## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{8}{27}(10\sqrt {10}-1)$
Formula to calculate the arc length is: $l=\int_p^q \sqrt {1+[f'(x)]^2} dx$ Now, $l=\int_0^4 \sqrt {1+[(3/2)x^{1/2})]^2}dx=\int_0^4 \sqrt {1+(9/4)x} dx$ Since, we know $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ Then, $\int_0^4 (1+(9/4)x)^{1/2} dx=(\dfrac{2}{3})(\dfrac{4}{9}) [1+(\dfrac{9}{4}) x)^{3/2}]_0^4=\dfrac{8}{27}(10\sqrt {10}-1)$