What is the integral from 1 to R of x^{-9/5} ?

The integral from 1 to R of x^{-9/5} is -5/(4R^{4/5)}+5/4

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integral from 1 to R of x^{-9/5}

\int_{1}^{R} x^{- \frac{9}{5}} d x

- \frac{5}{4 R ^{\frac{4}{5}}} + \frac{5}{4}

Solution steps

\int_{1}^{R} x^{- \frac{9}{5}} d x

Apply the Power Rule

= [- \frac{5}{4 x ^{\frac{4}{5}}}]_{1}^{R}

Compute the boundaries:

- \frac{5}{4 R ^{\frac{4}{5}}} + \frac{5}{4}

= - \frac{5}{4 R ^{\frac{4}{5}}} + \frac{5}{4}

\int \frac{4}{x} - \frac{4 x}{x ^{2} + 1} d x

(2 x + 1)^{^{'}}

\int t^{2} sin (t) d t

(x + 2)^{2} y^{^{'}} + 3 (x + 2) y = 4

\frac{d y}{d x} = 2 y + y^{5}, y (0) = 1

What is the integral from 1 to R of x^{-9/5} ?
The integral from 1 to R of x^{-9/5} is -5/(4R^{4/5)}+5/4