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

Arc length
`=\int_0^(\pi) √(1+[(4.5sin(4.5x))]^2)\ dx`

Arc length
`=9.75053`

Explanation:

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

`\int_0^(\pi) √(1+[(4.5sin(4.5x))]^2)\ dx`

After solving, Arc length
`=9.75053`