RaySix

Post

at November 6th 2023, 9:19:56 pm.

Question:

Find the length of the curve represented by the equation $y = \frac{1}{2}x^2$ over the interval $[0, 4]$ .

Answer:

To find the length of the curve, we will use the formula for the length of a curve given by:

L = \int_a^b \sqrt{1 + (f'(x))^2}\,dx

Here, $a$ and $b$ represent the interval of the curve, and $f'(x)$ is the derivative of the function representing the curve.

First, let's find $f'(x)$ :

f(x) = \frac{1}{2}x^2

f'(x) = x

Next, we can calculate the length of the curve using the formula:

L = \int_0^4 \sqrt{1 + (x)^2}\,dx

To evaluate this integral, we can use a substitution by letting $u = (1 + x^2)$ .

du = 2x \,dx

\frac{du}{2x} = dx

Substituting these values into the integral, we get:

L = \int_0^4 \sqrt{u} \cdot \frac{du}{2x}

L = \frac{1}{2} \int_0^4 \frac{\sqrt{u}}{x}\,du

Next, we need to express $x$ in terms of $u$ by solving $u = 1 + x^2$ for $x$ .

u - 1 = x^2

x = \sqrt{u - 1}

Substituting this value into the integral:

L = \frac{1}{2} \int_0^4 \frac{\sqrt{u}}{\sqrt{u - 1}}\,du

We can simplify the integrand by rationalizing the denominator:

L = \frac{1}{2} \int_0^4 \sqrt{u(u - 1)} \cdot \frac{\sqrt{u}}{\sqrt{u - 1}}\,du

L = \frac{1}{2} \int_0^4 \sqrt{u^2}\,du

L = \frac{1}{2} \int_0^4 u\,du

L = \frac{1}{2} \left[\frac{u^2}{2}\right]_0^4

L = \frac{1}{2} \cdot \frac{4^2}{2} - \frac{1}{2} \cdot \frac{0^2}{2}

L = \frac{1}{2} \cdot 8

L = 4

Therefore, the length of the curve represented by the equation $y = \frac{1}{2}x^2$ over the interval $[0, 4]$ is equal to 4.