Problem 11 - 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 9-16. $$ \int_{0}^{2}[2 f(x)+g(x)] d x $$

Step-by-Step Solution

Answer

The value of \(\int_{0}^{2} [2f(x) + g(x)] \, dx\) is 14.

Step 1: Apply Linearity of Integration

Recall that the linearity of integration allows us to distribute the integral across terms and constant factors. We can decompose the given integral as follows:\[\int_{0}^{2} [2f(x) + g(x)] \, dx = \int_{0}^{2} 2f(x) \, dx + \int_{0}^{2} g(x) \, dx\]

Step 2: Factor Out Constants

Apply the property of integrals that allows factoring out constants. The integral becomes:\[\int_{0}^{2} 2f(x) \, dx = 2 \cdot \int_{0}^{2} f(x) \, dx\]

Step 3: Use Interval Additivity for f(x)

We do not have the value of \(\int_{0}^{2} f(x) \, dx\) directly, but we can find it using the additivity property:\[\int_{0}^{2} f(x) \, dx = \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx = 2 + 3 = 5\]

Step 4: Substitute the Calculated Values

Substitute the known values into the equation derived from Step 1:\[2 \cdot \int_{0}^{2} f(x) \, dx = 2 \cdot 5 = 10\]\[\int_{0}^{2} g(x) \, dx = 4\]

Step 5: Combine the Results

Add the results from each integral split to find the total value:\[\int_{0}^{2} [2f(x) + g(x)] \, dx = 10 + 4 = 14\]

Key Concepts

Linearity of Integration

The linearity of integration is a powerful property that aids in simplifying complex integrals. This rule allows us to separate integrals when we have a sum or difference of terms inside the integral. For example: \

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• If we have an integral, \( \int (a(x) + b(x)) \, dx \), it can be split into \( \int a(x) \, dx + \int b(x) \, dx \).
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• Moreover, if a term is multiplied by a constant, such as \( \int c \cdot f(x) \, dx \), the constant \( c \) can be factored out as \( c \cdot \int f(x) \, dx \).
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\In our given problem where we calculate \( \int_{0}^{2} [2f(x) + g(x)] \, dx \), the linearity property lets us decompose this into two integrals: \\[\int_{0}^{2} 2f(x) \, dx + \int_{0}^{2} g(x) \, dx\]\This decomposition simplifies our calculation considerably, as we can focus on each integral separately.

Interval Additivity

Interval additivity is an essential concept in integration that allows us to break integrals over larger intervals into smaller sections. This means that if you know the integrals over smaller sub-intervals, you can easily calculate the integral over a larger interval that is the sum of these sub-intervals. \

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• For instance, if \( \int_{a}^{b} f(x) \, dx \) and \( \int_{b}^{c} f(x) \, dx \) are known, then \( \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \).
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\In our problem, we applied this property to determine \( \int_{0}^{2} f(x) \, dx \) by summing \( \int_{0}^{1} f(x) \, dx = 2 \) and \( \int_{1}^{2} f(x) \, dx = 3 \). Therefore, \\[\int_{0}^{2} f(x) \, dx = 2 + 3 = 5\]\With the interval additivity, you are reassured that integrals over sub-intervals can be added together to form the integral over the entire interval.

Integral Properties

Integration comes with a set of properties that makes solving integrals easier and more structured. Beyond linearity and interval additivity, there are other essential rules to note: \

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• **Zero width interval**: If the upper and lower limits are equal, then \( \int_{a}^{a} f(x) \, dx = 0 \).
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• **Reversal of limits**: \( \int_{b}^{a} f(x) \, dx = -\int_{a}^{b} f(x) \, dx \), helping in reversing the bounds while adjusting the integral's sign.
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• **Consistency with addition**: The result of an integral is consistently distributed over the sum of functions, reflecting how areas under curves can be considered independently.
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\By understanding and applying these properties, the integral \( \int_{0}^{2} [2f(x) + g(x)] \, dx \) was transformed into a simpler problem, ultimately calculated using these systematic steps to achieve a result of 14. This process showcases how integral properties streamline solving otherwise complex mathematical challenges.