What is the integral from 1 to R of x^{-7/5} ?

The integral from 1 to R of x^{-7/5} is -5/(2R^{2/5)}+5/2

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integral from 1 to R of x^{-7/5}

\int_{1}^{R} x^{- \frac{7}{5}} d x

- \frac{5}{2 R ^{\frac{2}{5}}} + \frac{5}{2}

Solution steps

\int_{1}^{R} x^{- \frac{7}{5}} d x

Apply the Power Rule

: [- \frac{5}{2 x ^{\frac{2}{5}}}]_{1}^{R}

= [- \frac{5}{2 x ^{\frac{2}{5}}}]_{1}^{R}

Compute the boundaries:

- \frac{5}{2 R ^{\frac{2}{5}}} + \frac{5}{2}

= - \frac{5}{2 R ^{\frac{2}{5}}} + \frac{5}{2}

\frac{d}{d x} ((6 x^{5} - 4 x^{4} + 7 x) (x^{3} - x^{2} - 1))

\frac{\partial}{\partial x} (e^{x + y + 5})

\frac{\partial}{\partial z} (\frac{x - y}{x + z})

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{t}}{1 + t ^{2}}

\int_{0}^{π} sin^{4} (3 x) d x

What is the integral from 1 to R of x^{-7/5} ?
The integral from 1 to R of x^{-7/5} is -5/(2R^{2/5)}+5/2