Displacement, total distance and changing inflow and outflow rates.
Example 1
Find the area between y=4x and y=x2 for 0≤x≤4.
y
01234051015 x
y=4xy=x2Show worked answer
Answer: 664
- The line is above the parabola inside the interval.
- Area =∫04(4x−x2)dx.
- An antiderivative is 24x2−3x3. Evaluating gives 264−364=664.
Example 2
A particle has velocity v(t)=2t−2m/s. Find its total distance travelled from t=0 to t=2.
v(m/s)
00.511.52-202 t(s)
Velocity versus timeShow worked answer
Answer: 2
- Velocity changes sign at t=1. Split there to integrate ∣v∣.
- Each half is a triangle with base 1 and height 2.
- Total distance =2× ½ ×1×2=2m.
Example 3
Water enters a tank at r(t)=6t+3litres/minute. Initially there are 50 litres. How much is present after 2 minutes, with no outflow?
r(minL)
012051015 t(min)
r(t)=6t+3Show worked answer
Answer: 68
- Accumulated inflow =∫02(6t+3)dt.
- This is 18 litres.
- Add the initial amount: 50+18=68.