SNBT Zone : Integral Tentu [2]
Table of Contents
Tipe:
No.
Hasil dari \(\displaystyle\intop_{10}^{12}\dfrac{dx}{\sqrt{28x+8x\sqrt{x}}}=\)- \({\sqrt2+\sqrt3-2+\sqrt5}\)
- \({\sqrt2+\sqrt3+2-\sqrt5}\)
- \({-\sqrt2+\sqrt3+2-\sqrt5}\)
- \({-\sqrt2+\sqrt3-2+\sqrt5}\)
- \({-\sqrt2+\sqrt3+2+\sqrt5}\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\displaystyle\intop_{10}^{12}\dfrac{dx}{\sqrt{28x+8x\sqrt{x}}}&=\displaystyle\intop_{10}^{12}\dfrac{dx}{\sqrt{4x\left(7+2\sqrt{x}\right)}}\\[8pt]
&=\displaystyle\intop_{10}^{12}\dfrac{dx}{2\sqrt{x}\sqrt{7+2\sqrt{x}}}
\end{aligned}\)
Misal
\(\begin{aligned} u&=7+2\sqrt{x}\\ du&=\dfrac2{2\sqrt{x}}\ dx\\[8pt] du&=\dfrac1{\sqrt{x}}\ dx \end{aligned}\)
\(\begin{aligned} \displaystyle\intop_{10}^{12}\dfrac{dx}{2\sqrt{x}\sqrt{7+2\sqrt{x}}}&=\displaystyle\intop_{10}^{12}\dfrac{du}{2\sqrt{u}}\\ &=\displaystyle\intop_{10}^{12}\dfrac12u^{-\frac12}\ du\\ &=\left[u^{\frac12}\right]_{10}^{12}\\ &=\left[\sqrt{u}\right]_{10}^{12}\\ &=\left[\sqrt{7+2\sqrt{x}}\right]_{10}^{12}\\ &=\left(\sqrt{7+2\sqrt{12}}\right)-\left(\sqrt{7+2\sqrt{10}}\right)\\ &=\left(\sqrt{4+3+2\sqrt{4\cdot3}}\right)-\left(\sqrt{2+5+2\sqrt{2\cdot5}}\right)\\ &=\left(\sqrt4+\sqrt3\right)-\left(\sqrt2+\sqrt5\right)\\ &=2+\sqrt3-\sqrt2+\sqrt5\\ &=\boxed{\boxed{-\sqrt2+\sqrt3+2+\sqrt5}} \end{aligned}\)
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
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 \({\displaystyle\intop_1^3f(x)\ dx=}\) ....- −4
- −3
- −2
- 3
- 4
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\displaystyle\intop_{-3}^3f(x)\left(\sin x+1\right)\ dx&=10\\
\displaystyle\intop_{-3}^3\left(f(x)\sin x+f(x)\right)\ dx&=10\\
\displaystyle\intop_{-3}^3f(x)\sin x\ dx+\displaystyle\intop_{-3}^3f(x)\ dx&=10
\end{aligned}\)
f(x) fungsi genap,sin x fungsi ganjil, sehingga f(x) sin x fungsi ganjil.
\(\displaystyle\intop_{-3}^3f(x)\sin x\ dx=0\)
\(\displaystyle\intop_{-3}^3f(x)\ dx=2\displaystyle\intop_0^3f(x)\ dx\) \(\begin{aligned} 0+2\displaystyle\intop_0^3f(x)\ dx&=10\\ \displaystyle\intop_0^3f(x)\ dx&=5 \end{aligned}\)
\(\begin{aligned} \displaystyle\intop_{-1}^0f(x)\ dx+\displaystyle\intop_0^3f(x)\ dx&=\displaystyle\intop_{-1}^3f(x)\ dx\\ \displaystyle\intop_{-1}^0f(x)\ dx+5&=7\\ \displaystyle\intop_{-1}^0f(x)\ dx&=2\\ \displaystyle\intop_0^1f(x)\ dx&=2 \end{aligned}\)
\(\begin{aligned} \displaystyle\intop_0^1f(x)\ dx+\displaystyle\intop_1^3f(x)\ dx&=\displaystyle\intop_0^3f(x)\ dx\\ 2+\displaystyle\intop_1^3f(x)\ dx&=5\\ \displaystyle\intop_1^3f(x)\ dx&=\boxed{\boxed{3}} \end{aligned}\)
f(x) fungsi genap,
\(\displaystyle\intop_{-3}^3f(x)\sin x\ dx=0\)
\(\displaystyle\intop_{-3}^3f(x)\ dx=2\displaystyle\intop_0^3f(x)\ dx\) \(\begin{aligned} 0+2\displaystyle\intop_0^3f(x)\ dx&=10\\ \displaystyle\intop_0^3f(x)\ dx&=5 \end{aligned}\)
\(\begin{aligned} \displaystyle\intop_{-1}^0f(x)\ dx+\displaystyle\intop_0^3f(x)\ dx&=\displaystyle\intop_{-1}^3f(x)\ dx\\ \displaystyle\intop_{-1}^0f(x)\ dx+5&=7\\ \displaystyle\intop_{-1}^0f(x)\ dx&=2\\ \displaystyle\intop_0^1f(x)\ dx&=2 \end{aligned}\)
\(\begin{aligned} \displaystyle\intop_0^1f(x)\ dx+\displaystyle\intop_1^3f(x)\ dx&=\displaystyle\intop_0^3f(x)\ dx\\ 2+\displaystyle\intop_1^3f(x)\ dx&=5\\ \displaystyle\intop_1^3f(x)\ dx&=\boxed{\boxed{3}} \end{aligned}\)
Jadi, \({\displaystyle\intop_1^3f(x)\ dx=3}\).
JAWAB: D
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- 2
- 3
- 4
- 5
- 6
ALTERNATIF PENYELESAIAN
Misal u = x2 + 1
\(\begin{aligned}
du&=2x\ dx\\
x\ dx&=\dfrac12du
\end{aligned}\)
x = 0 ⟶ u = 02 + 1 = 1
x = 1 ⟶ u = 12 + 1 = 2
\(\begin{aligned}
\displaystyle\intop^2_0x\ f\left(x^2+1\right) dx&=\displaystyle\intop^2_1\ f\left(u\right) du\\
&=\boxed{\boxed{4}}
\end{aligned}\)
Jadi, \(\displaystyle\intop^1_0x\ f\left(x^2+1\right) dx=4\).
JAWAB: C
JAWAB: C
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