SOAL SBMPTN INTEGRAL FUNGSI ALJABAR DAN PEMBAHASAN - Kikiholmes

1. 
   
   [Kem-IPA UM UNDIP 2011]

   1. $\frac{1}{2}$
   2. 1
   3. $1\frac{1}{2}$
   4. 2
   5. 3
JAWABAN A

$$\begin{align*} &= \int_{1}^{4}\dfrac{\left(\sqrt{x}-1 \right )^3}{\sqrt{x}} dx \\ &= \int_{1}^{4}\dfrac{\left(x^{\tfrac{1}{2}}-1 \right )^3}{\sqrt{x}}dx\\ &= \int_{1}^{4}\dfrac{x^{\tfrac{3}{2}}-3x+3x^{\tfrac{1}{2}}-1}{\sqrt{x}}dx\\ &= \int_{1}^{4}x-3x^{\tfrac{1}{2}}+3-x^{\tfrac{1}{2}}dx\\ &= \left [ \dfrac{1}{2}x^2-2x^{\tfrac{3}{2}}+3x-2x^{\tfrac{1}{2}} \right ]_{1}^{4}\\ &=\left( 8-16+12-4 \right )-\left(\dfrac{1}{2}-2+3-2 \right )\\ &=\dfrac{1}{2} \end{align*}$$




2. Diketahui fungsi f dan fungsi g yang memenuhi sistem
   
   $\left\{\begin{matrix} \displaystyle \int_{0}^{2}g\left(x \right)dx =\alpha\\ f\left(x \right) = 3x^2+4x+\alpha \end{matrix}\right.$

   1. -6
   2. -3
   3. 0
   4. 3
   5. 6
JAWABAN E

$$\displaystyle \begin{align*} \text{Misalkan } \int_0^2 g\left(x \right )dx=\alpha\\ f\left(x \right )=3x^2+4x+\alpha\\ \int_0^1f\left(x \right )dx &=\int_0^1 \left(3x^2+4x+\alpha \right )dx\\ &=\left [ x^3+2x^2+\alpha x \right ]\\ &=\left(1+2+\alpha \right )-\left(0 \right )=3+\alpha\\ \\ \int_0^1 f\left(x \right )dx+\left (\int_0^2g\left(x \right ) dx \right )^2=3\\ \left(3+\alpha \right )+\alpha^2=3\\ \alpha^2+\alpha=0\\ \alpha\left(\alpha+1 \right )\\ \alpha=0\ \ \text{tidak memenuhi} \cup \alpha=-1\\ \\ \text{Diperoleh} f\left(x \right )=3x^2+4x-1\\ \text{sehingga}f\left(1 \right )=3\left(1 \right )^2+4\left(1 \right )-1=6 \end{align*}$$





3. Nilai dari $\displaystyle\int_1^4\dfrac{\left(\sqrt{x}-1 \right )^3}{\sqrt{x}}dx=\cdots$
   
   [Kem-IPA UM UNDIP 2011]

   1. $\frac{1}{2}$
   2. 1
   3. $1\frac{1}{2}$
   4. 2
   5. 3
JAWABAN A

$$\small\begin{align*} \int_1^4\dfrac{\left(\sqrt{x}-1 \right )^3}{\sqrt{x}}dx &=\int_1^4\dfrac{\left(x^{\tfrac{1}{2}}-1 \right )^3}{\sqrt{x}}dx\\ &=\int_1^4\dfrac{x^{\tfrac{3}{2}}-3x+3x^{\tfrac{1}{2}}-1}{\sqrt{x}}dx\\ &=\int_1^4x-x^{\tfrac{1}{2}}+3-x^{-\tfrac{1}{2}}dx\\ &=\left[ \dfrac{1}{2}x^2-2x^{\tfrac{3}{2}}+3x-2x^{\tfrac{1}{2}}\right ]_1^4\\ &=\left(8-16+12-4 \right )-\left(\dfrac{1}{2}-2+3-2 \right )\\ &=\dfrac{1}{2} \end{align*}$$





4. Gradien garis singgung suatu kurva di titik $\left(x, y\right)$ sama dengan $\displaystyle 2x+5$. Jika kurva ini melalui titik $\left(2, 20 \right)$, maka kurva tersebut memotong sumbu x di ...
   
   [UM UGM 2008}

   1. $\left(2, 0 \right)$ dan $\left(3, 0 \right)$
   2. $\left(-2, 0 \right)$ dan $\left(-3, 0 \right)$
   3. $\left(-4, 0 \right)$ dan $\left(-3, 0 \right)$
   4. $\left(-1, 0 \right)$ dan $\left(-3, 0 \right)$
   5. $\left(2, 0 \right)$ dan $\left(-4, 0 \right)$
JAWABAN B

Ingat kembali materi turunan fungsi, dimana jika suatu fungsi $\displaystyle f\left(x \right )$ diturunkan, maka akan menjadi gradien. Sehingga untuk mencari fungsi awal dari 2x+5, kita menggunakan integral.
$$\small \displaystyle \begin{align*} \text{Gradien }\rightarrow y'&=2x+5\\ y &=\int2x+5dx \\ &=x^2+5x+C\\ \text{Kurva melalui (2, 20) sehingga }\\ y&= x^2+5x+C\\ 20&=\left(x \right )^2+5\left(x \right )+C\\ 20&=14+C\\ C&=6\\ \\ \text{Diperoleh persamaan }\\ y&=x^2+5x+6\\ \\ \text{Kurva memotong sumbu x, artinya y=0 }\\ x^2+5x+6=0\\ \left(x+3 \right )\ \ \left(x+2 \right )\\ x&=3 \ \cup x=-2\\ \\ \text{Jadi, koordinat titik potong} \\ \text{dengan sumbu x adalah (-2,0) dan (-3,0)} \end{align*}$$




5. Jika $\displaystyle \int_0^1\frac{x}{1+x}dx$, maka $\displaystyle \int_0^1\frac{1}{1+x}dx=\cdots$
   
   [UMB PTN 2009]

   1. a
   2. 1-a
   3. 2a
   4. a-$\frac{1}{2}$
   5. 5a
JAWABAN B

$$\small \displaystyle \begin{align*} \int_0^1\frac{1+x}{1+x}dx &=\int_0^1\frac{1}{1+x}dx+\int_0^1\frac{x}{1+x}dx\\ &\text{Misalkan }\boxed{\int_0^1\frac{x}{1+x}dx\text{ adalah a, maka}}\\\\ \int_0^11\ dx&=\int_0^1\frac{1}{1+x}dx+a\\ \left [ x \right ]_0^1&=\int_0^1\frac{1}{1+x}dx+a\\ 1&=\int_0^1\frac{1}{1+x}dx+a\\ \int_0^1\frac{1}{1+x}dx&=1-a \end{align*}$$




6. Diketahui $\displaystyle\int f\left ( x \right )dx=ax^2+bx+c$, dan $\displaystyle a\neq 0$. Jika a, f$\left(a \right)$, dan 2b membentuk barisan aritmatika, dan $f\left(b \right)=6$, maka $\displaystyle \int_0^1f\left(x \right )dx=\cdots$

   1. $\frac{17}{4}$
   2. $\frac{21}{4}$
   3. $\frac{25}{4}$
   4. $\frac{13}{4}$
   5. $\frac{11}{4}$
JAWABAN A

$$\int f \left( x \right)dx=ax^2+bx+c\\f\left( x \right)=2ax+b$$ a, f$\left( a \right)$, 2b membentuk barisan aritmatika, sehingga:
suku pertama = a
suku terakhir $\displaystyle U_n$ = 2b
Suku tengah $\displaystyle U_t$ = $f\left (a \right)$.

Ingat kembali rumus suku tengah:
$$\boxed{suku\ tengah =U_t=\dfrac{a+U_n}{2}}$$
$$\small \displaystyle \begin{align*} U_t&=\dfrac{a+U_n}{2}\\ 2\cdot U_t&=a+U_n\\ 2f\left(a \right )&=a+2b\\ 2\left(2\cdot a^2+b \right )&=a+2b\\ 4a^2+2b&=a+2b\\ 4a^2-a&=0\\ a&=0 \cup a=\frac{1}{4}\\ \end{align*}$$
Nilai a yang memenuhi adalah $\dfrac{1}{4}$
$$\small \displaystyle \begin{align*} f\left(b \right )&=6\\ 2ab+b&=6\\ 2\cdot \dfrac{1}{4} \cdot b+b=6\\ b=4\\ Jadi,\ f\left(x \right )=2\cdot \dfrac{1}{4}x+4&=\dfrac{1}{2}x+4 \int_0^1f\left(x \right )=\int_0^1\dfrac{1}{2}x+4dx\\&=\left[\dfrac{1}{4}x^2+4 \right ]_0^1\\ &=\frac{1}{4}+4\\ &=\frac{17}{4} \end{align*}$$





7. $\displaystyle \int_{-3}^3\left | x^2-2x-3 \right | dx= \cdots$
   
   [Mat-IPA UMPTN 2004]

   1. 0
   2. 18
   3. $ \frac{68}{3}$
   4. $ \frac{64}{3}$
   5. 9
JAWABAN D







8. $\displaystyle \int\dfrac{2x^2}{\sqrt[7]{\left ( 2x^3-5 \right )^3}}dx=\cdots\\$

   1. $\dfrac{3}{7}\sqrt[7]{\left ( 2x^3-5 \right )^3}+C$
   2. $\dfrac{6}{7}\sqrt[6]{\left ( 2x^3-5 \right )^7}+C$
   3. $\dfrac{6}{7}\sqrt[6]{\left ( 2x^3-5 \right )^6}+C$
   4. $\dfrac{7}{6}\sqrt[7]{\left ( 2x^3-5 \right )^2}+C$
   5. $\dfrac{7}{6}\sqrt[7]{\left ( 2x^3-5 \right )^7}+C$
JAWABAN D

$$\small \begin{align*} \int\dfrac{2x^2}{\sqrt[7]{\left ( 2x^3-5 \right )^3}}dx&=\int2x^2\left ( 2x^3-5 \right )^{-\tfrac{5}{7}}\dfrac{d\left ( 2x^3-5 \right )}{6x^2}\\ &=\dfrac{1}{3}\int\left ( 2x^3-5 \right )^{-\tfrac{5}{7}}\ d\left ( 2x^3-5 \right )\\ &=\dfrac{1}{3}\left ( \dfrac{7}{2} \left ( 2x^3-5 \right )^{\tfrac{2}{7}} \right )+C\\ &=\dfrac{7}{6}\sqrt[7]{\left ( 2x^3-5 \right )^2}+C \end{align*}$$





9. Jika pada $\displaystyle \int_0^{\tfrac{1}{2}}\dfrac{\sqrt{x}}{\sqrt{1-x}}dx$ disubstitusikan $\displaystyle \sqrt{x}=\sin y$, maka menghasilkan...
   
   [Mat-IPA SNMPTN 2009]

   1. $\displaystyle \int_0^{\tfrac{1}{2}}\sin^2 x\ dx$
   2. $\displaystyle 2 \int_0^{\tfrac{1}{2}}\dfrac{\sin^2 y}{\cos y}\ dy$
   3. $\displaystyle \int_0^{\tfrac{\pi}{4}}2 \sin^2 x\ dx$
   4. $\displaystyle \int_0^{\tfrac{\pi}{4}}\sin^2 y\ dy$
   5. $\displaystyle \int_0^{\tfrac{\pi}{6}}\sin^2 x\ dx$
JAWABAN C

Misalkan $$\displaystyle \small \begin{align*} \sqrt{x}=\sin y\\ \left ( \sqrt{x} \right )^2=\left ( \sin y \right )^2\\ x&=\sin^2 y\\ dx&=2\sin y \cos y\ dy\\ \text{Merubah batas: }\\ \text{Untuk }x&=0\rightarrow \sqrt{0}=\sin y\rightarrow y= 0 \\ \text{Untuk }x&=\tfrac{1}{2}\rightarrow\sqrt{\dfrac{1}{2}}=\sin y\rightarrow y=\dfrac{\pi}{4}\\ \int_0^{\tfrac{1}{2}}\dfrac{\sqrt{x}}{\sqrt{1-x}}\ dx&=\int_0^{\tfrac{\pi}{4}}\dfrac{\sin y}{\sqrt{1-\sin^2 y}}\ 2\sin y \cos y\ dy\\ \boxed{ingat\ \sin^2 y+\cos^2 y=1\rightarrow \cos^2 y=1-\sin^2 y}\\ &=\int_0^{\tfrac{\pi}{4}}\dfrac{\sin y}{\sqrt{\cos^2 y}}\ dy\\ &=\int_0^{\tfrac{\pi}{4}}\dfrac{\sin y}{\cos^2 y}\ 2\sin y\cos y\ dy\\ &=\int_0^{\tfrac{\pi}{4}}2 \sin^2 y\ dy \end{align*}$$






10. Nilai $\displaystyle \int_0^\pi\sin \left ( x \right )\left | \sin\left ( x \right ) \right |\ dx=\cdots$
    
    [Kem-IPA UM UNDIP 2011}

    1. $\displaystyle \frac{\pi}{3}$
    2. $\displaystyle \frac{\pi}{2}$
    3. $\pi$
    4. $\displaystyle \frac{3\pi}{3}$
    5. $2\pi$
JAWABAN B

Karena batas integralnya $\displaystyle 0\leq x \leq \pi$
$\rightarrow \left | \sin\left ( x \right ) \right |= \sin \left ( x \right )$
Ingat bahwa nilai sin untuk $\displaystyle 0\leq x \leq \pi$ adalah positif, maka nilai mutlaknya bernilai positif juga.

$$\small \begin{align*} \int_0^\pi\sin \left ( x \right )\left | \sin\left ( x \right ) \right |\ dx&=\int_0^\pi\sin \left ( x \right ) \sin \left ( x \right )\ dx\\ &=\int_0^\pi\sin^2 \left ( x \right )\ dx&\boxed{^*\tiny {pembuktian\ lihat\ di\ bawah}}\\ &=\int_0^\pi\dfrac{1}{2}-\dfrac{1}{2}\cos 2x\ dx\\ &=\left [ \dfrac{1}{2}x-\dfrac{1}{4}\sin 2x \right ]_0^\pi\\ &=\left ( \frac{1}{2}\pi-0 \right )- \left(0\right)\\ &=\dfrac{1}{2}\pi \end{align*}$$


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