# Caluculus

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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?

• Caluculus -

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.

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