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Using Accumulation Functions and Definite Integrals

AP Calculus AB· difficulty 2/5

Water leaks out of a tank at r(t)=5−tr(t) = 5 - t L/min for t∈[0,5]t \in [0, 5]. Total water lost is

  • A

    5 L5~\text{L}

  • B

    10 L10~\text{L}

  • C

    25 L25~\text{L}

  • D

    252 L\tfrac{25}{2}~\text{L}

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Explanation

∫05(5−t) dt=[5t−t2/2]05=25−12.5=12.5\int_0^5 (5 - t)\,dt = [5t - t^2/2]_0^5 = 25 - 12.5 = 12.5.

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