Question

Cum se integreaza aceasta functie?

Answer 1

$\begin{array}{rcl}\displaystyle \int_{0}^1\Big(e^x-\frac{x^2}{2}-1\Big)\,dx &=& \displaystyle \int_{0}^1 e^x\,dx -\int_{0}^1\frac{x^2}{2}\, dx -\int_{0}^11\,dx \\ &=& \displaystyle \int_{0}^1 e^x\,dx -\frac{1}{2}\int_{0}^1x^2\, dx -\int_{0}^11\,dx \\ &=&e^x\Big|_{0}^1-\dfrac{1}{2}\cdot \left \Big(\dfrac{x^3}{3}\Big)\right|_{0}^1 - x\big|_{0}^1 \end{array}$

$=(e^1-e^0) - \dfrac{1}{2}\cdot \Big(\dfrac{1^3}{3}-\dfrac{0^3}{3}\Big)-(1-0) \\ \\ =(e-1)-\dfrac{1}{6}-1 \\ \\ = e-\dfrac{1}{6}-2 \\ \\ = e - \dfrac{13}{6}$