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

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

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

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

- 1 + \frac{1}{t}

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

- 1 + \frac{1}{t}

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

x \to 8 lim (cos (\frac{π x}{3}))

y = \frac{x ^{4}}{f ( x )}

f (x) = 3 x^{2} - 4 x

y^{^{''}} + 2 y^{^{'}} + 26 y = 0, y (0) = 4, y^{^{'}} (0) = - 2

\frac{d}{d x} (x \cdot ln (x))

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