find-the-area-bounded-by-the-curve-y-2-x-3-and-the-lines-x-0-y-1-and-y-2-

Question Number 113111 by gopikrishnan last updated on 11/Sep/20

f i n d t h e a r e a b o u n d e d b y t h e c u r v e y^{2} = x^{3} a n d t h e l i n e s x = 0 y = 1 a n d y = 2

Answered by 1549442205PVT last updated on 11/Sep/20

y^{2} = x^{3} \Leftrightarrow y = \sqrt{x^{3}} . We find the

intersection points of y = \sqrt{x^{3}} with

y = 1 and y = 2 . We get

A (1, 1), B (^{3} \sqrt{4}, 2) . Hence,

S = \int_{0}^{1} (2 - 1) dx + \int_{1}^{^{3} \sqrt{4}} (2 - \sqrt{x^{3}}) dx

= x ∣_{0}^{1} + 2 x ∣_{1}^{^{3} \sqrt{4}} - \frac{2}{5} x^{\frac{5}{2}} ∣_{1}^{^{3} \sqrt{4}} =

= 1 + 2 (^{3} \sqrt{4} - 1) - \frac{2}{5} (^{6} \sqrt{4^{5}} - 1)

= - \frac{3}{5} + 2^{3} \sqrt{4} - \frac{2}{5}^{6} \sqrt{4^{5}} \approx 1.304881262

Commented by gopikrishnan last updated on 11/Sep/20

T h a n k u s i r

Commented by 1549442205PVT last updated on 12/Sep/20

You are welcome .