Exercises — Net change theorem

What we do: the theorem says net change $={\int }_1^4{F}^{\prime }(x)dx$. We are handed $F^{\prime }(x)=6x$, so integrate it. Why this and nothing else: we are asked for the change in $F$, and the theorem is exactly "integral of the rate = net change." We do not need to know $F(1)$ or any $+C$. ${\int }_1^{4}6xdx=[3x^2{]}_1^4=3(16)-3(1)=48-3=45.$ Net change $=45$.

What we do: velocity is the rate of change of position, so displacement $={\int }_0^{7}v(t)dt$. Why: $v=s^{\prime }$, and the theorem gives ${\int }_0^7{s}^{\prime }dt=s(7)-s(0)$ = net change in position = displacement. ${\int }_0^{7}4dt=[4t{]}_0^7=28-0=28\text{ m}.$ Displacement $=28$ m. (Since $v$ never turns negative, distance is the same $28$ m.)

What we do: $r$ is the rate of change of volume $V$, so net change $={\int }_0^{6}r(t)dt$. ${\int }_0^6(12-3t)dt=[12t-\frac{3}{2}{t}^2{]}_0^6=(72-54)-0=18.$ Net change $=+18$ L. (b) The inflow rate is positive for $t<4$ (water entering) and negative for $t>4$ (water leaving). The net result is still $+18$ L, i.e. the tank ends with 18 L more than it started, even though water drained out during the last two minutes.

Where does $v$ change sign? Set $2t-6=0\Rightarrow t=3$. On $[0,3)$, $v<0$ (moving backward); on $(3,5]$, $v>0$ (forward).

(a) Displacement $={\int }_0^5(2t-6)dt=[t^2-6t{]}_0^5=(25-30)-0=-5$ m. Net: the particle ends 5 m behind the start.

(b) Total distance — put the bars inside, split at $t=3$: ${\int }_0^3|v|dt+{\int }_3^5|v|dt={\int }_0^3(6-2t)dt+{\int }_3^5(2t-6)dt.$ First piece: $[6t-t^2{]}_0^3=18-9=9$. Second piece: $[t^2-6t{]}_3^5=(25-30)-(9-18)=(-5)-(-9)=4$. Total distance $=9+4=13$ m. (Bigger than $|-5|$ — good sanity check.)

What we do: marginal cost is $C^{\prime }$, so the extra cost is the net change ${\int }_{20}^{50}{C}^{\prime }(x)dx$. ${\int }_{20}^{50}(4x+30)dx=[2x^2+30x{]}_{20}^{50}=(5000+1500)-(800+600)=6500-1400=5100.$ Extra cost $=$ ₹5100. Why no baseline: the true cost function is $C(x)=2x^2+30x+C_0$ where $C_0$ is the fixed start-up cost. The change $C(50)-C(20)$ subtracts $C_0$ away. A difference is blind to the constant.

What we do: the theorem gives the change, then we add it to the known start. $P(2)-P(0)={\int }_0^{2}100e^{-0.5t}dt.$ Integrate ($\int e^{kt}dt=\frac{1}{k}{e}^{kt}$, here $k=-0.5$): $=100\cdot [\frac{1}{-0.5}{e}^{-0.5t}{]}_0^2=-200[e^{-0.5t}{]}_0^2=-200(e^{-1}-1)=200(1-e^{-1}).$ Numerically $e^{-1}\approx 0.3679$, so change $\approx 200(0.6321)=126.42$. $P(2)=800+200(1-e^{-1})\approx 800+126.42=926.42\approx 926$ animals. Why add $800$ here (but not in L3.1): we were asked for an absolute amount $P(2)$, not a change — so we take the net change and stack it on top of the known starting value.

Look at the figure below: $\sin t$ makes three humps on $[0,3\pi ]$ — up on $[0,\pi ]$, down on $[\pi ,2\pi ]$, up on $[2\pi ,3\pi ]$. Sign changes at $t=\pi$ and $t=2\pi$.

(a) Displacement $={\int }_0^{3\pi }\sin tdt=[-\cos t{]}_0^{3\pi }=(-\cos 3\pi )-(-\cos 0)=-(-1)-(-1)=1+1=2$ m.

(b) Total distance — bars inside, split at $\pi$ and $2\pi$. By symmetry each hump has area ${\int }_0^{\pi }\sin tdt=[-\cos t{]}_0^{\pi }=(1)-(-1)=2.$ All three humps have the same size $2$, so $\text{distance}=2+2+2=6\text{ m}.$ Displacement $=2$ m, distance $=6$ m. (Displacement $2=$ hump1 $-$ hump2 $+$ hump3 $=2-2+2$ — the middle hump cancels; distance counts all three.)

What we do: the net rate of change of volume is inflow minus outflow: $V^{\prime }(t)=8-2t$. The net change is ${\int }_0^6{V}^{\prime }(t)dt$. Why one integral: volume responds to the combined rate; adding two integrals $\int \text{in}-\int \text{out}$ gives the same thing, so we may merge them first. ${\int }_0^6(8-2t)dt=[8t-t^2{]}_0^6=(48-36)-0=12.$ Net change $=+12$ L. (Inflow beats outflow overall, even though after $t=4$ the outflow $2t>8$ drains the reservoir.)

(a) Position via net change. For any $t$, $s(t)-s(0)={\int }_0^t(3{\tau }^2-12)d\tau =[{\tau }^3-12\tau {]}_0^t=t^3-12t.$ Add the start: $s(t)=5+t^3-12t$. Why this works: the theorem hands us the change from $0$ to $t$; stacking it on $s(0)=5$ recovers the actual position — no separate $+C$ hunt needed.

(b) Sign of $v$: $3t^2-12=0\Rightarrow t^2=4\Rightarrow t=2$ (on $[0,3]$). On $[0,2)$, $v<0$; on $(2,3]$, $v>0$.

Displacement $=s(3)-s(0)=(5+27-36)-5=-4-5=\dots$ wait — compute directly: ${\int }_0^3(3t^2-12)dt=[t^3-12t{]}_0^3=(27-36)-0=-9$ m. Displacement $=-9$ m (ends 9 m behind start; check: $s(3)=5-9=-4$, and $-4-5=-9$ ✓).

Total distance — split at $t=2$: ${\int }_0^2|3t^2-12|dt+{\int }_2^3|3t^2-12|dt={\int }_0^2(12-3t^2)dt+{\int }_2^3(3t^2-12)dt.$ First: $[12t-t^3{]}_0^2=24-8=16$. Second: $[t^3-12t{]}_2^3=(27-36)-(8-24)=(-9)-(-16)=7$. Total distance $=16+7=23$ m. (Sanity: $23>|-9|$ ✓.)

What we do: net change $={\int }_0^4(a-t)dt$; set it to $0$ and solve for $a$. ${\int }_0^4(a-t)dt=[at-\frac{t^2}{2}{]}_0^4=4a-8.$ Set $4a-8=0\Rightarrow a=2$. $a=2$. Interpretation: with $a=2$, the inflow starts strong ($R(0)=2$) and later goes negative ($R(4)=-2$), and over the full interval the gains and losses cancel exactly — net change zero, even though the tank was rising then falling the whole time.

Net change / displacement / cost increase / net volume ::: ${\int }_a^b{F}^{\prime }(x)dx$ (keep the signs). Total distance / total amount moved ::: ${\int }_a^b|F^{\prime }(x)|dx$ (split at zeros, add sizes). Actual final value $F(b)$ ::: $F(a)+{\int }_a^b{F}^{\prime }(x)dx$ (stack the change on the known start).