# Find the exact length of the curve. y = 1 + 8x^{3/2}, 0 ≤ x ≤ 1.

**Solution:**

The length of the curve y = f(x) a ≤ x ≤ b is given by:

L = \(\int_{a}^{b}\sqrt{1 + (\frac{\mathrm{d} y}{\mathrm{d} x})^{2}}dx\)

y = 1 + 8x^{3/2}

dy/dx = 8(3/2)x^{1/2}

= 12x^{1/2}

√1 + (dy/dx)^{2} = √1 + (6x^{1/2})^{2} = √1 + 144x

Length of the curve L = \(\int_{0}^{1}\sqrt{1 + 144x}dx\)

= \(\frac{2}{3}.\frac{1}{144}[(1+144x)^{3/2}]_{0}^{1}\)

= \(\frac{1}{216}[(1+144(1))^{3/2} - (1+144(0))^{3/2})]\)

= \(\frac{1}{216}[(145)^{3/2} - (1)^{3/2})]\)

= \(\frac{1}{216}[(145)^{3/2} - (1)]\)

## Find the exact length of the curve. y = 1 + 8x^{3/2}, 0 ≤ x ≤ 1.

**Summary:**

The length of the curve y = 1 + 8x^{3/2} is L = (1/216)[(145)^{3/2} - 1]

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