Question

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

• A. \( \frac{3}{10} \)
• B. \( \frac{5}{2} \)
• C. \( \frac{10}{3} \)
• D. \( \frac{2}{5} \)

Hint:

Review integration of power functions and apply limits carefully.

Answer 1

The Correct Option is B

To evaluate the integral \( \int_0^2 |2x - 3| \, dx \), we need to consider the absolute value function, which affects how the integral is computed over different intervals. Step 1: Determine where the expression inside the absolute value equals zero. \[ 2x - 3 = 0 \implies x = \frac{3}{2} \] This point divides the interval \([0, 2]\) into two sub-intervals: \[ [0, \frac{3}{2}] \quad \text{and} \quad [\frac{3}{2}, 2] \] Step 2: Rewrite the integral by removing the absolute value, considering the sign of \(2x - 3\) in each sub-interval. 1. For \(0 \leq x \leq \frac{3}{2}\): \[ 2x - 3 \leq 0 \implies |2x - 3| = -(2x - 3) = 3 - 2x \] 2. For \(\frac{3}{2} \leq x \leq 2\): \[ 2x - 3 \geq 0 \implies |2x - 3| = 2x - 3 \] Step 3: Express the original integral as the sum of two integrals over the determined sub-intervals. \[ \int_0^2 |2x - 3| \, dx = \int_0^{\frac{3}{2}} (3 - 2x) \, dx + \int_{\frac{3}{2}}^2 (2x - 3) \, dx \] Step 4: Evaluate each integral separately. 1. First Integral: \[ \int_0^{\frac{3}{2}} (3 - 2x) \, dx = \left[ 3x - x^2 \right]_0^{\frac{3}{2}} = \left( 3 \cdot \frac{3}{2} - \left( \frac{3}{2} \right)^2 \right) - (0 - 0) = \frac{9}{2} - \frac{9}{4} = \frac{9}{4} \] 2. Second Integral: \[ \int_{\frac{3}{2}}^2 (2x - 3) \, dx = \left[ x^2 - 3x \right]_{\frac{3}{2}}^2 = \left( 2^2 - 3 \cdot 2 \right) - \left( \left( \frac{3}{2} \right)^2 - 3 \cdot \frac{3}{2} \right) = (4 - 6) - \left( \frac{9}{4} - \frac{9}{2} \right)\]\[ = (-2) - \left( -\frac{9}{4} \right) = -2 + \frac{9}{4} = \frac{1}{4} \] Step 5: Sum the results of the two integrals to find the total value. \[ \int_0^2 |2x - 3| \, dx = \frac{9}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2} \] Final Answer: \[ \boxed{\frac{5}{2}} \]