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AP Calculus AB : Definite integral of the rate of change of a quantity over an interval interpreted as the change of the quantity over the interval

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Example Questions

Example Question #23 : Interpretations And Properties Of Definite Integrals

A pot of water begins at a temperature of $20^\circ C$ and is heated at a rate of $\tfrac{10}{(t+1)^2}$ degrees Celsius per minute. What will the temperature of the water be after 4 minutes?

Possible Answers:

$30.5^\circ C$

$32^\circ C$

$28^\circ C$

$25^\circ C$

$24^\circ C$

Correct answer:

$28^\circ C$

Explanation:

Let denote the temperature of the pot after minutes.

The first thing to realize is that the quantity $\frac{-10}{(t+1)^2}$ is the derivative of T(t) . Using the fundamental theorem of Calculus, we know that

$\int_{0}^{4}\frac{10}{(t+1)^2} = T(4) - T(0)$

From there, we just need to solve the integral. Letting u = t+1, du=dt, we have the following:

$\int_{0}^{4}\frac{10}{(t+1)^2}dx = 10\int_{0}^{4} \frac{1}{u^2}du = \frac{-10}{(t+1)}]_0^4 = -2 + 10 = 8$

Where the second equality follows by the power rule, and re-substituting t+1 = u.

Thus, we now have the equation 8 = T(4) - T(0) . Because we know that the water started at $20^\circ C$ , all we need to do is rearrange and substitute.

8 + T(0) = T(4)

8 + 20 = T(4)

28 = T(4)

Yielding our final answer, $28^\circ C$

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AP Calculus AB : Definite integral of the rate of change of a quantity over an interval interpreted as the change of the quantity over the interval

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Example Question #23 : Interpretations And Properties Of Definite Integrals

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