Contoh Soal 1 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{l}\\ 1.& \mathbf{\text{(UAN 2014)}}\\ & \text{Hasil}{\displaystyle {\int }_{-1}^2(x^3+3x^2+4x+5)dx=....}\\ & \begin{array}{ll}\text{a}.& 33{\displaystyle \frac{1}{4}}\\ \text{b}.& 33{\displaystyle \frac{3}{4}}\\ \text{c}.& 32{\displaystyle \frac{1}{4}}\\ \text{d}.& 31{\displaystyle \frac{3}{4}}\\ \text{e}.& 23{\displaystyle \frac{3}{4}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\int }_{-1}^2(x^3+3x^2+4x+5)dx\\ & ={\displaystyle \frac{x^4}{4}+x^3+2x^2+5x{|}_{-1}^2}\\ & =\left({\displaystyle \frac{2^4}{4}+2^3+{2.2}^2+5.2}\right)\\ & -(\frac{{(-1)}^4}{4}+{(-1)}^3+2.{(-1)}^2+5(-1))\\ & =33\frac{3}{4}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \mathbf{\text{(UAN 2003)}}\\ & \text{Jika}f(x)={(x-2)}^2-4\text{dan}g(x)=-f(x),\\ & \text{maka luas daerah yang di batasi kurva }\\ & f\text{dan}g\text{adalah}....\\ & \begin{array}{ll}\text{a}.& 10{\displaystyle \frac{2}{3}\text{satuan luas}}\\ \text{b}.& 21{\displaystyle \frac{1}{3}\text{satuan luas}}\\ \text{c}.& 22{\displaystyle \frac{2}{3}\text{satuan luas}}\\ \text{d}.& 42{\displaystyle \frac{2}{3}\text{satuan luas}}\\ \text{e}.& 45{\displaystyle \frac{1}{3}\text{satuan luas}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \end{array}$.

$.\begin{array}{rl}& \text{Alternatif 1}\\ & \begin{array}{rl}& {\displaystyle {\int }_0^4(g(x)-f(x))dx}\\ & ={\displaystyle {\int }_0^4(4x-x^2)-(x^2-4x)dx}\\ & ={\displaystyle {\int }_0^4(8x-2x^2)dx}\\ & ={\displaystyle {[4x^2-\frac{2}{3}{x}^3]}_0^4}\\ & ={\displaystyle ({4.4}^2-\frac{2}{3}{.4}^3)-({4.0}^2-\frac{2}{3}{.0}^3)}\\ & ={\displaystyle (64-\frac{2}{3}.64)-0}\\ & ={\displaystyle \frac{64}{3}=21\frac{1}{3}\text{satuan luas}}\end{array}\\ & \text{Alternatif 2}\\ & \text{Dengan menggunakan rumus}L={\displaystyle \frac{D\sqrt{D}}{6a^2}}\\ & \begin{array}{|c|}\hline \begin{array}{rl}& f(x)=g(x)\\ & f(x)=-f(x),\text{ingat}g(x)=-f(x)\\ & 2f(x)=0,\text{tidak boleh disederhanakan},\\ & 2\times ({(x-2)}^2-4)=0,\text{karena akan }\\ & 2\times (x^2-4x)=0\text{mempengaruhi hasil }\\ & 2x^2-8x=0,\text{akhir}\\ & \{\begin{array}{ll}a=2,b=-8& c=0\\ D=b^2-4ac,& D={(-8)}^2-4(2)(0)=64\end{array}\\ & L_{\mathbf{\text{daerah}}}={\displaystyle \frac{\mathbf{\text{D}}\sqrt{\mathbf{\text{D}}}}{6{\mathbf{\text{a}}}^2}}\\ & ={\displaystyle \frac{64\sqrt{64}}{6(2{)}^2}}\\ & ={\displaystyle \frac{64\times 8}{6\times 4}}\\ & ={\displaystyle \frac{64}{3}}\\ & =21{\displaystyle \frac{1}{3}}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \mathbf{\text{(UAN 2014)}}\\ & \text{Luas daerah yang diarsir pada gambar }\\ & \text{dapat dinyatakan dengan rumus}\end{array}$.

$.\begin{array}{l}\\ & \begin{array}{ll}\text{a}.& {\displaystyle {\int }_0^{4}4xdx-{\int }_2^4(2x-4)dx}\\ \text{b}.& {\displaystyle {\int }_0^{4}4xdx-{\int }_2^4(2x+4)dx}\\ \text{c}.& {\displaystyle {\int }_0^{4}2\sqrt{2}dx-{\int }_2^4(2x-4)dx}\\ \text{d}.& {\displaystyle {\int }_0^{4}2\sqrt{2}dx-{\int }_2^4(4-2x)dx}\\ \text{e}.& {\displaystyle {\int }_0^{4}2\sqrt{x}dx-{\int }_2^4(4+2x)dx}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\text{Luas}}_{\text{arsiran}}={\displaystyle {\int }_0^4{y}_{1}dx-{\int }_2^4{y}_{2}dx}\\ & ={\displaystyle {\int }_0^4\sqrt{4x}dx-{\int }_2^4(2x-4)dx}\\ & ={\displaystyle {\int }_0^{4}2\sqrt{x}dx-{\int }_2^4(2x-4)dx.}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \mathbf{\text{(UAN 2014)}}\\ & \text{Volume benda putar yang terbentuk dari daerah}\\ & \text{yang dikuadran I yang dibatasi oleh kurva}\\ & x=2\sqrt{3}{y}^2,\text{sumbu Y , dan lingkaran}{x}^2+y^2=1,\\ & \text{diputar mengelilingi sumbu Y adalah}....\\ & \begin{array}{ll}\text{a}.& {\displaystyle \frac{4}{60}\pi \text{satuan volum}}\\ \text{b}.& {\displaystyle \frac{17}{60}\pi \text{satuan volum}}\\ \text{c}.& {\displaystyle \frac{23}{60}\pi \text{satuan volum}}\\ \text{d}.& {\displaystyle \frac{44}{60}\pi \text{satuan volum}}\\ \text{e}.& {\displaystyle \frac{112}{60}\pi \text{satuan volum}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi berikut}\end{array}$.

$.\begin{array}{rl}& \text{Volumenya jika diputar mengelilingi Sumbu }\\ & \text{Y adalah}:\\ & V=\pi {\displaystyle {\int }_0^{\frac{1}{2}}{x}_1^{2}dy+\pi {\int }_{\frac{1}{2}}^1{x}_2^{2}dy}\\ & \iff V=\pi {\int }_0^{\frac{1}{2}}{\left(2\sqrt{3}{y}^2\right)}^{2}dy+\pi {\int }_{\frac{1}{2}}^1(1-y^2)dy\\ & \iff V=\frac{12}{5}{y}^5\pi {|}_0^{\frac{1}{2}}+(y-\frac{1}{3}{y}^3)\pi {|}_{\frac{1}{2}}^1\\ & \iff V=\pi (\frac{12}{5}\times \frac{1}{32}+(1-\frac{1}{3})-(\frac{1}{2}-\frac{1}{3}\times \frac{1}{8}))\\ & \iff V=\frac{17}{60}\pi .\end{array}$.

$\begin{array}{l}\\ 5.& \text{Jika}{\displaystyle \frac{d}{dx}g(x)=f(x)\text{di mana}f(x)}\\ & \text{kontinu dari a sampai b, }\\ & \text{maka}{\displaystyle {\int }_a^{b}f(x).g(x)dx}\\ & \begin{array}{l}\\ \text{a}.0\\ \text{b}.f(b)-f(a)\\ \text{c}.g(b)-g(a)\\ \text{d}.{\displaystyle \frac{{[f(b)]}^2-{[f(a)]}^2}{2}}\\ \text{e}.{\displaystyle \frac{{[g(b)]}^2-{[g(a)]}^2}{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle {\int }_a^{b}f(x).g(x)dx}\\ & ={\displaystyle {\int }_a^{b}g(x).f(x)dx}\\ & ={\displaystyle {\int }_a^{b}g(x).{\displaystyle \frac{d(g(x))}{dx}dx,}}\\ & \text{ingat bahwa}{\displaystyle \frac{d}{dx}g(x)=f(x),}\\ & \text{f(x) kontinu dari a sampai b}\\ & ={\displaystyle {\int }_a^{b}g(x)d(g(x))}\\ & ={\left[{\displaystyle \frac{{(g(x))}^2}{2}}\right]}_a^b\\ & ={\displaystyle \frac{{[g(b)]}^2-{[g(a)]}^2}{2}}\end{array}\end{array}$.