What is the integral from 1 to t of (27)/(x^3) ?

The integral from 1 to t of (27)/(x^3) is 27(-1/(2t^2)+1/2)

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integral from 1 to t of (27)/(x^3)

\int_{1}^{t} \frac{27}{x ^{3}} d x

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

Solution steps

\int_{1}^{t} \frac{27}{x ^{3}} d x

Take the constant out:

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

= 27 \cdot \int_{1}^{t} \frac{1}{x ^{3}} d x

Apply the Power Rule

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

= 27 [- \frac{1}{2 x ^{2}}]_{1}^{t}

Compute the boundaries:

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

= 27 (- \frac{1}{2 t ^{2}} + \frac{1}{2})

5 x - 4 y x^{2} + 1 \frac{d y}{d x} = 0

\frac{\partial}{\partial x} (2 x^{2} yz)

\int_{2}^{5} (4 - 2 x) d x

\frac{d}{d x} (ln (5 x^{2} + 6))

9 x^{3} - x^{2} y + y^{3} - 3

What is the integral from 1 to t of (27)/(x^3) ?
The integral from 1 to t of (27)/(x^3) is 27(-1/(2t^2)+1/2)