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

Lösung

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

- 1 + \frac{1}{t}

Schritte zur Lösung

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

Wende die Potenzregel an

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

Berechne die Grenzen:

- 1 + \frac{1}{t}

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

x \to 8 lim (cos (\frac{π x}{3}))

d er i v a t i v e 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))

\frac{\partial}{\partial x} (x^{2 + y^{2}})