Question

Red $f(x) = 4 + 2 \cdot cos(x)$ Represent the LENGTH of $f(x)$ for $-3 \leq x \leq 5$ as a definite integral LENGTH $= \int_{-3}^{5} \sqrt{1 + 4 \sin^{2}(x)} \, dx$ Then use your calculator to evaluate the integral as a decimal number. LENGTH = Use $sqrt(1+x^3)$ for $\sqrt{1 + x^3}$ Remember: for trig functions you must be in radian mode.

          Red $f(x) = 4 + 2 \cdot cos(x)$
Represent the LENGTH of $f(x)$ for $-3 \leq x \leq 5$
as a definite integral
LENGTH $= \int_{-3}^{5} \sqrt{1 + 4 \sin^{2}(x)} \, dx$
Then use your calculator to evaluate the integral
as a decimal number.
LENGTH = 
Use $sqrt(1+x^3)$ for $\sqrt{1 + x^3}$
Remember: for trig functions you must be in radian mode.

Red f(x) = 4 + 2 · cos(x)
Represent the LENGTH of f(x) for -3 ≤ x ≤ 5
as a definite integral
LENGTH = ∫-3^5√(1 + 4 sin^2(x)) dx
Then use your calculator to evaluate the integral
as a decimal number.
LENGTH =
Use sqrt(1+x^3) for √(1 + x^3)
Remember: for trig functions you must be in radian mode.

Added by Brian L.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Amelia Hardy

09/19/2024

Step 1

Then $(f'(x))^2 = (-2\sin(x))^2 = 4\sin^2(x)$. Therefore, the length of the curve from $x = -3$ to $x = 5$ is $$L = \int_{-3}^5 \sqrt{1 + 4\sin^2(x)} dx$$ Show more…

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Red $f(x) = 4 + 2 \cdot \cos(x)$ Represent the LENGTH of $f(x)$ for $-3 \le x \le 5$ as a definite integral LENGTH = $\int_{-3}^{5} \sqrt{1 + 4 \sin^2(x)} dx$ Then use your calculator to evaluate the integral as a decimal number. LENGTH = Use $\text{sqrt}(1+x^3)$ for $\sqrt{1+x^3}$ Remember: for trig functions you must be in radian mode.