What is the arc length of teh curve given by #f(x)=3x^6 + 4x# in the interval #x in [-2,184]#?

Question

Isaiah Allen · Accepted Answer

#L=1.16xx10^14#

To find arc length, we use the equation:

#L=int_a^b sqrt(1+f'(x)^2)dx#

Take the derivative of #f(x)#:

#f'(x)=18x^5+4#

Now plug into the formula with the interval:

#int_-2^184 sqrt(1+(18x^5+4)^2)dx#

Solve inside the parentheses:

#int_-2^184 sqrt(1+324x^10+144x^5+16)dx#

Simplify inside the square root:

#int_-2^184 sqrt(324x^10+144x^5+17)dx#

Solve the integral:

#L=1.16xx10^14#

Emma Aldridge · Answer

undefined To find the arc length of the curve given by $f(x) = 3x^6 + 4x$ in the interval $x \in [-2, 184]$, you can use the formula for arc length:

$$ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx $$

where $f'(x)$ is the derivative of $f(x)$. Calculate the derivative of $f(x)$, then substitute it into the formula. Finally, integrate the expression over the given interval $[-2, 184]$.

Evelyn Agard · Answer

undefined To find the arc length of the curve given by $ f(x) = 3x^6 + 4x $ in the interval $ x $ in $[-2, 184]$, you use the formula for arc length:

$$ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx $$

Where $ f'(x) $ is the derivative of $ f(x) $.

1. Find the derivative of $ f(x) $.
$$ f'(x) = 18x^5 + 4 $$

2. Compute $ \sqrt{1 + (f'(x))^2} $.
$$ \sqrt{1 + (f'(x))^2} = \sqrt{1 + (18x^5 + 4)^2} $$

3. Integrate $ \sqrt{1 + (f'(x))^2} $ from $-2$ to $184$.
$$ L = \int_{-2}^{184} \sqrt{1 + (18x^5 + 4)^2} \, dx $$

4. Calculate the definite integral to find the arc length.