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

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

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

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

\frac{1}{5} (- \frac{1}{t} + \frac{1}{2})

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

- \frac{1}{t} + \frac{1}{2}

= \frac{1}{5} (- \frac{1}{t} + \frac{1}{2})

\frac{d}{d x} ((x + 2)^{2} - 3 x)

\int \frac{8 x ^{2} - 7 x + 16}{x ^{3} + 4 x} d x

\int_{1}^{3} \frac{1}{x ^{3}} d x

x 3 - x^{2}

x \to 2 lim (\frac{1}{∣ 2 - x ∣})

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