Question

Write the integral that gives the length of the curve y = f (x) = ∫0 to 4.5x sin t dt on the interval ​[0,π​].

Answer 1

Answer:

Arc length $=\int_0^{\pi} \sqrt{1+[(4.5sin(4.5x))]^2}\ dx$

Arc length $=9.75053$

Step-by-step explanation:

The arc length of the curve is given by $\int_a^b \sqrt{1+[f'(x)]^2}\ dx$

Here, $f(x)=\int_0^{4.5x}sin(t) \ dt$ interval $[0, \pi]$

Now, $f'(x)=\frac{\mathrm{d} }{\mathrm{d} x} \int_0^{4.5x}sin(t) \ dt$

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( [-cos(t)]_0^{4.5x} \right )$

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( -cos(4.5x)+1 \right )$

$f'(x)=4.5sin(4.5x)$

Now, the arc length is $\int_0^{\pi} \sqrt{1+[f'(x)]^2}\ dx$

$\int_0^{\pi} \sqrt{1+[(4.5sin(4.5x))]^2}\ dx$

After solving, Arc length $=9.75053$