{1}{t ^{5}} d t

Apply the Power Rule

: [- \frac{1}{4 t ^{4}}]_{5}^{x^{2}}

= [- \frac{1}{4 t ^{4}}]_{5}^{x^{2}}

Compute the boundaries:

- \frac{1}{4 x ^{8}} + \frac{1}{2500}

= - \frac{1}{4 x ^{8}} + \frac{1}{2500}

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

\int_{1}^{2} \frac{x}{( x ^{2} + 4 ) ^{3}} d x

\int_{- 1}^{1} cos (\frac{π}{3} x) d x

\int_{0}^{2} 4 t - 3 t^{2} d t

\int_{0}^{2 - x} 2 - x - y d y

What is the integral from 5 to x^2 of 1/(t^5) ?
The integral from 5 to x^2 of 1/(t^5) is -1/(4x^8)+1/2500