Question

in exercises 23-28, use areas to 23. \\( \int_{0}^{b} x \\, dx, \quad b > 0 \\)

Explanation:

Step1: Recognize the graph of \( y = x \)

The function \( y = x \) is a straight line passing through the origin with a slope of 1. For the integral \( \int_{0}^{b} x \, dx \) where \( b>0 \), we are finding the area under the curve \( y = x \) from \( x = 0 \) to \( x = b \).

Step2: Identify the shape of the region

The region under \( y = x \) from \( x = 0 \) to \( x = b \) is a triangle. The base of the triangle is the length along the \( x \)-axis from 0 to \( b \), so the base \( = b \). The height of the triangle is the value of \( y \) when \( x = b \), which is \( y = b \) (since \( y = x \)).

Step3: Use the formula for the area of a triangle

The formula for the area of a triangle is \( A=\frac{1}{2}\times\text{base}\times\text{height} \). Substituting the base \( = b \) and height \( = b \) into the formula, we get:
\[
A=\frac{1}{2}\times b\times b=\frac{1}{2}b^{2}
\]
Since the integral \( \int_{0}^{b} x \, dx \) represents the area under the curve \( y = x \) from 0 to \( b \), the value of the integral is equal to this area.

Answer:

\(\frac{1}{2}b^{2}\)