Caluculus
posted by Alice on .
Can anyone help me with this problem?
Suppose that during a controlled experiment, the temperature in a beaker containing some chemical substance at time t is rising at a rate of (15/13)t^2(7t) degrees centigrade per minute. If the temperature after 1 minute is measured to be 18.6 degrees C, what is the temperature in the test tube after 10 minutes?

If you integrate the rate, you get temp.
Temp= INT (15/13 t^2 7t)dt
You know temp at 1 min is 18.6C. Solve for the constant of integration.
Then, you can find the temp after ten min.