Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that \(\int_{0}^{1} f(x) d x=2, \int_{1}^{2} f(x) d x=3, \int_{0}^{1} g(x) d x=-1\) and \(\int_{0}^{2} g(x) d x=4 .\) Use properties of definite integrals (linearity, interval additivity, and so on) to calculate each of the integrals in Problems. $$ \int_{0}^{2}[3 f(t)+2 g(t)] d t $$

Step-by-Step Solution

Answer

The value of the integral is 23.

Step 1: Applying Linearity of Integrals

The first step is to apply the linearity property of integrals, which states that \( \int (a \cdot f(t) + b \cdot g(t)) \, dt = a \int f(t) \, dt + b \int g(t) \, dt \). In this problem, we need to compute \( \int_{0}^{2} [3f(t) + 2g(t)] \, dt \). This can be rewritten as: \[ 3 \int_{0}^{2} f(t) \, dt + 2 \int_{0}^{2} g(t) \, dt \]

Step 2: Compute \(\int_{0}^{2} f(t) \, dt\) using Interval Additivity

The integral of \( f(t) \) from 0 to 2 can be found by adding the known integrals over subintervals, as \( \int_{0}^{2} f(t) \, dt = \int_{0}^{1} f(t) \, dt + \int_{1}^{2} f(t) \, dt \). Given \( \int_{0}^{1} f(t) \, dt = 2 \) and \( \int_{1}^{2} f(t) \, dt = 3 \), we have: \[ \int_{0}^{2} f(t) \, dt = 2 + 3 = 5 \]

Step 3: Simplify \(\int_{0}^{2} g(t) \, dt\)

We already know from the problem statement that \( \int_{0}^{2} g(t) \, dt = 4 \). This information is directly provided.

Step 4: Combine Results Using Linearity

Now, substitute the values we computed into the expression for linearity. So, we have: \[ 3 \int_{0}^{2} f(t) \, dt + 2 \int_{0}^{2} g(t) \, dt = 3(5) + 2(4) \]

Step 5: Calculate the Final Value

Calculate each part: \( 3(5) = 15 \) and \( 2(4) = 8 \). Now, add these results to get the final value: \[ 15 + 8 = 23 \]

Key Concepts

Linearity of Integrals

In integral calculus, the linearity of integrals is a key property that allows us to simplify complex integrals into simpler components. This property states that the integral of a sum of functions is equal to the sum of the integrals of the individual functions. Specifically, it can be expressed as: \( \int (a \cdot f(x) + b \cdot g(x)) \, dx = a \int f(x) \, dx + b \int g(x) \, dx \).
This rule is crucial because it helps in breaking down complicated integrals into more manageable parts by using constant factors. This means when you have some constants multiplying with the functions inside an integral (like 3 and 2 in this problem), you can "factor" them out, making the integral easier to evaluate.

• Allows decomposition of complex expressions into simpler terms.
• Facilitates handling of weighted functions by simply adjusting the constants outside the integral.

In our exercise, we applied linearity right away to separate and individually calculate \( \int_{0}^{2} f(t) \, dt \) and \( \int_{0}^{2} g(t) \, dt \), multiplied by their corresponding constants 3 and 2.

Interval Additivity

Interval additivity is another cornerstone in the understanding and computation of definite integrals. It states that if you want to find the integral over a larger interval, you can divide it into smaller, consecutive intervals. The integral over the whole interval is the sum of the integrals over these smaller intervals.
This is expressed mathematically as: \( \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \). This property is incredibly useful for solving problems where integrals over specific segments are provided.

• Allows calculation over larger intervals by summing smaller intervals.
• Helps split complex problems into parts we might already know.

In the problem, we utilized interval additivity to find \( \int_{0}^{2} f(t) \, dt \) by adding the integrals \( \int_{0}^{1} f(t) \, dt \) and \( \int_{1}^{2} f(t) \, dt \), which were given. This gave us the result of 5 for the integral over the entire interval from 0 to 2.

Integral Calculus

Integral calculus is aimed at understanding the calculation and application of integrals. An integral, conceptually, is like the accumulation of quantities, which can represent areas under curves, total distances, or any accumulative measurement.
Definite integrals are particularly significant because they provide exact values, thus enabling practical applications across various fields like physics, engineering, and statistics. The definite integral of a function \( f(x) \) from \( a \) to \( b \) is denoted as \( \int_{a}^{b} f(x) \, dx \), and it computes the net area under the curve from \( x = a \) to \( x = b \).

• Integral calculus unites differential calculus through the Fundamental Theorem of Calculus, linking derivatives and integrals.
• Encompasses techniques for finding and applying both definite and indefinite integrals.

In our example, integral calculus allowed us to use properties like linearity and interval additivity to compute the definite integral of the function \( 3f(t) + 2g(t) \) over the interval from 0 to 2, arriving at a final answer of 23 after all calculations.