Question

Find the exact length of the curve. y=1+6 x³/², 0 ≤ x ≤ 1

Answer 1

The exact length of the curve y=1+6x^(3/2), 0 ≤ x ≤ 1 is approximately 4.07 units.

Step-by-step explanation:

To find the exact length of the curve, we can use the formula for the arc length of a function y=f(x) on the interval [a, b], which is given by the integral from a to b of √(1 + (f’(x))²) dx. First, we need to find f’(x) by differentiating y=1+6x^(3/2) with respect to x.

The derivative of y with respect to x is y’ = 9x^(1/2). Then, we can plug this derivative into the arc length formula and integrate from 0 to 1: ∫[0 to 1] √(1 + (9x^(1/2))²) dx. This simplifies to ∫[0 to 1] √(1 + 81x) dx.

Next, we integrate √(1 + 81x) with respect to x. Using the substitution method where u = 1 + 81x, du = 81dx, and dx = du/81, we get ∫√u (1/81) du. This integrates to (1/81) (2/3)
`u^(^3^/^2^)`, which simplifies to (2/243) (1 +
`81x)^(^3^/^2^)`. Evaluating this from 0 to 1 gives us ((2/243)
`(82^(^3^/^2^)^)^)` - ((2/243)
`(1^(^3^/^2^)^)^)`, which simplifies to approximately 4.07 units.

Therefore, the exact length of the curve y=
`1+6x^(^3^/^2^)`, 0 ≤ x ≤ 1 is approximately 4.07 units.