What is the integral from 2 to t of 1/((5x^5)) ?

The integral from 2 to t of 1/((5x^5)) is 1/5 (-1/(4t^4)+1/64)

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integral from 2 to t of 1/((5x^5))

\int_{2}^{t} \frac{1}{( 5 x ^{5} )} d x

\frac{1}{5} (- \frac{1}{4 t ^{4}} + \frac{1}{64})

Solution steps

\int_{2}^{t} \frac{1}{5 x ^{5}} d x

Take the constant out:

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

= \frac{1}{5} \cdot \int_{2}^{t} \frac{1}{x ^{5}} d x

Apply the Power Rule

= \frac{1}{5} [- \frac{1}{4 x ^{4}}]_{2}^{t}

Compute the boundaries:

- \frac{1}{4 t ^{4}} + \frac{1}{64}

= \frac{1}{5} (- \frac{1}{4 t ^{4}} + \frac{1}{64})

\int \frac{1}{P - P ^{2}} d P

\int \frac{x}{x + 4} d x

t an g e n t f (x) = \frac{2 x}{x ^{2} - 7}, (3, 3)

(x^{2} + 1) y^{^{'}} + 9 x (y - 1) = 0

\int (cos (x))^{4} d x

What is the integral from 2 to t of 1/((5x^5)) ?
The integral from 2 to t of 1/((5x^5)) is 1/5 (-1/(4t^4)+1/64)