y=√x, y=√3-2x, y=0 chiziqlar bilan chegaralangan soha yuzini toping.

Ikkita egri chiziq va $Ox$ o‘qi hosil qilgan sohaning yuzini hisoblaymiz. Buning uchun funksiyalarning kesishish nuqtasini topib olamiz. $\sqrt{x}=\sqrt{3-2x} \Rightarrow x_0=1$ da kesishadi. Chizmadan ko‘rinadiki, 2 ta: $S_1$ va $S_2$ yuzalarning yig‘indisini hisoblashga to‘g‘ri keladi. Buning uchun $S=S_1+S_2=\int_0^{x_0}\sqrt{x}dx+\int_{x_0}^{\frac{3}{2}}\sqrt{3-2x}dx$ integralni hisoblaymiz. $S=\int_0^1\sqrt{x}dx+\int_1^{\frac{3}{2}}\sqrt{3-2x}dx=\frac{2}{3}\sqrt{x^3}\Big|_0^1-\frac{1}{3}\sqrt{(3-2x)^3}\Big|_1^{\frac{3}{2}}=\frac{2}{3}\left(\sqrt{1^3}-\sqrt{0^3}\right)-\frac{1}{3}\left(\sqrt{\left(3-2\cdot\frac{3}{2}\right)^3}-\sqrt{(3-2\cdot 1)^3}\right)=1$

Manba: UZBMB namunaviy test topshiriqlari (2021, mutaxassislik). M.A.Mirzaahmedov, Sh.N.Ismailov, A.Q.Amanov, B.Q.Xaydarov. Matematika I va II qism, umumiy o‘rta ta’lim maktablarining 11-sinfi va o‘rta maxsus, kasb-hunar ta’limi muassasalari o‘quvchilari uchun darslik. “ZAMIN NASHR”, Toshkent−2018