2-4-16-x-2-dx-x-4-

Question Number 11865 by ankhaaankhaa last updated on 03/Apr/17

\underset{2}{\int^{4}} \sqrt{16 - x^{2}} d x / x^{4} =

Answered by ajfour last updated on 03/Apr/17

I = \int_{2}^{4} \frac{\sqrt{16 - x^{2}}}{x^{4}} d x = \int_{2}^{4} \frac{\sqrt{\frac{16}{x^{2}} - 1}}{x^{3}} d x

n o w l e t \frac{16}{x^{2}} = t \Rightarrow \frac{- 32 d x}{x^{3}} = d t

F u r t h e r t = 1 w h e n x = 4

a n d t = 4 w h e n x = 2

T h e n I = - \frac{1}{32} \int_{4}^{1} \sqrt{t - 1} d t

= \frac{1}{32} \int_{1}^{4} \sqrt{t - 1} d t

= \frac{1}{32} \frac{{(t - 1)}^{3 / 2}}{(3 / 2)} ∣_{1}^{4}

= \frac{1}{48} (3 \sqrt{3}) = \frac{\sqrt{3}}{16} .

Facebook Tweet Pin