Question

The value of \(\int_0^3 |2x-6|dx\) is:

• A. \(-\frac{9}{4}\)
• B. \(\frac{19}{2}\)
• C. 9
• D. -3

Answer 1

The Correct Option is C

To find the value of \( \int_0^3 |2x - 6| \, dx \), we first identify where the function \( f(x) = |2x - 6| \) changes form. The expression \( 2x - 6 = 0 \) when \( x = 3 \). So, we define the absolute value piecewise:

\[ |2x - 6| = \begin{cases} -(2x - 6) = 6 - 2x, & \text{if } x < 3 \\ 2x - 6, & \text{if } x \geq 3 \end{cases} \]

Since the limits of integration are from 0 to 3, we are only concerned with \( x \in [0, 3] \), where \( |2x - 6| = 6 - 2x \).

The integral becomes:

\[ \int_0^3 |2x - 6| \, dx = \int_0^3 (6 - 2x) \, dx = \int_0^3 6 \, dx - \int_0^3 2x \, dx \]

Compute each integral:

• \[ \int_0^3 6 \, dx = 6x \Big|_0^3 = 6(3) - 6(0) = 18 \]
• \[ \int_0^3 2x \, dx = x^2 \Big|_0^3 = 3^2 - 0^2 = 9 \]

Therefore,

\[ \int_0^3 (6 - 2x) \, dx = 18 - 9 = 9 \]

Thus, the value of \( \int_0^3 |2x - 6| \, dx \) is 9.