Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.1.46a

Chapter 6, Problem 6.1.46a

Flow rates in the Spokane River The daily discharge of the Spokane River as it flows through Spokane, Washington, in April and June is modeled by the functions
r1(t) = 0.25t²+37.46t+722.47 (April) and
r2(t) = 0.90t²−69.06t+2053.12 (June), where the discharge is measured in millions of cubic feet per day, and t=0 corresponds to the beginning of the first day of the month (see figure).

a. Determine the total amount of water that flows through Spokane in April (30 days).

Verified step by step guidance

Identify the function that models the daily discharge for April, which is given as $r_1(t) = 0.25t^2 + 37.46t + 722.47$, where $t$ is the number of days from the start of April and $r_1(t)$ is the discharge in millions of cubic feet per day.

Understand that the total amount of water discharged over the 30 days of April is the integral of the discharge rate function $r_1(t)$ from $t=0$ to $t=30$. This is because integration sums up the continuous flow over time.

Set up the definite integral to find the total discharge: $\int_0^{30} (0.25t^2 + 37.46t + 722.47) \, dt$.

Integrate the function term-by-term: the integral of \$0.25t^2$ is \(\frac{0.25}{3}t^3$, the integral of \)37.46t$ is \(\frac{37.46}{2}t^2$, and the integral of $722.47$ is \)722.47t$.

Evaluate the resulting antiderivative at the upper limit $t=30$ and subtract the value at the lower limit $t=0$ to find the total volume of water discharged in April.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integral as Total Accumulation

The definite integral of a rate function over a time interval gives the total accumulated quantity during that period. In this problem, integrating the discharge rate r1(t) over 30 days calculates the total volume of water flowing through the river in April.

Interpreting Quadratic Functions

The discharge rates r1(t) and r2(t) are modeled by quadratic functions, which describe how the flow changes over time. Understanding the shape and behavior of these parabolas helps in setting up the integral limits and interpreting the physical meaning of the flow rates.

Units and Context in Integration

The discharge rates are given in millions of cubic feet per day, and time t is in days. When integrating, it is important to keep track of units to ensure the final result represents the total volume of water in millions of cubic feet over the specified time.

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