Question

write an integral expression that will give the length of the path given by f(x) = -5 cos(-2x⁶) + 4 from x = 3 to x = 9. answer attempt 1 out of 3 a = 3 b = 9 ∫₃⁹ √(1 + ) dx submit answer

Explanation:

Step1: Find the derivative of \( f(x) \)

Use the chain rule. If \( y = - 5\cos(-2x^{6})+4\), let \( u=-2x^{6}\), then \( y=-5\cos(u)+4\).
The derivative of \( y\) with respect to \( u\) is \( y_{u}' = 5\sin(u)\), and the derivative of \( u\) with respect to \( x\) is \( u_{x}'=-12x^{5}\).
By the chain rule \( f'(x)=y_{u}'\cdot u_{x}'=5\sin(-2x^{6})\cdot(- 12x^{5})=-60x^{5}\sin(-2x^{6}) = 60x^{5}\sin(2x^{6})\) (since \(\sin(-\alpha)=-\sin(\alpha)\)).

Step2: Apply the arc - length formula

The arc - length formula for a function \( y = f(x)\) from \( x=a\) to \( x = b\) is \(L=\int_{a}^{b}\sqrt{1+(f'(x))^{2}}dx\).
Substitute \(a = 3\), \(b = 9\) and \(f'(x)=60x^{5}\sin(2x^{6})\) into the formula.
We get \(L=\int_{3}^{9}\sqrt{1+(60x^{5}\sin(2x^{6}))^{2}}dx=\int_{3}^{9}\sqrt{1 + 3600x^{10}\sin^{2}(2x^{6})}dx\)

Answer:

\(\int_{3}^{9}\sqrt{1+3600x^{10}\sin^{2}(2x^{6})}dx\)