Question

Find the arc length of the curve below on the given interval by integrating with respect to x. y = 4x + 2; [0, 3] (Use calculus.) Which of the following is a simplified form of the integral? integral_0^3 Squareroot 1 + 16 x^2 dx integral_0^3 Squareroot 17 dx integral_0^3 Squareroot 5 dx integral_0^3 Squareroot 1 + 4 x^2 dx The length of the curve is (Type an exact answer, using radicals as needed.)

Answer 1

The simplified form of the integral is
`L=\int\limits^3_0 {√(17)} \, dx`.

The length of the curve is
`L=3√(17)`

Explanation:

THE ARC LENGTH DEFINITION

If
`f'` is continuous on
`[a, b]`, then the length of the curve
`y=f(x)`,
`a\leq x\leq b`, is

`L=\int\limits^b_a {\sqrt{1+((dy)/(dx) )^(2) } } \, dx`

We know that
`y = 4x + 2`; [0, 3]

Applying the above definition we get

`(dy)/(dx)=(d)/(dx)(4x+2)=(d)/(dx)\left(4x\right)+(d)/(dx)\left(2\right)=4`

`L=\int\limits^3_0 {\sqrt{1+(4 )^(2) } } \, dx\\L=\int\limits^3_0 {√(1+16) } \, dx\\L=\int\limits^3_0 {√(17)} \, dx`

`L=\int _0^3√(17)dx=\left[√(17)x\right]^3_0=3√(17)`