## University Calculus: Early Transcendentals (3rd Edition)

$1$
Solve $\int_{\ln (2)}^{\ln (3)} e^x dx$ This implies $\int_{\ln (2)}^{\ln (3)} e^x dx=[e^x]_{\ln (2)}^{\ln (3)}$ Thus, $\int_{\ln (2)}^{\ln (3)} e^x dx=[e^{ln (3)}-e^{ln (2)}]$ or, $\int_{\ln (2)}^{\ln (3)} e^x dx=3-2$ Hence, $\int_{\ln (2)}^{\ln (3)} e^x dx=1$