Question

Aria și volumul integralelor.
Am nevoie de ajutor urgent.

Answer 1

Răspuns:

Explicație pas cu pas:

x²-x=3x, ⇒x²-4x=0, ⇒x(x-4)=0, ⇒x=0 si x=4 sunt limitele de integrare.

$Aria=\int\limits^4_0 {(3x-(x^{2}-x))} \, dx= \int\limits^4_0 {(3x-x^{2}+x)} \, dx= \int\limits^4_0 {(4x-x^{2})} \, dx=(2x^{2}-\frac{x^{3}}{3})|_{0} ^{4}=2*4^{2}-\frac{4^{3}}{3}=\frac{16*6-16*4}{3}=\frac{16*2}{3} =10\frac{2}{3}$

$f(x)=x^{2},~g(x)=\sqrt{x} ,~~f(x)=g(x),~x^{2}=\sqrt{x},~x~{4}=x,~x~{4}-x=0,~x*(x^{3}-1)=0,~deci~x=0~si~x=1~sunt~limitele~de~integrare\\V=\pi \int\limits^1_0 {g^{2}(x)} \, dx -\pi \int\limits^1_0 {f^{2}(x)} \, dx=\pi*\int\limits^1_0 {(\sqrt{x}) ^{2}} \, dx-\pi *\int\limits^1_0 {(x^{2})^{2}} \, dx=\pi *\int\limits^1_0 {x} \, dx-\pi *\int\limits^1_0 {x^{4}} \, dx=\pi*(\frac{x^{2}}{2}-\frac{x^{5}}{5} )|_{0}^{1}=\pi *(\frac{1}{2}-\frac{1}{5})=\pi *\frac{3}{10}=0,3\pi$