What is the integral from 1 to 2 of x^{-3} ?

The integral from 1 to 2 of x^{-3} is 3/8

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integral from 1 to 2 of x^{-3}

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

\frac{3}{8}

Decimal Notation

0.375

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{3}{8}

= \frac{3}{8}

t y^{^{'}} + (t + 1) y = t

\int 6 x (3 x^{2} - 2)^{4} d x

\int_{0}^{\frac{1}{5}} 4 arcsin (5 y) d y

y = 3 x - 4 x^{2}, (- 2, - 22)

\frac{\partial}{\partial x} (e^{- x y} cos (y))

What is the integral from 1 to 2 of x^{-3} ?
The integral from 1 to 2 of x^{-3} is 3/8