Integral Tertentu Bersama Sifat-Sifatnya Beserta Contoh Soalnya

 

Integral Tertentu Bersama Sifat-Sifatnya Beserta Contoh Soalnya

Fransiska Oktaviana Putri (12)

 XI IPS 3

Selasa, 30 Maret 2021

1. Jika $f\left(x\right)=\int x\text{d}x+{\int }_0^{1}x\text{d}x+{\int }_1^{2}x\text{d}x$ dan $f\left(2\right)=4$, maka nilai $f\left(0\right)=\cdots \cdot$
A. $1$                     C. $3$                    E. $5$
B. $2$                     D. $4$

Cara Penyelesaian:

Integralkan terlebih dahulu, lalu kita substitusikan $x=2$ untuk mencari nilai konstanta integral tak tentu $C$.
$$\begin{array}{cc}f\left(x\right)& =\int x\text{d}x+{\int }_0^{1}x\text{d}x+{\int }_1^{2}x\text{d}x\\ & =(\frac{1}{2}{x}^2+C)+{\left[\frac{1}{2}{x}^2\right]}_0^1+{\left[\frac{1}{2}{x}^2\right]}_1^2\\ & =\frac{1}{2}{x}^2+C+\frac{1}{2}(1^2-0^2)+\frac{1}{2}(2^2-1^2)\\ & =\frac{1}{2}{x}^2+C+\frac{1}{2}+\frac{1}{2}\left(3\right)\\ & =\frac{1}{2}{x}^2+C+2\\ f\left(2\right)& =\frac{1}{2}(2{)}^2+C+2\\ 4& =2+C+2\\ C& =0\end{array}$$Dengan demikian, $f\left(x\right)=\frac{1}{2}{x}^2+2$, sehingga
$f\left(0\right)=\frac{1}{2}(0{)}^2+2=2$

Jawaban: B

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2. Diketahui $f\left(x\right)=4x^3+3x^2+2x+{\int }_0^{2}f\left(x\right)\text{d}x$, maka nilai dari ${\int }_0^2\left(f^{\prime \prime }\right(x)+f(2\left)\right)\text{d}x$ $=\cdots \cdot$
A. $92$                    C. $96$                    E. $100$
B. $94$                    D. $98$

Cara Penyelesaian:

Diketahui $f\left(x\right)=4x^3+3x^2+2x+{\int }_0^{2}f\left(x\right)\text{d}x$. Perhatikan bahwa ekspresi ${\int }_0^{2}f\left(x\right)\text{d}x$ merupakan suatu konstanta, kita notasikan saja dengan $C$.
Dengan demikian, diperoleh turunan pertama $f\left(x\right)$, yakni
$f^{\prime }\left(x\right)=12x^2+6x+2$,
dan turunan keduanya adalah
$f^{\prime \prime }=24x+6$.
Selanjutnya,
$$\begin{array}{cc}{\int }_0^{2}f\left(x\right)\text{d}x& ={\int }_0^2(4x^3+3x^2+2x+C)\text{d}x\\ C& ={[x^4+x^3+x^2+Cx]}_0^2\\ C& =\left(\right(2{)}^4+(2{)}^3+(2{)}^2+C\left(2\right))-0\\ C& =16+8+4+2C\\ C& =-28\end{array}$$Ini berarti, $f\left(x\right)=4x^3+3x^2+2x-28$, sehingga $f\left(2\right)=4(2{)}^3+3(2{)}^2+2\left(2\right)-28=20$.
Oleh karena itu, kita peroleh
$\begin{array}{cc}& {\int }_0^2\left(f^{\prime \prime }\right(x)+f(2\left)\right)\text{d}x\\ & ={\int }_0^2(24x+6+20)\text{d}x\\ & ={\int }_0^2(24x+26)\text{d}x\\ & ={[12x^2+26x]}_0^2\\ & =\left(12\right(2{)}^2+26\left(2\right))-0\\ & =48+52=100\end{array}$
Jadi, nilai dari integral tersebut adalah $100$
(Jawaban E)

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3. Diketahui fungsi $f\left(x\right)=x^3+3x^2-5x+$ ${\int }_{-1}^{1}f\left(x\right)\text{d}x$. Nilai $f\left(1\right)=\cdots \cdot$
A. $-3$                     C. $-1$                    E. $4$
B. $-2$                     D. $3$

Cara Penyelesaian:

$f\left(x\right)$ adalah fungsi kubik dengan konstanta ${\int }_{-1}^{1}f\left(x\right)\text{d}x=C$.
Dari sini, kita peroleh
$$\begin{array}{cc}{\int }_{-1}^{1}f\left(x\right)\text{d}x& ={\int }_{-1}^1(x^3+3x^2-5x+{\int }_{-1}^{1}f\left(x\right)\text{d}x)\text{d}x\\ C& ={\int }_{-1}^1(x^3+3x^2-5x+C)\text{d}x\\ C& ={[\frac{1}{4}{x}^4+x^3-\frac{5}{2}{x}^2+Cx]}_{-1}^1\\ C& =\left(\frac{1}{4}\right(1{)}^4+(1{)}^3-\frac{5}{2}(1{)}^2+C\left(1\right))-\left(\frac{1}{4}\right(-1{)}^4+(-1{)}^3-\frac{5}{2}(-1{)}^2+C(-1))\\ C& =(\frac{1}{4}+1-\frac{5}{2}+C)-(\frac{1}{4}-1-\frac{5}{2}-C)\\ C& =C+(1+1)+C\\ C& =-2\end{array}$$Kita peroleh bahwa $f\left(x\right)=x^3+3x^2-5x-2$.
Untuk itu, jika $x=1$, didapat
$\begin{array}{cc}f\left(1\right)& =(1{)}^3+3(1{)}^2-5\left(1\right)-2\\ & =1+3-5-2=-3\end{array}$
Jadi, nilai dari $f\left(1\right)=-3$

Jawaban: A

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 4. Jika diketahui $g\left(x\right)=\left({\int }_0^{2}g\left(x\right)\text{d}x\right)x^2+$ $\left({\int }_0^{1}g\left(x\right)\text{d}x\right)x+\left({\int }_0^{3}g\right(x\left)\text{d}x\right)+2$, maka nilai $g\left(5\right)$ adalah $\cdots \cdot$
A $-\frac{274}{39}$                        D. $-\frac{254}{13}$
B. $-\frac{274}{13}$                       E. $\frac{274}{39}$
C. $-\frac{254}{39}$

Cara Penyelesaian:

Diketahui $g\left(x\right)$ merupakan fungsi kuadrat. Misal $g\left(x\right)=a{x}^2+bx+c$, maka diperoleh integralnya terhadap $x$, yakni $G\left(x\right)=\frac{1}{3}a{x}^3+\frac{1}{2}b{x}^2+cx+D$ untuk suatu konstanta real $D$.
Dari sini, kita juga peroleh bahwa
$$\begin{array}{cccc}a& ={\int }_0^{2}g\left(x\right)\text{d}x& & \\ & =G\left(2\right)-G\left(0\right)& & \\ & =\frac{1}{3}a(2{)}^3+\frac{1}{2}b(2{)}^2+c\left(2\right)-0& & \\ & =\frac{8}{3}a+2b+2c& & (\cdots 1)\\ b& ={\int }_0^{1}g\left(x\right)\text{d}x& & \\ & =G\left(1\right)-G\left(0\right)& & \\ & =\frac{1}{3}a(1{)}^3+\frac{1}{2}b(1{)}^2+c\left(1\right)-0& & \\ & =\frac{1}{3}a+\frac{1}{2}b+c& & (\cdots 2)\\ c& ={\int }_0^{3}g\left(x\right)\text{d}x+2& & \\ & =G\left(3\right)-G\left(0\right)+2& & \\ & =\frac{1}{3}a(3{)}^3+\frac{1}{2}b(3{)}^2+c\left(3\right)-0+2& & \\ & =9a+\frac{9}{2}b+3c+2& & (\cdots 3)\end{array}$$Persamaan $\left(1\right)$, $\left(2\right)$, dan $\left(3\right)$ masing-masing dapat disederhanakan sehingga diperoleh sistem persamaan linear tiga variabel berikut.
$\{\begin{array}{cccc}5a+6b+6c& =0& & (\cdots 1)\\ 2a-3b+6c& =0& & (\cdots 2)\\ 18a+9b+4c& =-4& & (\cdots 3)\end{array}$
Eliminasi $c$ dari persamaan $\left(1\right)$ dan $\left(2\right)$ untuk memperoleh
$\begin{array}{}3a+9b=0\Rightarrow a+3b=0& \text{(4)}\end{array}$
Eliminasi $c$ dari persamaan $\left(2\right)$ dan $\left(3\right)$ untuk memperoleh
$\begin{array}{cccc}-100a-66b& =24& & \\ \Rightarrow 50a+33b& =-12& & (\cdots 5)\end{array}$
Dari persamaan $\left(4\right)$ dan $\left(5\right)$, diperoleh $a=-\frac{4}{13}$ dan $b=\frac{4}{39}$, sehingga $c=\frac{2}{13}$.
Jadi, $g\left(x\right)=-\frac{4}{13}{x}^2+\frac{4}{39}x+\frac{2}{13}$, berarti
$\begin{array}{cc}g\left(5\right)& =-\frac{100}{13}+\frac{20}{39}+\frac{2}{13}\\ & =\frac{-300+20+6}{39}=-\frac{274}{39}\end{array}$

Jawaban: A

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5. Nilai $y$ yang memenuhi persamaan ${\int }_0^{\sqrt{{}^2\log (5y+1)}}3x\sqrt{-2x^2+9}\text{d}x=13$ adalah $\cdots \cdot$
A. $1$                     C. $3$                    E. $5$
B. $2$                     D. $4$

Cara Penyelesaian:

Tanpa batas integral, kita akan mencari hasil dari $\int 3x\sqrt{-2x^2+9}\text{d}x$ terlebih dahulu.
Misalkan $u=-2x^2+9$, maka
$\begin{array}{cc}\text{d}u& =-4x\text{d}x\\ -\frac{1}{4}\text{d}u& =x\text{d}x\end{array}$
Dengan demikian, didapat
$$\begin{array}{cc}\int 3x\sqrt{-2x^2+9}\text{d}x& =\int 3\cdot (-\frac{1}{4})\cdot u^{1/2}\text{d}u\\ & =-\frac{3}{4}\cdot \frac{2}{3}\cdot u^{3/2}+C\\ & =-\frac{1}{2}{u}^{3/2}+C\end{array}$$Batas integrasi berubah untuk variabel $u$. Karena $u=-2x^2+9$, maka
$\begin{array}{cc}{u}_{\text{atas}}& =-2{\left(\sqrt{{}^2\log (5y+1)}\right)}^2+9\\ & =-2{(}^2\log (5y+1))+9\\ u_{\text{bawah}}& =-2(0{)}^2+9=9\end{array}$
Dengan demikian,
$$\begin{array}{cc}{\int }_0^{\sqrt{{}^2\log (5y+1)}}3x\sqrt{-2x^2+9}\text{d}x& =13\\ {[-\frac{1}{2}{u}^{3/2}]}_9^{-2\left({}^2\log \right(5y+1\left)\right)+9}& =13\\ {\left[u^{3/2}\right]}_9^{-2\left({}^2\log \right(5y+1\left)\right)+9}& =-26\\ {(-2\left({}^2\log \right(5y+1\left)\right)+9)}^{3/2}-9^{3/2}& =-26\\ {(-2\left({}^2\log \right(5y+1\left)\right)+9)}^{3/2}-27& =-26\\ {(-2\left({}^2\log \right(5y+1\left)\right)+9)}^{3/2}& =1\\ \text{Kedua ruas dipangkatkan}& \frac{2}{3}\\ -2\left({}^2\log \right(5y+1\left)\right)+9& =1\\ -2\left({}^2\log \right(5y+1\left)\right)& =-8\\ {}^2\log (5y+1)& =4\\ 5y+1& =2^4=16\\ 5y& =15\\ y& =3\end{array}$$Jadi, nilai dari $y$ adalah $3$

Jawaban: C