What is the integral from 1 to t of 39x^{-3} ?

The integral from 1 to t of 39x^{-3} is 39(-1/(2t^2)+1/2)

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integral from 1 to t of 39x^{-3}

\int_{1}^{t} 39 x^{- 3} d x

39 (- \frac{1}{2 t ^{2}} + \frac{1}{2})

Solution steps

\int_{1}^{t} 39 x^{- 3} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 39 \cdot \int_{1}^{t} x^{- 3} d x

Apply the Power Rule

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

= 39 [- \frac{1}{2 x ^{2}}]_{1}^{t}

Compute the boundaries:

- \frac{1}{2 t ^{2}} + \frac{1}{2}

= 39 (- \frac{1}{2 t ^{2}} + \frac{1}{2})

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What is the integral from 1 to t of 39x^{-3} ?
The integral from 1 to t of 39x^{-3} is 39(-1/(2t^2)+1/2)