Question

1. -/4 points evaluate the integral by interpreting it in terms of areas. \\( \int_{0}^{3} |2x - 3| \\, dx \\)

Explanation:

Step1: Find zero of the integrand

Set $2x-3=0$, solve for $x$:
$2x=3 \implies x=\frac{3}{2}$

Step2: Split the integral at $x=\frac{3}{2}$

For $0\leq x<\frac{3}{2}$, $|2x-3|=3-2x$; for $\frac{3}{2}\leq x\leq3$, $|2x-3|=2x-3$.
$$\int_{0}^{3} |2x-3|dx = \int_{0}^{\frac{3}{2}} (3-2x)dx + \int_{\frac{3}{2}}^{3} (2x-3)dx$$

Step3: Interpret as area of triangles

Each integral corresponds to a right triangle.

• First triangle: base $=\frac{3}{2}$, height $=3-2(0)=3$

Area: $\frac{1}{2} \times \frac{3}{2} \times 3 = \frac{9}{4}$

• Second triangle: base $=3-\frac{3}{2}=\frac{3}{2}$, height $=2(3)-3=3$

Area: $\frac{1}{2} \times \frac{3}{2} \times 3 = \frac{9}{4}$

Step4: Sum the two areas

Total area = $\frac{9}{4} + \frac{9}{4}$

Answer:

$\frac{9}{2}$