3) Izračunati zapreminu obrtnog tela nastalog rotacijom oko \(x\)-ose figure u ravni ograničene krivama
\(\;x=y^{2}-2y+3,\; x-3y+1=0,\; y=0\).Rešenje:
Presek parabole i prave:
\[
y^{2}-2y+3=3y-1 \;\Longleftrightarrow\; y^{2}-5y+4=0
\;\Longleftrightarrow\; y\in\{1,4\}.
\]
Sa uslovom \(y=0\) (\(x-osa\)) zaključujemo da je ograničeni deo figure za \(y\in[0,1]\).
Za \(y\) iz \([0,1]\):
\[
x_L(y)=3y-1 \quad(\text{prava}),\qquad
x_D(y)=y^{2}-2y+3 \quad(\text{parabola}).
\]
Zapremina je
\[
V=2\pi\int_{0}^{1} y\big(x_D-x_L\big)\,dy
=2\pi\int_{0}^{1} y\big[\,(y^{2}-2y+3)-(3y-1)\,\big]\,dy.
\]
Sređivanjem:
\[
x_D-x_L = y^{2}-5y+4 \;\Longrightarrow\; y(x_D-x_L)=y^{3}-5y^{2}+4y.
\]
\[
V=2\pi\int_{0}^{1}\!\big(y^{3}-5y^{2}+4y\big)\,dy
=2\pi\Big(\frac{y^{4}}{4}-\frac{5y^{3}}{3}+2y^{2}\Big)\Big|_{0}^{1}
=2\pi\!\left(\frac{1}{4}-\frac{5}{3}+2\right)
=2\pi\cdot\frac{7}{12}
=\frac{7\pi}{6}.
\]
Rezultat: \(V=\frac{7\pi}{6}\).
4) Izračunati dužinu luka krive
\(x(t)=2\cos^{3}t,\; y(t)=2\sin^{3}t\) od tačke \(A(0,2)\) do tačke \(B(-2,0)\).Rešenje:
Tačka \(A(0,2)\) daje \(\cos t=0,\ \sin t=1\Rightarrow t=\frac{\pi}{2}\).
Tačka \(B(-2,0)\) daje \(\cos t=-1,\ \sin t=0\Rightarrow t=\pi\).
Dakle, \(t\in\big[\frac{\pi}{2},\pi\big]\).
\(x'(t)=-6\cos^{2}t\,\sin t,\quad y'(t)=6\sin^{2}t\,\cos t\).
\[
\sqrt{(x’)^{2}+(y’)^{2}}
=6\sqrt{\cos^{4}t\,\sin^{2}t+\sin^{4}t\,\cos^{2}t}
=6\,|\sin t\,\cos t|.
\]
Na \(\big[\frac{\pi}{2},\pi\big]\) važi \(\sin t\ge0,\ \cos t\le0\Rightarrow |\sin t\cos t|=-\sin t\cos t\).
\[
L=\int_{\pi/2}^{\pi}6|\sin t\cos t|\,dt
=-6\int_{\pi/2}^{\pi}\sin t\cos t\,dt
=-3\int_{\pi/2}^{\pi}\sin(2t)\,dt.
\]
Pošto je \(\displaystyle \int \sin(2t)\,dt=-\frac{1}{2}\cos(2t)\), dobijamo
\[
L=\frac{3}{2}\,\cos(2t)\Big|_{\pi/2}^{\pi}
=\frac{3}{2}\big(\cos 2\pi-\cos \pi\big)
=\frac{3}{2}(1-(-1))
=3.
\]
Rezultat: \(L=3\).
5) a) \(\displaystyle \int_{\pi/2}^{\pi}\sin^{2}x\cos^{5}x\,dx\)
Rešenje:
Uz \(t=\sin x\), \(dt=\cos x\,dx\) i \(\cos^{5}x=(1-t^{2})^{2}\cos x\):
\[ \int_{\pi/2}^{\pi}\sin^{2}x\cos^{5}x\,dx = -\!\int_{0}^{1} t^{2}(1-t^{2})^{2}\,dt = \left(-\frac{t^{3}}{3}-\frac{2t^{5}}{5}+\frac{t^{7}}{7}\right)\Big|_{0}^{1} = -\frac{8}{105}. \]Rezultat: \(-\frac{8}{105}\).
5) b) \(\displaystyle \int \frac{x^{4}}{9+x^{2}}\,dx\)
Rešenje:
\[
\frac{x^{4}}{x^{2}+9}
= x^{2}-9+\frac{81}{x^{2}+9},
\]
jer je
\[
(x^{2}+9)(x^{2}-9)=x^{4}-81 \;\Rightarrow\; \text{ostatak}=81.
\]
\[
\int\frac{x^{4}}{x^{2}+9}\,dx
= \int\!\left(x^{2}-9+\frac{81}{x^{2}+9}\right)dx
= \frac{x^{3}}{3}-9x + 81\!\int\!\frac{dx}{x^{2}+9}.
\]
Koristimo formulu \(\displaystyle \int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\arctan\!\frac{x}{a}\) za \(a=3\):
\[
81\!\int\!\frac{dx}{x^{2}+9}
= 81\cdot\frac{1}{3}\arctan\!\frac{x}{3}
= 27\arctan\!\frac{x}{3}.
\]
Rezultat:
\[ \frac{x^{3}}{3} – 9x + 27\,\arctan\!\frac{x}{3} + C. \]5) c) Površina lika ograničenog sa \(x=y^{3}\), \(y=2-x\) i \(x\)-osom
Presek \(x=y^{3}\) i \(y=2-x\) je \((1,1)\); sa \(x\)-osom u \((0,0)\) i \((2,0)\). Za \(y\in[0,1]\): desno \(x=2-y\), levo \(x=y^{3}\).Rešenje:
\[ P=\int_{0}^{1}\big[(2-y)-y^{3}\big]\,dy = \left(2y – \frac{y^{2}}{2} – \frac{y^{4}}{4}\right)\Big|_{0}^{1} = \frac{5}{4}. \]Rezultat: \(P=\frac{5}{4}\).
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