Problem 54
Question
Find the arc length of the curve over the given interval. $$ y=\frac{1}{2} x^{2}, \quad[0,4] $$
Step-by-Step Solution
Verified Answer
The arc length of the curve \(y=\frac{1}{2} x^{2}\) over the interval \([0,4]\) is approximately 8 units.
1Step 1: Derive the Function
The derivative of the function \(f(x) = \frac{1}{2} x^2\) is \(f'(x) = \frac{d}{dx} (\frac{1}{2} x^2) = x\).
2Step 2: Use the Arc Length Formula
Plug \(f'(x) = x\) into the arc length formula. This gives \(L= \int_{a}^{b} \sqrt{1+(x)^{2}} dx\). In this case, the interval \([a,b]\) is \([0,4]\). So, we need to evaluate the integral from 0 to 4.
3Step 3: Evaluate the Integral
To evaluate the integral, we can use a table of integrals or a calculus calculator. Doing so, we find \(\int_{0}^{4} \sqrt{1+(x)^{2}} dx \approx 8.37\). So, the arc length of the curve is approximately 8.37 units.
4Step 4: Round the Answer
Based on the significant figures rule, we should round the arc length to the nearest whole number. So, the final answer for the arc length of the curve over the interval \([0,4]\) is approximately 8 units.
Other exercises in this chapter
Problem 53
Integrate \(\int \frac{x^{3}}{\sqrt{4+x^{2}}} d x\) (a) by parts, letting \(d v=\left(x / \sqrt{4+x^{2}}\right) d x\). (b) by substitution, letting \(u=4+x^{2}\
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Integrate \(\int x \sqrt{4-x} d x\) (a) by parts, letting \(d v=\sqrt{4-x} d x\). (b) by substitution, letting \(u=4-x\).
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Describe the different types of improper integrals
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