What is the integral from 1 to t of (17)/(x^3) ?

The integral from 1 to t of (17)/(x^3) is 17(-1/(2t^2)+1/2)

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integral from 1 to t of (17)/(x^3)

\int_{1}^{t} \frac{17}{x ^{3}} d x

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

Solution steps

\int_{1}^{t} \frac{17}{x ^{3}} d x

Take the constant out:

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

= 17 \cdot \int_{1}^{t} \frac{1}{x ^{3}} d x

Apply the Power Rule

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

= 17 [- \frac{1}{2 x ^{2}}]_{1}^{t}

Compute the boundaries:

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

= 17 (- \frac{1}{2 t ^{2}} + \frac{1}{2})

3 - \frac{1}{2} x, 3 \leq x \leq 15

\int_{0}^{2} (x + 2 - x^{2}) d x

e^{x} + e^{- x}

\int \frac{3}{x ^{2} x ^{2} + 25} d x

x \to 0 lim (2 cos (x))

What is the integral from 1 to t of (17)/(x^3) ?
The integral from 1 to t of (17)/(x^3) is 17(-1/(2t^2)+1/2)