The Value of $\int_1^3 |2x - 1|dx$ equal to

The absolute value function requires determining where $2x - 1$ is positive or negative.

Setting $2x - 1 = 0$:

$2x = 1$

$x = \frac{1}{2}$

For $x < \frac{1}{2}$, the expression $2x - 1$ is negative.

For $x > \frac{1}{2}$, the expression $2x - 1$ is positive.

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The integration limits are from $1$ to $3$.

Since both $x = 1$ and $x = 3$ are greater than $\frac{1}{2}$, the expression $2x - 1$ is positive throughout the entire interval $[1, 3]$.

Therefore, $|2x - 1| = 2x - 1$ for all $x \in [1, 3]$.

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$\int_1^3 |2x - 1|dx = \int_1^3 (2x - 1)dx$

$= \left[x^2 - x\right]_1^3$

$= (3^2 - 3) - (1^2 - 1)$

$= (9 - 3) - (1 - 1)$

$= 6 - 0$

$= 6$

Therefore, the value of $\int_1^3 |2x - 1|dx = 6$.