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

Answer:

The simplified form of the integral is $L=\int\limits^3_0 {\sqrt{17}} \, dx$.

The length of the curve is $L=3\sqrt{17}$

Step-by-step 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

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

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

Applying the above definition we get

$\frac{dy}{dx}=\frac{d}{dx}(4x+2)=\frac{d}{dx}\left(4x\right)+\frac{d}{dx}\left(2\right)=4$

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

$L=\int _0^3\sqrt{17}dx=\left[\sqrt{17}x\right]^3_0=3\sqrt{17}$