Questions: Find the arclength of y=3x+2 on 0 ≤ x ≤ 1

Solution Steps

To find the arc length of the function \( y = 3x + 2 \) over the interval \( 0 \leq x \leq 1 \), we use the arc length formula for a function \( y = f(x) \):

\[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]

1. Compute the derivative \( \frac{dy}{dx} \) of the function \( y = 3x + 2 \).
2. Substitute \( \frac{dy}{dx} \) into the arc length formula.
3. Evaluate the integral from \( x = 0 \) to \( x = 1 \).

Given the function \( y = 3x + 2 \), we compute the derivative with respect to \( x \):

\[ \frac{dy}{dx} = 3 \]

The arc length \( L \) of a function \( y = f(x) \) from \( x = a \) to \( x = b \) is given by:

\[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]

Substituting \( \frac{dy}{dx} = 3 \) into the formula, we get:

\[ L = \int_{0}^{1} \sqrt{1 + 3^2} \, dx = \int_{0}^{1} \sqrt{10} \, dx \]

Since \( \sqrt{10} \) is a constant, the integral simplifies to:

\[ L = \sqrt{10} \int_{0}^{1} dx = \sqrt{10} \left[ x \right]_{0}^{1} = \sqrt{10} (1 - 0) = \sqrt{10} \]

The numerical value of \( \sqrt{10} \) is approximately:

\[ \sqrt{10} \approx 3.1623 \]

Final Answer