What is the integral from 1 to t of (13)/(x^3) ?

The integral from 1 to t of (13)/(x^3) is 13(-1/(2t^2)+1/2)

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integral from 1 to t of (13)/(x^3)

\int_{1}^{t} \frac{13}{x ^{3}} d x

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

Solution steps

\int_{1}^{t} \frac{13}{x ^{3}} d x

Take the constant out:

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

= 13 \cdot \int_{1}^{t} \frac{1}{x ^{3}} d x

Apply the Power Rule

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

= 13 [- \frac{1}{2 x ^{2}}]_{1}^{t}

Compute the boundaries:

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

= 13 (- \frac{1}{2 t ^{2}} + \frac{1}{2})

\frac{d}{d x} (e^{x} sin (2 x) + 2 e^{x} cos (2 x))

\frac{x}{x + 17}

y = 9 - x^{2}, y = 4 x

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

(2 x + 5)^{2}

What is the integral from 1 to t of (13)/(x^3) ?
The integral from 1 to t of (13)/(x^3) is 13(-1/(2t^2)+1/2)