\(\int_1^\4 \mathrm{-x}^{2}\,\mathrm{d}x\)Wie berechne ich dieses Integral (schriftlich)?
+26367 \(\int \limits_{1}^{4} -x^2\ dx\) Wie berechne ich dieses Integral (schriftlich)?
\(\small{ \begin{array}{rcl} \int \limits_{1}^{4} -x^2\ dx &=& -\int \limits_{1}^{4} x^2\ dx \\ &=& - \left[ \frac{x^3} {3} \right]_{1}^{4} \\ &=& - \left( \frac{4^3} {3}- \frac{1^3} {3} \right)\\ &=& - \left( \frac{64} {3}- \frac{1} {3} \right)\\ &=& - \frac{63} {3}\\ &=& - 21\\ \mathbf{ \int \limits_{1}^{4} -x^2\ dx } & \mathbf{=} & \mathbf{- 21 }\\ \end{array} }\)
+26367 \(\int \limits_{1}^{4} -x^2\ dx\) Wie berechne ich dieses Integral (schriftlich)?
\(\small{ \begin{array}{rcl} \int \limits_{1}^{4} -x^2\ dx &=& -\int \limits_{1}^{4} x^2\ dx \\ &=& - \left[ \frac{x^3} {3} \right]_{1}^{4} \\ &=& - \left( \frac{4^3} {3}- \frac{1^3} {3} \right)\\ &=& - \left( \frac{64} {3}- \frac{1} {3} \right)\\ &=& - \frac{63} {3}\\ &=& - 21\\ \mathbf{ \int \limits_{1}^{4} -x^2\ dx } & \mathbf{=} & \mathbf{- 21 }\\ \end{array} }\)
