SBMPTN Zone : Integral Tentu [2]

Berikut ini adalah kumpulan soal mengenai Integral Tentu. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

No.

Hasil dari \displaystyle\intop_{10}^{12}\dfrac{dx}{\sqrt{28x+8x\sqrt{x}}}=

1. {\sqrt2+\sqrt3-2+\sqrt5}
2. {\sqrt2+\sqrt3+2-\sqrt5}
   
3. {-\sqrt2+\sqrt3+2-\sqrt5}
   

4. {-\sqrt2+\sqrt3-2+\sqrt5}
5. {-\sqrt2+\sqrt3+2+\sqrt5}
   

Facebook

\begin{aligned} \displaystyle\intop_{10}^{12}\dfrac{dx}{\sqrt{28x+8x\sqrt{x}}}&=\displaystyle\intop_{10}^{12}\dfrac{dx}{\sqrt{4x\left(7+2\sqrt{x}\right)}}\\[8pt] &=\displaystyle\intop_{10}^{12}\dfrac{dx}{2\sqrt{x}\sqrt{7+2\sqrt{x}}} \end{aligned}

Misal
\begin{aligned} u&=7+2\sqrt{x}\\ du&=\dfrac2{2\sqrt{x}}\ dx\\[8pt] du&=\dfrac1{\sqrt{x}}\ dx \end{aligned}

\begin{aligned} \displaystyle\intop_{10}^{12}\dfrac{dx}{2\sqrt{x}\sqrt{7+2\sqrt{x}}}&=\displaystyle\intop_{10}^{12}\dfrac{du}{2\sqrt{u}}\\ &=\displaystyle\intop_{10}^{12}\dfrac12u^{-\frac12}\ du\\ &=\left[u^{\frac12}\right]_{10}^{12}\\ &=\left[\sqrt{u}\right]_{10}^{12}\\ &=\left[\sqrt{7+2\sqrt{x}}\right]_{10}^{12}\\ &=\left(\sqrt{7+2\sqrt{12}}\right)-\left(\sqrt{7+2\sqrt{10}}\right)\\ &=\left(\sqrt{4+3+2\sqrt{4\cdot3}}\right)-\left(\sqrt{2+5+2\sqrt{2\cdot5}}\right)\\ &=\left(\sqrt4+\sqrt3\right)-\left(\sqrt2+\sqrt5\right)\\ &=2+\sqrt3-\sqrt2+\sqrt5\\ &=\boxed{\boxed{-\sqrt2+\sqrt3+2+\sqrt5}} \end{aligned}

Jadi, \displaystyle\intop_{10}^{12}\dfrac{dx}{\sqrt{28x+8x\sqrt{x}}}=-\sqrt2+\sqrt3+2+\sqrt5. JAWAB: E

No.

Jika {\displaystyle\intop_{-3}^3f(x)\left(\sin x+1\right)\ dx=10} dengan f(x) fungsi genap dan {\displaystyle\intop_{-1}^3f(x)\ dx=7}, maka {\displaystyle\intop_1^3f(x)\ dx=} ....

1. -4
2. -3
   
3. -2
   

4. 3
5. 4
   

Facebook

\begin{aligned} \displaystyle\intop_{-3}^3f(x)\left(\sin x+1\right)\ dx&=10\\ \displaystyle\intop_{-3}^3\left(f(x)\sin x+f(x)\right)\ dx&=10\\ \displaystyle\intop_{-3}^3f(x)\sin x\ dx+\displaystyle\intop_{-3}^3f(x)\ dx&=10 \end{aligned}
f(x) fungsi genap, \sin x fungsi ganjil, sehingga f(x)\sin x fungsi ganjil.
\displaystyle\intop_{-3}^3f(x)\sin x\ dx=0
\displaystyle\intop_{-3}^3f(x)\ dx=2\displaystyle\intop_0^3f(x)\ dx

\begin{aligned} 0+2\displaystyle\intop_0^3f(x)\ dx&=10\\ \displaystyle\intop_0^3f(x)\ dx&=5 \end{aligned}

\begin{aligned} \displaystyle\intop_{-1}^0f(x)\ dx+\displaystyle\intop_0^3f(x)\ dx&=\displaystyle\intop_{-1}^3f(x)\ dx\\ \displaystyle\intop_{-1}^0f(x)\ dx+5&=7\\ \displaystyle\intop_{-1}^0f(x)\ dx&=2\\ \displaystyle\intop_0^1f(x)\ dx&=2 \end{aligned}

\begin{aligned} \displaystyle\intop_0^1f(x)\ dx+\displaystyle\intop_1^3f(x)\ dx&=\displaystyle\intop_0^3f(x)\ dx\\ 2+\displaystyle\intop_1^3f(x)\ dx&=5\\ \displaystyle\intop_1^3f(x)\ dx&=\boxed{\boxed{3}} \end{aligned}

Jadi, {\displaystyle\intop_1^3f(x)\ dx=3}. JAWAB: D

No.

Jika nilai \displaystyle\intop_1^2f( x) dx= 4, maka nilai \displaystyle\intop^1_0x\ f\left(x^2+1\right) dx adalah

1. 2
2. 3
3. 4

4. 5
5. 6

SKMP Juli 2020

Misal u=x^2+1
\(\eqalign{ du&=2x\ dx\\ x\ dx&=\dfrac12du }\)

{x=0\to u=0^2+1=1}
{x=1\to u=1^2+1=2}

\(\eqalign{ \displaystyle\intop^2_0x\ f\left(x^2+1\right) dx&=\displaystyle\intop^2_1\ f\left(u\right) du\\ &=\boxed{\boxed{4}} }\)

Jadi, \displaystyle\intop^1_0x\ f\left(x^2+1\right) dx=4. JAWAB: C