Calculus: average temperature

a piece of steel at 1500 degrees F is removed from the oven and placed in a room at 70 degrees F. The temperature T of the steel, t minutes after it starts cooling, is given by T = 70 + 1430e^(-0.3t). Find, to the nearest degree, the average temperature of the steel over the first hour.

i don't know how to approach this/what formula to use. I'm thinking of Newton's Law of Cooling, but that's not exactly what "average temperature" is. please help, thanks.

Avg of f(x) over interval [a,b]
= 1/(b-a) ∫ₐᵇ f(x) dx

Avg temperature T(t) over first hour (from t = 0 to t = 60)
Avg = 1/60 ∫₀⁶⁰ (70 + 1430 e^(-0.3t)) dt
Avg = 1/60 (70t - 14300/3 e^(-0.3t)) |₀⁶⁰
Avg = 1/60 (70(60-0) - 14300/3 (e⁻¹⁸ - e⁰))
Avg = 70 - 715/9 (e⁻¹⁸ - 1)
Avg = 149.444443235

Average temperature (to nearest degree) = 149 degree

Integrate the temperature as a function of time over the interval. Then divide the result by the length of the interval.That will give you the average temperature over the interval.

Interval is 1 hour, temperature is a function of minutes, so integrate the temperature from 0 to 60 minutes and divide the result by 60.

formula? there is nothing to integrate - you are given temperature equation (result, not the cooling rate). since everything is given, just plug in numbers:
T = 70 + 1430e^(-0.3t)

for example:
T(0)=70 + 1430e^(0)=1500F (correct, this is initial temperature)

then
T(60)=70 + 1430e^(-0.3*60)
T(60)=70+1430e^(-18)
T(60)=70.00002177887103493905866367538 F