Question

The length of the curve $f(x) = x^{\frac{1}{4}}$ on the interval $0 \le x \le 16$ is (a) $\int_0^{16} \sqrt{1 + y^2} dy$. (b) $\int_0^2 \sqrt{1 + 16y^6} dy$. (c) $\int_0^{16} \sqrt{1 + 16y^2} dy$.

          The length of the curve $f(x) = x^{\frac{1}{4}}$ on the interval $0 \le x \le 16$ is

(a) $\int_0^{16} \sqrt{1 + y^2} dy$.
(b) $\int_0^2 \sqrt{1 + 16y^6} dy$.
(c) $\int_0^{16} \sqrt{1 + 16y^2} dy$.

The length of the curve f(x) = x^(1)/(4) on the interval 0 ≤ x ≤ 16 is

(a) ∫0^16√(1 + y^2) dy.
(b) ∫0^2 √(1 + 16y^6) dy.
(c) ∫0^16√(1 + 16y^2) dy.

Added by Jacqueline B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

06/09/2023

Step 1

Step 1: To find the length of the curve f(x) = x^(1/4) on the interval 0 <= x <= 16, we can use the formula for arc length of a curve given by L = ∫[a,b] √(1 + (f'(x))^2) dx. Show more…

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The length of the curve f(x) = x^(1/4) on the interval 0 <= x <= 16 is: (a) ∫₀¹⁶ √(1 + y^(1/2)) dy. (b) ∫₀² √(1 + 16y^6) dy. (c) ∫₀¹⁶ √(1 + 16y^2) dy. The length of the curve f(x) = x on the interval 0 <= x <= 16 is: (a) ∫₀¹⁶ √(1 + y^2) dy (b) ∫₀² √(1 + 16y^6) dy (c) ∫₀¹⁶ √(1 + 16y^2) dy