Calcolare i seguenti integrali definiti:

Esercizio 1 \[ \int_{0}^{1}\left(3x^{2}-x+2\right)dx \] Soluzione

Calcoliamo l’integrale indefinito \[ F\left(x\right)=\int\left(3x^{2}-x+2\right)dx=x^{3}-\frac{x^{2}}{2}+2x+C \] Utilizziamo ora la formula fondamentale del calcolo integrale: \[ \int_{a}^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right) \] ovvero \[ \int_{0}^{1}\left(3x^{2}-x+2\right)dx=F\left(1\right)-F\left(0\right) \] \[ \int_{0}^{1}\left(3x^{2}-x+2\right)dx=\left(1^{3}-\frac{1^{2}}{2}+2\cdot1+C\right)-\left(0^{3}-\frac{0^{2}}{2}+2\cdot0+C\right) \] \[ \int_{0}^{1}\left(3x^{2}-x+2\right)dx=1-\frac{1}{2}+2=\frac{5}{2} \] Quindi otteniamo: \[ \int_{0}^{1}\left(3x^{2}-x+2\right)dx=\frac{5}{2} \] Esercizio 2 \[ \int_{-\frac{1}{2}}^{1}\left(3x^{3}-4x^{2}+3x-1\right)dx \] Soluzione

Calcoliamo l’integrale indefinito \[ F\left(x\right)=\int\left(3x^{3}-4x^{2}+3x-1\right)dx=\frac{3}{4}x^{4}-\frac{4}{3}x^{3}+\frac{3}{2}x^{2}-x+C \] Utilizziamo ora la formula fondamentale del calcolo integrale: \[ \int_{a}^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right) \] ovvero \[ \int_{-\frac{1}{2}}^{1}\left(3x^{3}-4x^{2}+3x-1\right)dx=F\left(1\right)-F\left(-\frac{1}{2}\right) \] \[ \int_{-\frac{1}{2}}^{1}\left(3x^{3}-4x^{2}+3x-1\right)dx= \] \[ =\left(\frac{3}{4}\cdot1^{4}-\frac{4}{3}\cdot1^{3}+\frac{3}{2}\cdot1^{2}-1+C\right)-\left[\frac{3}{4}\left(-\frac{1}{2}\right)^{4}-\frac{4}{3}\left(-\frac{1}{2}\right)^{3}+\frac{3}{2}\left(-\frac{1}{2}\right)^{2}+\frac{1}{2}+C\right]= \] \[ =\frac{3}{4}-\frac{4}{3}+\frac{3}{2}-1+C-\frac{3}{64}-\frac{1}{6}-\frac{3}{8}-\frac{1}{2}-C \] \[ \int_{-\frac{1}{2}}^{1}\left(3x^{3}-4x^{2}+3x-1\right)dx= \] \[ =\frac{144-256+288-192-9-32-72-96}{192}=-\frac{225}{192}=-\frac{75}{64} \] Quindi otteniamo: \[ \int_{-\frac{1}{2}}^{1}\left(3x^{3}-4x^{2}+3x-1\right)dx=-\frac{75}{64} \] Esercizio 3 \[ \int_{-1}^{1}\left(\frac{3}{4}x^{2}-\frac{1}{2}x^{\frac{1}{3}}-2\right)dx \] Soluzione

Calcoliamo l’integrale indefinito \[ F\left(x\right)=\int\left(\frac{3}{4}x^{2}-\frac{1}{2}x^{\frac{1}{3}}-2\right)dx=\frac{1}{4}x^{3}-\frac{3}{8}x^{\frac{4}{3}}-2x+C \] \[ F\left(x\right)=\int\left(\frac{3}{4}x^{2}-\frac{1}{2}x^{\frac{1}{3}}-2\right)dx=\frac{1}{4}x^{3}-\frac{3}{8}\cdot\sqrt[3]{x^{4}}-2x+C \] Utilizziamo ora la formula fondamentale del calcolo integrale: \[ \int_{a}^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right) \] ovvero \[ \int_{-1}^{1}\left(\frac{3}{4}x^{2}-\frac{1}{2}x^{\frac{1}{3}}-2\right)dx=F\left(1\right)-F\left(-1\right) \] \[ \int_{-1}^{1}\left(\frac{3}{4}x^{2}-\frac{1}{2}x^{\frac{1}{3}}-2\right)dx=\left(\frac{1}{4}\cdot1^{3}-\frac{3}{8}\cdot\sqrt[3]{1^{4}}-2+C\right)-\left[\frac{1}{4}\left(-1\right)^{3}-\frac{3}{8}\cdot\sqrt[3]{\left(-1\right)^{4}}+2+C\right] \] \[ \int_{-1}^{1}\left(\frac{3}{4}x^{2}-\frac{1}{2}x^{\frac{1}{3}}-2\right)dx=\frac{1}{4}-\frac{3}{8}-2+\frac{1}{4}+\frac{3}{8}-2=-\frac{7}{2} \] Quindi otteniamo: \[ \int_{-1}^{1}\left(\frac{3}{4}x^{2}-\frac{1}{2}x^{\frac{1}{3}}-2\right)dx=-\frac{7}{2} \]