Using Accumulation Functions and Definite Integrals Practice

Sample questions

4 of 4 — sign in to practice the rest with adaptive difficulty and mastery tracking.

1. Sample 1difficulty 2/5

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

   • A

     $5~\text{L}$

   • B

     $10~\text{L}$

   • C

     $25~\text{L}$

   • D

     $\tfrac{25}{2}~\text{L}$

     check_circle

   Why

   $\int_0^5 (5 - t)\,dt = [5t - t^2/2]_0^5 = 25 - 12.5 = 12.5$.

   View full solution & statsarrow_forward

2. Sample 2difficulty 2/5

   A car's velocity is $v(t) = 60 - 10t$ mph for $t \in [0, 6]$. The total displacement (mi) is

   • A

     $120$

   • B

     $360$

   • C

     $0$

   • D

     $180$

     check_circle

   Why

   $\int_0^6 (60 - 10t)\,dt = 360 - 180 = 180$.

   View full solution & statsarrow_forward

3. Sample 3difficulty 3/5

   By the net change theorem, $\int_a^b f'(t)\,dt$ equals:

   • A

     $f(b) + f(a)$

   • B

     $f(b) - f(a)$

     check_circle
   • C

     $\dfrac{f(b)+f(a)}{2}$

   • D

     $f'(b) - f'(a)$

   Why

   Net change theorem: $\int_a^b f'(t)\,dt = f(b) - f(a)$.

   View full solution & statsarrow_forward

4. Sample 4difficulty 3/5

   Water flows in at rate $f(t) = 100$ L/h and flows out at rate $g(t) = 30 + 5 t$ L/h on $[0, 4]$. Net change in tank:

   • A

     $160$

   • B

     $240$

     check_circle
   • C

     $280$

   • D

     $320$

   Why

   Net rate $= 70 - 5t$. $\int_0^4 (70 - 5t)\,dt = 280 - 40 = 240$.

   View full solution & statsarrow_forward