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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

Compute the boundaries:

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

= 3 (- \frac{1}{2 t ^{2}} + \frac{1}{2})

\int \frac{1}{9 x ^{2} + 4} d x

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

\frac{d}{d x} (2 x^{3} - 4 x + 5)

d er i v a t i v e \frac{t}{t - 1}

(1 + cos (t)) y^{^{'}} = (1 + e - y) sin (t)

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