AP Calculus AB:: Applications Of Integration

If \(v(t)\) is velocity, which integral gives the object's DISPLACEMENT over \([a,b]\)?

- **A)** \(\int_a^b |v(t)|\,dt\)
- **B)** \(\int_a^b v(t)\,dt\)
- **C)** \(v(b) - v(a)\)
- **D)** \(\int_a^b v'(t)\,dt\)

Which integral gives the TOTAL DISTANCE traveled by an object with velocity \(v(t)\) over \([a,b]\)?

- **A)** \(\int_a^b a(t)\,dt\)
- **B)** \(\int_a^b v(t)\,dt\)
- **C)** \(v(b)-v(a)\)
- **D)** \(\int_a^b |v(t)|\,dt\)

Given \(v(t) = t^2 - 4\) on \([0,3]\) (negative for \(t2\)), which expression correctly computes total distance traveled?

- **A)** \(\left|\int_0^2 (t^2-4)\,dt\right| + \int_2^3 (t^2-4)\,dt\)
- **B)** \(\int_0^2 (t^2-4)\,dt + \int_2^3 (t^2-4)\,dt\)
- **C)** \(\int_0^3 (t^2-4)\,dt\)
- **D)** \(v(3) - v(0)\)

If \(a(t)\) is acceleration and \(v(0)=5\), how do you find \(v(t)\) for \(t>0\)?

- **A)** \(v(t) = a(t) + 5\)
- **B)** \(v(t) = 5 + \int_0^t a(s)\,ds\)
- **C)** \(v(t) = \int_0^t a(s)\,ds\)
- **D)** \(v(t) = 5 \cdot \int_0^t a(s)\,ds\)

A particle has velocity \(v(t) = 3t^2 - 6t\) and starts at position \(s(0)=2\). Find \(s(t)\).

- **A)** \(s(t)=6t-6+2\)
- **B)** \(s(t)=t^3-3t^2+2t\)
- **C)** \(s(t) = t^3 - 3t^2 + 2\)
- **D)** \(s(t) = t^3-3t^2\)

If \(R(t)\) is the rate (in liters/min) that water enters a reservoir, and the reservoir starts with 500 liters, which expression gives the total amount of water at time \(T\)?

- **A)** \(500 + \int_0^T R(t)\,dt\)
- **B)** \(500 \cdot R(T)\)
- **C)** \(R(T) - R(0)\)
- **D)** \(\int_0^T R(t)\,dt\)

For the area between two curves \(y=f(x)\) (on top) and \(y=g(x)\) (below), where they intersect at \(x=a\) and \(x=b\), the area is given by:

- **A)** \(\int_a^b f(x)\,dx \cdot \int_a^b g(x)\,dx\)
- **B)** \(\int_a^b [f(x)-g(x)]\,dx\)
- **C)** \(\int_a^b [g(x)-f(x)]\,dx\)
- **D)** \(f(b)-g(a)\)

Find the area between \(y=x^2\) and \(y=x\) on \([0,1]\) (where \(y=x\) lies above \(y=x^2\)).

- **A)** \(1\)
- **B)** \(\dfrac{1}{6}\)
- **C)** \(\dfrac{1}{2}\)
- **D)** \(\dfrac{1}{3}\)