Question:

The value of \(\displaystyle\int_{-5}^{5}(4-|x|dx)dx\) is equal to

Updated On: Apr 4, 2025
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Solution and Explanation

We are given the integral \( \int_{-5}^{5} (4 - |x|) \, dx \) and are asked to find the value of the integral. 

Since the absolute value function \( |x| \) is piecewise, we need to split the integral at \( x = 0 \), because the function behaves differently for \( x \geq 0 \) and \( x < 0 \).

The integral becomes:

\( \int_{-5}^{5} (4 - |x|) \, dx = \int_{-5}^{0} (4 - (-x)) \, dx + \int_{0}^{5} (4 - x) \, dx \).

Now simplify each integral:

\( \int_{-5}^{0} (4 + x) \, dx \) and \( \int_{0}^{5} (4 - x) \, dx \).

First, compute \( \int_{-5}^{0} (4 + x) \, dx \):

\( \int (4 + x) \, dx = 4x + \frac{x^2}{2} \), so:

\( \left[ 4x + \frac{x^2}{2} \right]_{-5}^{0} = (0 + 0) - (-20 + \frac{25}{2}) = 20 - \frac{25}{2} = \frac{40}{2} - \frac{25}{2} = \frac{15}{2}. \)

Next, compute \( \int_{0}^{5} (4 - x) \, dx \):

\( \int (4 - x) \, dx = 4x - \frac{x^2}{2} \), so:

\( \left[ 4x - \frac{x^2}{2} \right]_{0}^{5} = (20 - \frac{25}{2}) - (0) = 20 - \frac{25}{2} = \frac{40}{2} - \frac{25}{2} = \frac{15}{2}. \)

Adding the results of the two integrals, we get:

\( \frac{15}{2} + \frac{15}{2} = 15. \)

The correct answer is 15.

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