Problem 16 - 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. $$ \int_{0}^{2}[\sqrt{3} f(t)+\sqrt{2} g(t)+\pi] d t $$

Step-by-Step Solution

Answer

The integral evaluates to \(5\sqrt{3} + 4\sqrt{2} + 2\pi\).

Step 1: Applying Linearity to Separate Terms

The integral is given as \( \int_{0}^{2} [\sqrt{3} f(t) + \sqrt{2} g(t) + \pi] \, dt \). Using the linearity property, we can separate and multiply the terms: \[\int_{0}^{2} \sqrt{3} f(t) \, dt + \int_{0}^{2} \sqrt{2} g(t) \, dt + \int_{0}^{2} \pi \, dt\]

Step 2: Evaluate \(\int_{0}^{2} \sqrt{3} f(t) \, dt\)

Here, we can factor the constant \(\sqrt{3}\) out of the integral: \[\sqrt{3} \int_{0}^{2} f(t) \, dt\]We need to find \(\int_{0}^{2} f(t) \, dt\), using the additivity property: \[\int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx = 2 + 3 = 5\]Thus, \(\int_{0}^{2} f(t) \, dt = 5\), and substituting back, we have: \[\sqrt{3} \times 5 = 5\sqrt{3} \]

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

Similarly, factor the constant \(\sqrt{2}\) out:\[\sqrt{2} \int_{0}^{2} g(t) \, dt\]We already know from the problem statement that \(\int_{0}^{2} g(t) \, dt = 4\). Therefore, substitute back: \[\sqrt{2} \times 4 = 4\sqrt{2}\]

Step 4: Evaluate \(\int_{0}^{2} \pi \, dt\)

Since \(\pi\) is a constant, we simply multiply \(\pi\) by the interval length, which is \(2 - 0 = 2\):\[\pi \times 2 = 2\pi\]

Step 5: Combine All Evaluated Integrals

Now, combine the results from Steps 2, 3, and 4:\[5\sqrt{3} + 4\sqrt{2} + 2\pi\]Thus, the value of the integral \( \int_{0}^{2} [\sqrt{3} f(t) + \sqrt{2} g(t) + \pi] \, dt \) is: \[5\sqrt{3} + 4\sqrt{2} + 2\pi\]

Key Concepts

Linearity of Integrals

The linearity of integrals is a fundamental concept, which allows us to simplify the evaluation of definite integrals involving sums or constant multiples of functions. It states that the integral of a sum of functions is equal to the sum of the integrals of each function separately. Similarly, if a function is multiplied by a constant, this constant can be factored out of the integral.

• For example, if you have an expression like \( \int (a \cdot f(x) + b \cdot g(x)) \, dx \), you can split it as: \( a \int f(x) \, dx + b \int g(x) \, dx \).
• This property is particularly useful because it breaks down complex integrals into simpler parts that can be dealt with individually.

In the exercise, we used linearity to separate the terms \( \sqrt{3}f(t) \), \( \sqrt{2}g(t) \), and the constant \( \pi \) from the given integral. This allows each part to be evaluated independently, making the problem much more manageable. Remember, applying linearity is crucial when dealing with functions that are added together and scaled.

Properties of Integrals

Understanding the properties of integrals helps in effortlessly handling continuous functions over a given interval. These properties include linearity, constant factors, and the effects of reversing limits. Let's delve into these:

• Linearity: This is like playing with Lego blocks—separate and rearrange functions inside the integral.
• Constant Multiple: If a function is multiplied by a constant inside the integral, you can pull that constant outside. E.g., \( \int c \cdot f(x) \, dx = c \int f(x) \, dx \).
• Sign Change on Limits: Reversing the limits of integration changes the sign. For example, \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \).

Utilizing these properties simplifies the process, making it easier to solve integrals. When you comprehend how these work, long and daunting-looking calculations become simple, step-by-step operations. Like in the problem, knowledge of these properties helps treat each component separately, reducing complexities and allowing a total summation towards the end.

Interval Additivity

Interval additivity is another key property of definite integrals. It means that if an interval is divided into parts, the integral over the entire interval is the sum of the integrals over those parts. Mathematically, this is expressed as: \[ \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \]. This property is quite intuitive if you envision it as summing up various sections of an area under a curve.

• This concept is invaluable when dealing with piecewise functions or if given values of integrals over sub-intervals.
• In our exercise, this property was crucial in determining \( \int_{0}^{2} f(t) \, dt \), where we added \( \int_{0}^{1} f(x) \, dx \) and \( \int_{1}^{2} f(x) \, dx \).

Think of it as stitching together smaller segments to get a complete picture. This approach cuts through overwhelming data into digestible pieces, a handy strategy for problem-solving in mathematics.