Areas, Integrals, and Antiderivatives

The antiderivative method of evaluating definite integrals can also be used when we need to find an "area", and it is useful for solving applied problems.

Example 5: A robot has been programmed so that when it starts to move, its velocity after $t$ seconds will be $3 t^{2}$ feet/second.

(a) How far will the robot travel during its first 4 seconds of movement?

(b) How far will the robot travel during its next 4 seconds of movement?

(c) How many seconds before the robot is 729 feet from its starting place?

Solution:

(a) The distance during the first 4 seconds will be the area under the graph (Fig. 8) of velocity , $\mathrm{f}(t)=t$ from $t= 0$ to $t = 4$, and that area is the definite integral $\int_{0}^{4} 3 t^{2} \mathrm{dt}$. An antiderivative of $3 t^{2} \text { is } t^{3} \text { so } \int_{0}^{4} 3 t^{2} \mathrm{dt}=\left.t^{3}\right|_{0} ^{4}=(4)^{3}-(0)^{3}=64$ feet.

(b) $\int_{4}^{8} 3 t^{2} \mathrm{dt}=\left.t^{3}\right|_{4} ^{8}=(8)^{3}-(4)^{3}=512-64=448$ feet.

(c) This part is different from the other two parts. Here we are told the lower integration endpoint, $t = 0$, and the total distance, 729 feet, and we are asked to find the upper endpoint. Calling the upper endpoint $T$, we know that $729=\int_{0}^{\mathrm{T}} 3 t^{2} \mathrm{dt}=\left.t^{3}\right|_{0} ^{\mathrm{T}}=(\mathrm{T})^{3}-(0)^{3}=\mathrm{T}^{3}, \text { so } \mathrm{T}=\sqrt[3]{729}=9$ seconds.

Practice 4: (a) How far will the robot move between $t = 1$ second and $t = 5$ seconds? (b) How many seconds before the robot is 343 feet from its starting place?

Example 6: Suppose that $t$ minutes after putting 1000 bacteria on a Petri plate the rate of growth of the population is 6t bacteria per minute. (a) How many new bacteria are added to the population during the first 7 minutes? (b) What is the total population after 7 minutes? (c) When will the total population be 2200 bacteria?

Solution: 

(a) The number of new bacteria is the area under the rate of growth graph (Fig. 9), and one antiderivative of $6t$ is $3 t^{2}$ (check that $\mathrm{D}\left(3 t^{2}\right)=6 t$) so new bacteria = $\int_{0}^{7} 6 t \mathrm{dt}=\left.3 t^{2}\right|_{0} ^{7}=3(7)^{2}-3(0)^{2}=147$.

(b) The new population = {old population} + {new bacteria} = 1000 + 147 = 1147 bacteria

(c) If the total population is 2200 bacteria, then there are 2200 – 1000 = 1200 new bacteria, and we need to find the time T needed for that many new bacteria to occur.

1200 new bacteria = $\int_{0}^{\mathrm{T}} 6 t \mathrm{dt}=\left.3 t^{2}\right|_{0} ^{\mathrm{T}}=3(\mathrm{~T})^{2}-3(0)^{2}=3 \mathrm{~T}^{2} \text { so } \mathrm{T}^{2}=400$ and $T = 20$ minutes. After 20 minutes, the total bacteria population will be 1000 + 1200 = 2200.

Practice 5: (a) How many new bacteria will be added to the population between $t = 4$ and $t = 8$ minutes? (b) When will the total population be 2875 bacteria? (Hint: How many are new?)