Solving Definite Integral with Absolute Value
The problem asks us to evaluate the definite integral \( \mathop \smallint \limits_0^{\frac{\pi }{2}} |\sin x - \cos x|dx \). This involves an absolute value function, which requires us to determine the sign of the expression inside the absolute value within the given interval.
Understanding the Absolute Value
The absolute value \(|f(x)|\) is defined as:
\(f(x)\) if \(f(x) \ge 0\)
\(-f(x)\) if \(f(x) < 0\)
In this case, \(f(x) = \sin x - \cos x\). We need to find the values of \(x\) in the interval \( [0, \frac{\pi}{2}] \) where \(\sin x - \cos x\) is positive, negative, or zero.
Finding the Critical Point
The expression \(\sin x - \cos x\) changes sign when \(\sin x - \cos x = 0\), which means \(\sin x = \cos x\). Dividing by \(\cos x\) (assuming \(\cos x \neq 0\)), we get \(\tan x = 1\).
Within the interval \( [0, \frac{\pi}{2}] \), the only solution for \(\tan x = 1\) is \(x = \frac{\pi}{4}\).
Analyzing the Sign of \(\sin x - \cos x\)
We split the interval \( [0, \frac{\pi}{2}] \) at the critical point \( x = \frac{\pi}{4} \):
For \( x \in [0, \frac{\pi}{4}] \): In this interval, \(\sin x \le \cos x\). For example, at \( x=0 \), \(\sin 0 = 0\) and \(\cos 0 = 1\), so \(\sin x - \cos x = 0 - 1 = -1 < 0\). Thus, \(|\sin x - \cos x| = -(\sin x - \cos x) = \cos x - \sin x\) for \( x \in [0, \frac{\pi}{4}] \).
For \( x \in [\frac{\pi}{4}, \frac{\pi}{2}] \): In this interval, \(\sin x \ge \cos x\). For example, at \( x=\frac{\pi}{2} \), \(\sin \frac{\pi}{2} = 1\) and \(\cos \frac{\pi}{2} = 0\), so \(\sin x - \cos x = 1 - 0 = 1 > 0\). Thus, \(|\sin x - \cos x| = \sin x - \cos x\) for \( x \in [\frac{\pi}{4}, \frac{\pi}{2}] \).
Splitting the Integral
Based on the sign analysis, we split the definite integral into two parts:
\( \mathop \smallint \limits_0^{\frac{\pi }{2}} |\sin x - \cos x|dx = \mathop \smallint \limits_0^{\frac{\pi }{4}} (\cos x - \sin x)dx + \mathop \smallint \limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} (\sin x - \cos x)dx \)
Evaluating the Integrals
Now we evaluate each part of the integral:
First Integral: \( \mathop \smallint \limits_0^{\frac{\pi }{4}} (\cos x - \sin x)dx \)
The antiderivative of \( \cos x - \sin x \) is \( \sin x - (-\cos x) = \sin x + \cos x \). Applying the limits of integration:
\( [\sin x + \cos x]_0^{\frac{\pi }{4}} = (\sin \frac{\pi}{4} + \cos \frac{\pi}{4}) - (\sin 0 + \cos 0) \)
\( = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) \)
\( = \frac{2}{\sqrt{2}} - 1 \)
\( = \sqrt{2} - 1 \)
Second Integral: \( \mathop \smallint \limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} (\sin x - \cos x)dx \)
The antiderivative of \( \sin x - \cos x \) is \( -\cos x - \sin x \). Applying the limits of integration:
\( [-\cos x - \sin x]_{\frac{\pi }{4}}^{\frac{\pi }{2}} = (-\cos \frac{\pi}{2} - \sin \frac{\pi}{2}) - (-\cos \frac{\pi}{4} - \sin \frac{\pi}{4}) \)
\( = (0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) \)
\( = -1 - (-\frac{2}{\sqrt{2}}) \)
\( = -1 - (-\sqrt{2}) \)
\( = \sqrt{2} - 1 \)
Adding the Results
The total value of the definite integral is the sum of the two parts:
\( (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1) \)
Final Answer
The value of the definite integral \( \mathop \smallint \limits_0^{\frac{\pi }{2}} |\sin x - \cos x|dx \) is \( 2(\sqrt{2} - 1) \).
| Integral Section | Interval | Function | Result |
|---|---|---|---|
| First Part | \( [0, \frac{\pi}{4}] \) | \( \cos x - \sin x \) | \( \sqrt{2} - 1 \) |
| Second Part | \( [\frac{\pi}{4}, \frac{\pi}{2}] \) | \( \sin x - \cos x \) | \( \sqrt{2} - 1 \) |
| Total | \( [0, \frac{\pi}{2}] \) | \( |\sin x - \cos x| \) | \( 2(\sqrt{2} - 1) \) |
Revision Table: Key Concepts for Definite Integrals
| Concept | Description |
|---|---|
| Definite Integral | Represents the signed area under a curve between two limits. Calculated using the Fundamental Theorem of Calculus: \( \mathop \smallint \limits_a^b f(x)dx = F(b) - F(a) \), where \( F(x) \) is the antiderivative of \( f(x) \). |
| Absolute Value Function | \( |u| = u \) if \( u \ge 0 \) and \( |u| = -u \) if \( u < 0 \). Requires checking the sign of the expression inside the absolute value over the integration interval. |
| Splitting Integrals | If the integrand changes definition (e.g., due to absolute value) within the interval \( [a, b] \) at point \( c \), the integral is split: \( \mathop \smallint \limits_a^b f(x)dx = \mathop \smallint \limits_a^c f(x)dx + \mathop \smallint \limits_c^b f(x)dx \). |
| Trigonometric Values | Knowing key trigonometric values (like \(\sin x\) and \(\cos x\) at 0, \(\frac{\pi}{4}\), \(\frac{\pi}{2}\)) is crucial for evaluating definite integrals involving trigonometric functions. |
Additional Information: Integrals and Absolute Values
When dealing with definite integrals involving absolute values, the most important step is to correctly identify the intervals where the expression inside the absolute value is positive or negative. This determines how the absolute value is removed, either as the expression itself or its negative.
For trigonometric functions like \( \sin x \) and \( \cos x \), comparing their values or using identities can help. For instance, the inequality \( \sin x > \cos x \) is equivalent to \( \tan x > 1 \) (for \(\cos x > 0\)), which occurs for \( x \in (\frac{\pi}{4}, \frac{5\pi}{4}) \) and so on, considering the periodicity.
Visualizing the graphs of \( y = \sin x \) and \( y = \cos x \) can also help understand where one is greater than the other in a given interval.
Always remember to split the integral at every point within the interval where the expression inside the absolute value becomes zero or undefined.
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Important Questions from Definite Integrals
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