Problem 16 - Calculus

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

Step-by-Step Solution

Answer

The value is \(5\sqrt{3} + 4\sqrt{2} + 2\pi\).

Step 1: Apply Linearity of Integration

The linearity of integration allows us to separate the integral of a sum into the sum of integrals multiplied by their respective constants. Thus, we have: \[ \int_{0}^{2}[\sqrt{3} f(t)+\sqrt{2} g(t)+\pi] d t = \sqrt{3} \int_{0}^{2} f(t) d t + \sqrt{2} \int_{0}^{2} g(t) d t + \int_{0}^{2} \pi d t \]

Step 2: Calculate the Integral of f(t) from 0 to 2

We use interval additivity. Knowing \(\int_{0}^{1} f(x) d x=2\) and \(\int_{1}^{2} f(x) d x=3\), we find:\[ \int_{0}^{2} f(t) d t = \int_{0}^{1} f(x) d x + \int_{1}^{2} f(x) d x = 2 + 3 = 5\]

Step 3: Use Given Values for g(t)

We already know \(\int_{0}^{2} g(t) d t = 4\). Thus, we can directly substitute this value.

Step 4: Calculate the Integral of a Constant Function

For a constant function \(c\), the integral over an interval \([a, b]\) is given by \(c(b-a)\). Since the integral of \(\pi\) from \(0\) to \(2\) is calculated as:\[ \int_{0}^{2} \pi d t = \pi \times (2-0) = 2\pi \]

Step 5: Substitute All Values and Simplify

Substitute the values obtained into the expression from Step 1:\[ \sqrt{3} \int_{0}^{2} f(t) d t + \sqrt{2} \int_{0}^{2} g(t) d t + \int_{0}^{2} \pi d t = \sqrt{3} \times 5 + \sqrt{2} \times 4 + 2\pi = 5\sqrt{3} + 4\sqrt{2} + 2\pi \]

Step 6: Final Answer

The value of the integral is \(5\sqrt{3} + 4\sqrt{2} + 2\pi\).

Key Concepts

Linearity of Integration

The property of linearity in integration is a fundamental concept that helps us easily manipulate and solve integrals. When you have a combination of functions inside an integral, the linearity allows you to split them up. This means you can break a complex integral into smaller, more manageable parts.
For example, consider the integral \(\int_{a}^{b} [k_1 f(x) + k_2 g(x)] \, dx \). Using linearity, this can be separated as follows:
\[\int_{a}^{b} [k_1 f(x) + k_2 g(x)] \, dx = k_1 \int_{a}^{b} f(x) \, dx + k_2 \int_{a}^{b} g(x) \, dx\]

The constants \(k_1\) and \(k_2\) can be pulled out of their respective integrals because of linearity. This greatly simplifies calculations, especially when dealing with different functions and constants combined within the same integral.

Interval Additivity

Interval additivity is another key property of definite integrals. It allows us to split an integral over a larger interval into parts that can be evaluated separately. This is particularly useful for sums or differences of functions observed over adjacent intervals.

• If you know that \(\int_{a}^{c} h(x) \, dx\) and \(\int_{c}^{b} h(x) \, dx\) are individual integrals, you can combine them to obtain the integral over the entire range:

\[\int_{a}^{b} h(x) \, dx = \int_{a}^{c} h(x) \, dx + \int_{c}^{b} h(x) \, dx\]
This property simplifies problems where individual parts of the interval are already known. For instance, in the exercise, knowing \(\int_{0}^{1} f(x) \, dx\) and \(\int_{1}^{2} f(x) \, dx\) enabled us to find \(\int_{0}^{2} f(x) \, dx\) easily.

Integration of Constant Functions

Integrating a constant function is simpler than it seems. When you integrate a constant over a specified interval, you multiply the constant by the length of the interval.

• For a constant \(c\) and an interval \([a, b]\), the integral is given by:

\[\int_{a}^{b} c \, dx = c(b-a)\]
This is because the area under a horizontal line (which is what a constant function is) is simply the length of the interval times the height of the line. In our example, integrating \(\pi\) from 0 to 2 was calculated as \(\pi \times (2-0) = 2\pi\). This straightforward approach is very practical for dealing with constant functions in integrals.