a student blows a balloon and its volume increases at a rate of \( \pi(20 - t^2)\text{ cm}^3\text{ s}^{-1} \) after t seconds. If the initial volume is 0 cm...

The volume of the balloon increases at a rate of π(20-t^2) cm^3S^-1 after t seconds. If we want to find the volume of the balloon after 2 seconds, we need to integrate the given rate with respect to time from 0 to 2, since we want to know the change in volume from the initial volume of 0 cm^3 after 2 seconds. Integrating the given rate with respect to time, we get: ∫[0,2] π(20-t^2) dt = π[20t - (t^3/3)] from 0 to 2 Plugging in the values, we get: π[20(2) - (2^3/3)] - π[20(0) - (0^3/3)] = π[40 - 8/3] = π[120/3 - 8/3] = π[112/3] = 37.33π Therefore, the volume of the balloon after 2 seconds is approximately 37.33π cubic centimeters.