Question
Find the exact length of the curve.$$y=\frac{x^{3}}{3}+\frac{1}{4 x}, \quad 1 \leqslant x \leqslant 2$$
Step 1
Step 1: The length of a curve $y=f(x)$ from $x=a$ to $x=b$ is given by the formula: $$L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x$$ So, first we need to find the derivative of the function $y=\frac{x^{3}}{3}+\frac{1}{4 x}$. Show more…
Show all steps
Your feedback will help us improve your experience
Rakesh Kumar Sharma and 62 other Calculus 2 / BC educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Find the exact length of the curve. $ y = \displaystyle \frac{x^3}{3} + \displaystyle \frac{1}{4x} $ , $ 1 \le x \le 2 $
Further Applications of Integration
Arc Length
Find the exact length of the curve. $$ y=\frac{2}{3}\left(1+x^{2}\right)^{3 / 2}, \quad 0 \leq x \leq 1 $$
Find the exact length of the curve. y = 3 + 2x^3/2, 0 ≤ x ≤ 1
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD