Calculadora de integrales definidas e impropias
Calcula integrales definidas e impropias paso a paso
La calculadora intentará evaluar la integral definida (es decir, con límites), incluida la impropia, con los pasos que se muestran.
Solution
Your input: calculate $$$\int_{- t}^{0}\left( z \left(1 + \frac{z}{t}\right) \right)dz$$$
First, calculate the corresponding indefinite integral: $$$\int{z \left(1 + \frac{z}{t}\right) d z}=\frac{z^{2} \left(\frac{t}{2} + \frac{z}{3}\right)}{t}$$$ (for steps, see indefinite integral calculator)
According to the Fundamental Theorem of Calculus, $$$\int_a^b F(x) dx=f(b)-f(a)$$$, so just evaluate the integral at the endpoints, and that's the answer.
$$$\left(\frac{z^{2} \left(\frac{t}{2} + \frac{z}{3}\right)}{t}\right)|_{\left(z=0\right)}=0$$$
$$$\left(\frac{z^{2} \left(\frac{t}{2} + \frac{z}{3}\right)}{t}\right)|_{\left(z=- t\right)}=\frac{t^{2}}{6}$$$
$$$\int_{- t}^{0}\left( z \left(1 + \frac{z}{t}\right) \right)dz=\left(\frac{z^{2} \left(\frac{t}{2} + \frac{z}{3}\right)}{t}\right)|_{\left(z=0\right)}-\left(\frac{z^{2} \left(\frac{t}{2} + \frac{z}{3}\right)}{t}\right)|_{\left(z=- t\right)}=- \frac{t^{2}}{6}$$$
Answer: $$$\int_{- t}^{0}\left( z \left(1 + \frac{z}{t}\right) \right)dz=- \frac{t^{2}}{6}$$$