# Total Change using Definite Integral

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If given a **rate of change** function, we can find the **total change** over a given period of time by integrating.

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If the rate of change of the function is given by: .. $\dfrac{df}{dx}$

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Then the **total change** between $x = a$ and $x = b$ is given by

… … $\displaystyle{ \int\limits_a^b \; \dfrac{df}{dx} \; dx }$

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### Example 1

A water tank is leaking such that the volume changes according to:

… … $\dfrac{dV}{dt} = -4e^{-0.5t} \quad t \geqslant 0$

… … … where **t** is measured in minutes and **V** is measured in cm^{3}

Find the total amount of water that leaks from the tank in the first **6** minutes (correct to 1 decimal place).

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**Solution:**

… … **Change in Volume** from $t = 0$ to $t = 6$

… … $= \displaystyle{ \int\limits_0^6 \; -4e^{-0.5t} \; dt }$

… … $= \Big[ 8e^{-0.5t} \Big]_0^6$

… … $= \Big( 8e^{-3} \Big) - \Big( 8e^0 \Big)$

… … $= -7.6$ cm^{3}

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… … Hence $7.6$ cm^{3} of water leaks from the tank in the first 6 minutes.

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### Example 2

{still to come}

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