Question

Could someone show me a step by step process on how to do this problem? Calculus 2

Answer 1

The arc length is given by the definite integral

$\displaystyle \int_1^3 \sqrt{1 + \left(y'\right)^2} \, dx = \int_1^3 \sqrt{1+9x} \, dx$

since by the power rule for differentiation,

$y = 2x^{3/2} \implies y' = \dfrac32 \cdot 2x^{3/2-1} = 3x^{1/2} \implies \left(y'\right)^2 = 9x$

To compute the integral, substitute

$u = 1+9x \implies du = 9\,dx$

so that by the power rule for integration and the fundamental theorem of calculus,

$\displaystyle \int_{x=1}^{x=3} \sqrt{1+9x} \, dx = \frac19 \int_{u=10}^{u=28} u^{1/2} \, du = \frac19\times\frac23 u^{1/2+1} \bigg|_{10}^{28} = \boxed{\frac2{27}\left(28^{3/2} - 10^{3/2}\right)}$