Question

Find the length of the curve $y = x^2$ over the interval $[0, \frac{1}{2}]$. (2 points available if integral table used, 5 points available if the integral table is not used). Hint: $\int sec^3(\theta) \,d\theta = \frac{1}{2} sec(\theta)tan(\theta) + \frac{1}{2} ln|sec(\theta) + tan(\theta)| + C.$

          Find the length of the curve $y = x^2$ over the interval $[0, \frac{1}{2}]$.
(2 points available if integral table used, 5 points available if the integral table is not used).
Hint: $\int sec^3(\theta) \,d\theta = \frac{1}{2} sec(\theta)tan(\theta) + \frac{1}{2} ln|sec(\theta) + tan(\theta)| + C.$

Find the length of the curve y = x^2 over the interval [0, (1)/(2)].
(2 points available if integral table used, 5 points available if the integral table is not used).
Hint: ∫ sec^3(θ) dθ = (1)/(2) sec(θ)tan(θ) + (1)/(2) ln|sec(θ) + tan(θ)| + C.

Added by Miguel -Ngel D.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert William Semus

10/08/2023

Step 1

Step 1: The arc length of a curve y = f(x) over the interval [a, b] is given by: $L = \int_a^b \sqrt{1 + (f'(x))^2} dx$ Show more…

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Find the length of the curve y = $x^2$ over the interval [0, $\frac{1}{2}$]. (2 points available if integral table used, 5 points available if the integral table is not used). Hint: $\int sec^3(\theta) d\theta = \frac{1}{2}sec(\theta)tan(\theta) + \frac{1}{2}ln|sec(\theta) + tan(\theta)| + C$.