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

The integral from 1 to 2 of 2/(x^3) is 3/4

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integral from 1 to 2 of 2/(x^3)

\int_{1}^{2} \frac{2}{x ^{3}} d x

\frac{3}{4}

Solution steps

\int_{1}^{2} \frac{2}{x ^{3}} d x

Take the constant out:

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

= 2 \cdot \int_{1}^{2} \frac{1}{x ^{3}} d x

Apply the Power Rule

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

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

Compute the boundaries:

\frac{3}{8}

= 2 \cdot \frac{3}{8}

Simplify

= \frac{3}{4}

9 (t + 1) \frac{d y}{d t} - 8 y = 8 t

y = \frac{1}{x} + cos (2 x)

x \to 4 + lim (16 - x^{2})

\int_{0}^{1} 6 e^{2 x} d x

What is the integral from 1 to 2 of 2/(x^3) ?
The integral from 1 to 2 of 2/(x^3) is 3/4