What is the integral from 1 to 16 of 2x^{-3/2} ?

The integral from 1 to 16 of 2x^{-3/2} is 3

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integral from 1 to 16 of 2x^{-3/2}

\int_{1}^{16} 2 x^{- \frac{3}{2}} d x

3

Solution steps

\int_{1}^{16} 2 x^{- \frac{3}{2}} d x

Take the constant out:

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

= 2 \cdot \int_{1}^{16} x^{- \frac{3}{2}} d x

Apply the Power Rule

: [- \frac{2}{x ^{\frac{1}{2}}}]_{1}^{16}

= 2 [- \frac{2}{x ^{\frac{1}{2}}}]_{1}^{16}

Compute the boundaries:

\frac{3}{2}

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

Simplify

= 3

\int_{1}^{4} x x^{2} + 1 d x

\int_{1}^{4} \frac{1}{6 x} d x

\int_{2}^{3} t d t

\int_{0}^{2} (8 - x^{3}) d x

\int_{π}^{2 π} 1 d x

What is the integral from 1 to 16 of 2x^{-3/2} ?
The integral from 1 to 16 of 2x^{-3/2} is 3