Question

Evaluate the integral: $\int_{0}^{1} |5x - 3| dx$

• A. $\frac{13}{10}$
• B. $1$
• C. $\frac{3}{10}$
• D. $\frac{1}{2}$

Hint:

Integrals of linear absolute value functions $|mx + c|$ can be evaluated geometrically without calculus! The graph forms two triangles above the $x$-axis. The total area is simply the sum of the areas of these two triangles: $\text{Area} = \frac{1}{2} \times \text{base}_1 \times \text{height}_1 + \frac{1}{2} \times \text{base}_2 \times \text{height}_2$. Here, $\frac{1}{2}(\frac{3}{5})(3) + \frac{1}{2}(\frac{2}{5})(2) = \frac{9}{10} + \frac{4}{10} = \frac{13}{10}$.

Answer 1

The Correct Option is A

Step 1: Spot the absolute value.
We need $\int_0^1 |5x-3|\,dx$. The expression inside changes sign, so we cannot integrate blindly.
Step 2: Find where the sign flips.
Set $5x-3 = 0$, giving $x = \dfrac{3}{5}$. This point lies inside $[0,1]$, so we split there.
Step 3: Decide the sign on each piece.
For $0 \le x < \dfrac{3}{5}$, $5x-3$ is negative so $|5x-3| = 3-5x$. For $\dfrac{3}{5} < x \le 1$, it is positive so $|5x-3| = 5x-3$.
Step 4: Integrate the first piece.
$\int_0^{3/5}(3-5x)\,dx = \left[3x - \tfrac{5x^2}{2}\right]_0^{3/5} = \tfrac{9}{5} - \tfrac{9}{10} = \tfrac{9}{10}$.
Step 5: Integrate the second piece.
$\int_{3/5}^{1}(5x-3)\,dx = \left[\tfrac{5x^2}{2} - 3x\right]_{3/5}^{1} = \left(-\tfrac{1}{2}\right) - \left(-\tfrac{9}{10}\right) = \tfrac{4}{10}$.
Step 6: Add the pieces.
Total $= \dfrac{9}{10} + \dfrac{4}{10} = \dfrac{13}{10}$, which is option 1 and agrees with the key.
\[ \boxed{\dfrac{13}{10}} \]