What is the integral from 1 to t of (33)/(x^3) ?

The integral from 1 to t of (33)/(x^3) is 33(-1/(2t^2)+1/2)

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integral from 1 to t of (33)/(x^3)

\int_{1}^{t} \frac{33}{x ^{3}} d x

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

Solution steps

\int_{1}^{t} \frac{33}{x ^{3}} d x

Take the constant out:

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

= 33 \cdot \int_{1}^{t} \frac{1}{x ^{3}} d x

Apply the Power Rule

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

= 33 [- \frac{1}{2 x ^{2}}]_{1}^{t}

Compute the boundaries:

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

= 33 (- \frac{1}{2 t ^{2}} + \frac{1}{2})

\frac{d}{d x} (lo g_{6} (2 x^{2} + 4))

\int (\frac{10 x}{1 + 5 x ^{2}}) d x

\frac{2 x ^{3}}{3}

\int_{1}^{3} (x^{2} + 2 x - 5) d x

\frac{d}{d x} (\frac{e ^{x}}{3 x ^{2}})

What is the integral from 1 to t of (33)/(x^3) ?
The integral from 1 to t of (33)/(x^3) is 33(-1/(2t^2)+1/2)