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For a bounded function, is the integral of the function from -infinity to +infinity defined and finite?
- a)Yes
- b)Never
- c)Not always
- d)None of the mentioned
Correct answer is option 'C'. Can you explain this answer?
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For a bounded function, is the integral of the function from -infinity...
If the bounded function, is say y = 2, then the integral ceases to hold. Similarly, if it is just the block square function, it is finite. Hence, it depends upon the spread of the signal on either side. If the spread is finite, the integral will be finite.
Most Upvoted Answer
For a bounded function, is the integral of the function from -infinity...
The correct answer is option 'C' - Not always.
Explanation:
A bounded function is a function that has a finite range. In other words, the function does not take on values that go to infinity or negative infinity.
When we talk about the integral of a bounded function from -infinity to infinity, we are referring to the improper integral. The improper integral is defined as the limit of the definite integral as the limits of integration approach infinity and negative infinity.
In some cases, the integral of a bounded function from -infinity to infinity may be defined and finite. This occurs when the function has a well-behaved behavior as the limits of integration go to infinity and negative infinity.
However, there are cases when the integral of a bounded function from -infinity to infinity is not defined or is infinite. This happens when the function has oscillations, or when it approaches infinity or negative infinity as the limits of integration are reached.
For example, consider the function f(x) = sin(x) / x. This function is bounded because the range of sin(x) is between -1 and 1, and dividing by x does not change the boundedness. However, the integral of this function from -infinity to infinity is not defined because it oscillates and does not converge to a finite value.
Therefore, the integral of a bounded function from -infinity to infinity is not always defined and finite. It depends on the behavior of the function as the limits of integration approach infinity and negative infinity.
Explanation:
A bounded function is a function that has a finite range. In other words, the function does not take on values that go to infinity or negative infinity.
When we talk about the integral of a bounded function from -infinity to infinity, we are referring to the improper integral. The improper integral is defined as the limit of the definite integral as the limits of integration approach infinity and negative infinity.
In some cases, the integral of a bounded function from -infinity to infinity may be defined and finite. This occurs when the function has a well-behaved behavior as the limits of integration go to infinity and negative infinity.
However, there are cases when the integral of a bounded function from -infinity to infinity is not defined or is infinite. This happens when the function has oscillations, or when it approaches infinity or negative infinity as the limits of integration are reached.
For example, consider the function f(x) = sin(x) / x. This function is bounded because the range of sin(x) is between -1 and 1, and dividing by x does not change the boundedness. However, the integral of this function from -infinity to infinity is not defined because it oscillates and does not converge to a finite value.
Therefore, the integral of a bounded function from -infinity to infinity is not always defined and finite. It depends on the behavior of the function as the limits of integration approach infinity and negative infinity.
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For a bounded function, is the integral of the function from -infinity to +infinity defined and finite?a)Yesb)Neverc)Not alwaysd)None of the mentionedCorrect answer is option 'C'. Can you explain this answer?
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