Riemann Sums: Videos & Practice Problems

$\frac{1}{100}{\int }_0^{100}T\left(t\right)dt\approx 60.4$

This value means that the average temperature of the soup over the first $100$ seconds is about $60.4{}^{\circ }\text{C}$

$\frac{1}{100}{\int }_0^{100}T\left(t\right)dt\approx 63.8$

This value means that the average temperature of the soup over the first $100$ seconds is about $63.8{}^{\circ }\text{C}$

$\frac{1}{100}{\int }_0^{100}T\left(t\right)dt\approx 60.4$

This value means that the exact temperature of the soup after $50$ seconds is $60.4{}^{\circ }\text{C}$

$\frac{1}{100}{\int }_0^{100}T\left(t\right)dt\approx 63.8$

This value means that the exact temperature of the soup after $50$ seconds is $63.8{}^{\circ }\text{C}$