Problem 38 - Biocalculus Calculus for the Life Sciences

Question

If $$\int_{1}^{5} f(x) d x=12$$ and $$\int_{4}^{5} f(x) d x=3.6$$, find $$\int_{1}^{4} f(x) d x$$.

Step-by-Step Solution

Answer

\( \int_{1}^{4} f(x) \, dx = 8.4 \)

Step 1: Understanding the Problem

We are given two definite integrals: one from 1 to 5 and another from 4 to 5. We are asked to find the definite integral from 1 to 4. Our goal is to use the information given to determine this unknown value.

Step 2: Using the Property of Integrals

We know that the integral from 1 to 5 can be split into two parts: from 1 to 4, and from 4 to 5. Thus, we can write: \[ \int_{1}^{5} f(x) \, dx = \int_{1}^{4} f(x) \, dx + \int_{4}^{5} f(x) \, dx \]

Step 3: Substituting Known Values

Substitute the known values into the equation: \[ 12 = \int_{1}^{4} f(x) \, dx + 3.6 \] This reflects the total area under the curve from 1 to 5 as being split into two segments.

Step 4: Solving for the Unknown Integral

Now, solve for the integral from 1 to 4 by isolating it:\[ \int_{1}^{4} f(x) \, dx = 12 - 3.6 \] Perform the subtraction to find the answer.

Step 5: Final Calculation

Calculate the difference: \[ \int_{1}^{4} f(x) \, dx = 8.4 \] This is the value of the integral over the interval from 1 to 4.

Key Concepts

Definite Integrals

A definite integral is a fundamental concept in calculus. It represents the signed area under a curve over a specific interval. In simpler terms, when you calculate \( \int_{a}^{b} f(x) \, dx \), you're finding the total area between the curve of the function \( f(x) \) and the \( x \)-axis, from \( x = a \) to \( x = b \).
These integrals are crucial not only for determining areas but also for solving real-world problems related to accumulated quantities like distance and volume.
In the example provided, you're given the integral from 1 to 5 as 12, which means the total area under \( f(x) \) from \( x = 1 \) to \( x = 5 \) is 12. Similarly, the area from 4 to 5 is 3.6.
It's important to understand that the values of definite integrals can be manipulated using calculations and algebra, as they were in this problem, to determine unknowns in the relationships between different intervals.

Properties of Integrals

The properties of integrals are powerful tools in solving calculus problems. They allow you to break down complex integrals into simpler parts, and also to piece them together again. One essential property used often, including in the given exercise, is the additivity property of integrals.
This property states that for any function \( f(x) \) and numbers \( a \), \( b \), and \( c \) (where \( a < b < c \)), you can write:

• \( \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \)

This is precisely the property utilized in solving the problem at hand. By knowing the integral from 1 to 5 and the part from 4 to 5, the problem applies this property to find the missing segment from 1 to 4.
Integrals also share properties with sums and can be manipulated in similar ways, involving shifting intervals and adjusting integration limits to solve problems effectively. Recognizing and applying these properties can simplify and solve complex integrals efficiently.

Integration Techniques

Integration techniques are methods for calculating integrals that go beyond basic application of definitions. They help solve integrals quickly and accurately.
While the given problem involves basic arithmetic to split integrals, many problems require advanced techniques such as:

• Substitution: A technique similar to the chain rule for differentiation, where you change variables to simplify the integral.
• Integration by Parts: Utilized when integrals involve products of functions, extending the product rule for derivatives.
• Partial Fractions: Used for rational functions, breaking them into simpler fractions before integrating.

Mastering these techniques equips you with the ability to solve complex integration problems efficiently. In more advanced applications, understanding when and how to use each technique becomes crucial.
For simpler problems, as demonstrated in the exercise where splitting and combining known integrals suffices, these techniques are often not necessary. However, building a strong foundation in these methods ensures preparedness for any integral calculus challenge.