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Chapter 12: Problem 1005

The value of the integral \({ }^{5} \int_{-5}(\mathrm{x}-[\mathrm{x}]) \mathrm{dx}\) is \(\ldots \ldots\) (a) 0 (b) 5 (c) 10 (d) 15

Short Answer

Expert verified
The value of the integral \(\int_{-5}^5 (\mathrm{x}-[\mathrm{x}])\mathrm{dx}\) is 5, which corresponds to answer choice (b).

Step by step solution

01

Break The Integral Into Two Intervals

Since the function inside the integral behaves differently for negative and positive x values, we can break the given integral into two separate integrals, one for the negative x values and the other for the positive x values. So, the integral will look like this: \[ \int_{-5}^5 (\mathrm{x}-[\mathrm{x}])\mathrm{dx} = \int_{-5}^0 (\mathrm{x}-[\mathrm{x}])\mathrm{dx} + \int_{0}^5 (\mathrm{x}-[\mathrm{x}])\mathrm{dx} \]
02

Handle The Negative Interval

For the negative interval, the floor function will round down to the nearest integer. As a result, it will "jump up" by 1 at each integer value. In this case, the function will be equal to \(x - (x-1) = 1\) between each integer, with the exception of 0. Thus, the integral becomes: \[ \int_{-5}^0 (\mathrm{x}-[\mathrm{x}])\mathrm{dx} = \int_{-5}^{-4} 1\mathrm{dx} + \int_{-4}^{-3} 1\mathrm{dx} + \dots + \int_{-1}^{0} 1\mathrm{dx} \] Now, we can evaluate each individual integral and sum them up.
03

Evaluate The Negative Interval

All the integrals in the negative interval are of the form \( \int_a^b 1\mathrm{dx} \), which evaluates to \(b - a\). Hence, we can sum them up: \[ \int_{-5}^0 (\mathrm{x}-[\mathrm{x}])\mathrm{dx} = (-4 - (-5)) + (-3 - (-4)) + \dots + (0 - (-1)) = 1 + 1 + \dots + 1 = 5 \]
04

Handle The Positive Interval

For the positive interval, the floor function will "round down" the given number to the nearest integer, resulting in \(x - x = 0\). Hence, the integral becomes: \[ \int_{0}^5 (\mathrm{x}-[\mathrm{x}])\mathrm{dx} = \int_{0}^5 0\mathrm{dx} \]
05

Evaluate The Positive Interval

The integral in the positive interval is of the form \( \int_a^b 0\mathrm{dx} \), which evaluates to 0, as there is no area underneath a function that is always zero. \[ \int_{0}^5 (\mathrm{x}-[\mathrm{x}])\mathrm{dx} = 0 \]
06

Combine Both Intervals

Finally, we combine the results from both intervals: \[ \int_{-5}^5 (\mathrm{x}-[\mathrm{x}])\mathrm{dx} = \int_{-5}^0 (\mathrm{x}-[\mathrm{x}])\mathrm{dx} + \int_{0}^5 (\mathrm{x}-[\mathrm{x}])\mathrm{dx} = 5 + 0 = 5 \] So, the value of the integral is 5, which corresponds to the answer choice (b).

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\((1 / e) \int_{1}[(\log t) /(1+t)] d t+e \int_{1}[(\log t) /(1+t)] d t\) is equal to...... (a) \(\mathrm{e}\) (b) \((1 / e)\) (c) 2 (d) \((1 / 2)\)

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