SBMPTN Zone : Integral Tentu [2]

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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}
   

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$$\begin{array}{rl}{\displaystyle \underset{10}{\overset{12}{\int }}{\displaystyle \frac{dx}{\sqrt{28x+8x\sqrt{x}}}}}& ={\displaystyle \underset{10}{\overset{12}{\int }}{\displaystyle \frac{dx}{\sqrt{4x(7+2\sqrt{x})}}}}\\ & ={\displaystyle \underset{10}{\overset{12}{\int }}{\displaystyle \frac{dx}{2\sqrt{x}\sqrt{7+2\sqrt{x}}}}}\end{array}$$

Misal
$$\begin{array}{rl}u& =7+2\sqrt{x}\\ du& ={\displaystyle \frac{2}{2\sqrt{x}}}dx\\ du& ={\displaystyle \frac{1}{\sqrt{x}}}dx\end{array}$$

$$\begin{array}{rl}{\displaystyle \underset{10}{\overset{12}{\int }}{\displaystyle \frac{dx}{2\sqrt{x}\sqrt{7+2\sqrt{x}}}}}& ={\displaystyle \underset{10}{\overset{12}{\int }}{\displaystyle \frac{du}{2\sqrt{u}}}}\\ & ={\displaystyle \underset{10}{\overset{12}{\int }}{\displaystyle \frac{1}{2}}{u}^{-\frac{1}{2}}du}\\ & ={\left[u^{\frac{1}{2}}\right]}_{10}^{12}\\ & ={\left[\sqrt{u}\right]}_{10}^{12}\\ & ={\left[\sqrt{7+2\sqrt{x}}\right]}_{10}^{12}\\ & =\left(\sqrt{7+2\sqrt{12}}\right)-\left(\sqrt{7+2\sqrt{10}}\right)\\ & =\left(\sqrt{4+3+2\sqrt{4\cdot 3}}\right)-\left(\sqrt{2+5+2\sqrt{2\cdot 5}}\right)\\ & =(\sqrt{4}+\sqrt{3})-(\sqrt{2}+\sqrt{5})\\ & =2+\sqrt{3}-\sqrt{2}+\sqrt{5}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -\sqrt{2}+\sqrt{3}+2+\sqrt{5}}}}}\end{array}$$

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
   

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$$\begin{array}{rl}{\displaystyle \underset{-3}{\overset{3}{\int }}f(x)(\sin x+1)dx}& =10\\ {\displaystyle \underset{-3}{\overset{3}{\int }}(f(x)\sin x+f(x))dx}& =10\\ {\displaystyle \underset{-3}{\overset{3}{\int }}f(x)\sin xdx+{\displaystyle \underset{-3}{\overset{3}{\int }}f(x)dx}}& =10\end{array}$$
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

$$\begin{array}{rl}0+2{\displaystyle \underset{0}{\overset{3}{\int }}f(x)dx}& =10\\ {\displaystyle \underset{0}{\overset{3}{\int }}f(x)dx}& =5\end{array}$$

$$\begin{array}{rl}{\displaystyle \underset{-1}{\overset{0}{\int }}f(x)dx+{\displaystyle \underset{0}{\overset{3}{\int }}f(x)dx}}& ={\displaystyle \underset{-1}{\overset{3}{\int }}f(x)dx}\\ {\displaystyle \underset{-1}{\overset{0}{\int }}f(x)dx+5}& =7\\ {\displaystyle \underset{-1}{\overset{0}{\int }}f(x)dx}& =2\\ {\displaystyle \underset{0}{\overset{1}{\int }}f(x)dx}& =2\end{array}$$

$$\begin{array}{rl}{\displaystyle \underset{0}{\overset{1}{\int }}f(x)dx+{\displaystyle \underset{1}{\overset{3}{\int }}f(x)dx}}& ={\displaystyle \underset{0}{\overset{3}{\int }}f(x)dx}\\ 2+{\displaystyle \underset{1}{\overset{3}{\int }}f(x)dx}& =5\\ {\displaystyle \underset{1}{\overset{3}{\int }}f(x)dx}& =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$$

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
$\begin{array}{rl}du& =2xdx\\ xdx& ={\displaystyle \frac{1}{2}}du\end{array}$

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

$\begin{array}{rl}{\displaystyle \underset{0}{\overset{2}{\int }}xf(x^2+1)dx}& ={\displaystyle \underset{1}{\overset{2}{\int }}f\left(u\right)du}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4}}}}\end{array}$

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