Exercise Zone : Integral Tentu [2]
Table of Contents
Tipe:
No.
Jika nilai \({\displaystyle\intop_1^2f(x)\ dx=10}\), maka nilai \({\displaystyle\intop_0^1x\cdot f\left(x^2+1\right)\ dx}\) adalah- 3
- 4
- 5
- 6
- 7
ALTERNATIF PENYELESAIAN
Misal u = x2 + 1
\(\begin{aligned} du&=2x\ dx\\ \dfrac12du&=x\ dx \end{aligned}\)
x = 0 ⟶ u = 02 + 1 = 1
x = 1 ⟶ u = 12 + 1 = 2
\(\begin{aligned} \displaystyle\intop_0^1x\cdot f\left(x^2+1\right)\ dx&=\dfrac12\displaystyle\intop_1^2f(u)\ du\\ &=\dfrac12(10)\\ &=\boxed{\boxed{5}} \end{aligned}\)
\(\begin{aligned} du&=2x\ dx\\ \dfrac12du&=x\ dx \end{aligned}\)
x = 0 ⟶ u = 02 + 1 = 1
x = 1 ⟶ u = 12 + 1 = 2
\(\begin{aligned} \displaystyle\intop_0^1x\cdot f\left(x^2+1\right)\ dx&=\dfrac12\displaystyle\intop_1^2f(u)\ du\\ &=\dfrac12(10)\\ &=\boxed{\boxed{5}} \end{aligned}\)
Jadi, \(\displaystyle\intop_0^1x\cdot f\left(x^2+1\right)\ dx=5\).
JAWAB: C
JAWAB: C
No.
Hasil dari \({\displaystyle\intop_2^42x^3+6x^2-2x-5\ dx=}\) ....ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\displaystyle\intop_2^42x^3+6x^2-2x-5\ dx&=\left[\dfrac24x^4+\dfrac63x^3-\dfrac22x^2-5x\right]_2^4\\
&=\left[\dfrac12x^4+2x^3-x^2-5x\right]_2^4\\
&=\left(\dfrac12(4)^4+2(4)^3-(4)^2-5(4)\right)-\left(\dfrac12(2)^4+2(2)^3-(2)^2-5(2)\right)\\
&=\left(8+128-16-20\right)-\left(8+16-4-10\right)\\
&=100-20\\
&=\boxed{\boxed{80}}
\end{aligned}\)
Jadi, \(\displaystyle\intop_2^42x^3+6x^2-2x-5\ dx=80\).
No.
Nilai \({\displaystyle\intop_1^4\left(3\sqrt{x}-2\right)\ dx=}\) ....- 2
- 4
- 6
- 7
- 8
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\displaystyle\intop_1^4\left(3\sqrt{x}-2\right)\ dx&=\displaystyle\intop_1^4\left(3x^{\frac12}-2\right)\ dx\\
&=\left.3\cdot\dfrac23x^{\frac32}-2x\right|_1^4\\
&=\left.2x\sqrt{x}-2x\right|_1^4\\
&=\left(2(4)\sqrt4-2(4)\right)-\left(2(1)\sqrt1-2(1)\right)\\
&=\left(8(2)-8\right)-\left(2(1)-2\right)\\
&=\left(16-8\right)-\left(2-2\right)\\
&=8-0\\
&=\boxed{\boxed{8}}
\end{aligned}\)
Jadi, \(\displaystyle\intop_1^4\left(3\sqrt{x}-2\right)\ dx=8\).
JAWAB: E
JAWAB: E
No.
Nilai \({\displaystyle\intop_1^4\dfrac2{x\sqrt{x}}\ dx=}\) ....- −12
- −4
- −3
- 2
- \(\dfrac32\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\displaystyle\intop_1^4\dfrac2{x\sqrt{x}}\ dx&=\displaystyle\intop_1^4\dfrac2{x\cdot x^{\frac12}}\ dx\\
&=\displaystyle\intop_1^4\dfrac2{x^{\frac32}}\ dx\\
&=\displaystyle\intop_1^42x^{-\frac32}\ dx\\
&=\left.2\left(-\dfrac21x^{-\frac12}\right)\right|_1^4\\
&=\left.-\dfrac4{x^{\frac12}}\right|_1^4\\
&=\left.-\dfrac4{\sqrt{x}}\right|_1^4\\
&=\left(-\dfrac4{\sqrt4}\right)-\left(-\dfrac4{\sqrt1}\right)\\
&=-\dfrac42+\dfrac41\\
&=-2+4\\
&=\boxed{\boxed{2}}
\end{aligned}\)
Jadi, \(\displaystyle\intop_1^4\dfrac2{x\sqrt{x}}\ dx=2\).
JAWAB: D
JAWAB: D
No.
Hasil \({\displaystyle\intop_1^3\left(x^2+\dfrac16\right)\ dx=}\) ....- \(9\dfrac13\)
- 9
- 8
- \(\dfrac{10}3\)
- 3
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\displaystyle\intop_1^3\left(x^2+\dfrac16\right)\ dx&=\left.\dfrac13x^3+\dfrac16x\right|_1^3\\
&=\left(\dfrac13(3)^3+\dfrac16(3)\right)-\left(\dfrac13(1)^3+\dfrac16(1)\right)\\
&=\left(9+\dfrac12\right)-\left(\dfrac13+\dfrac16\right)\\
&=9+\dfrac12-\dfrac36\\
&=9+\dfrac12-\dfrac12\\
&=\boxed{\boxed{9}}
\end{aligned}\)
Jadi, \(\displaystyle\intop_1^3\left(x^2+\dfrac16\right)\ dx=9\).
JAWAB: B
JAWAB: B
No.
Hasil \({\displaystyle\intop_0^2x^2\left(x+2\right)\ dx=}\) ....- 6
- \(6\dfrac13\)
- \(6\dfrac23\)
- \(9\dfrac13\)
- 20
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\displaystyle\intop_0^2x^2\left(x+2\right)\ dx&=\displaystyle\intop_0^2\left(x^3+2x^2\right)\ dx\\
&=\left.\dfrac14x^4+\dfrac23x^3\right|_0^2\\
&=\left(\dfrac14(2)^4+\dfrac23(2)^3\right)-\left(\dfrac14(0)^4+\dfrac23(0)^3\right)\\
&=\left(4+\dfrac{16}3\right)-0\\
&=4+5\dfrac13\\
&=\boxed{\boxed{9\dfrac13}}
\end{aligned}\)
Jadi, \(\displaystyle\intop_0^2x^2\left(x+2\right)\ dx=9\dfrac13\).
JAWAB: D
JAWAB: D
No.
Hasil dari \(\displaystyle\intop_{-1}^03x\sqrt[7]{1+x}\ dx=\)- \(-\dfrac{120}{147}\)
- \(-\dfrac{157}{120}\)
- \(-\dfrac{147}{120}\)
- \(\dfrac{147}{120}\)
- \(\dfrac{120}{147}\)
ALTERNATIF PENYELESAIAN
CARA 1: SUBSTITUSI
Misal u = 1 + x ⟶ x = u − 1du = dx
\(\begin{aligned} \displaystyle\intop_{-1}^03x\sqrt[7]{1+x}\ dx&=\displaystyle\intop_{-1}^03(u-1)\sqrt[7]{u}\ du\\ &=\displaystyle\intop_{-1}^03(u-1)u^{\frac17}\ du\\ &=\displaystyle\intop_{-1}^0\left(3u^{\frac87}-3u^{\frac17}\right)\ du\\ &=\left[3\cdot\dfrac7{15}u^{\frac{15}7}-3\cdot\dfrac78u^{\frac87}\right]_{-1}^0\\ &=\left[\dfrac75(1+x)^{\frac{15}7}-\dfrac{21}8(1+x)^{\frac87}\right]_{-1}^0\\ &=\left[\dfrac75(1+0)^{\frac{15}7}-\dfrac{21}8(1+0)^{\frac87}\right]-\left[\dfrac75(1+(-1))^{\frac{15}7}-\dfrac{21}8(1+(-1))^{\frac87}\right]\\ &=\left[\dfrac75(1)^{\frac{15}7}-\dfrac{21}8(1)^{\frac87}\right]-\left[\dfrac75(0)^{\frac{15}7}-\dfrac{21}8(0)^{\frac87}\right]\\ &=\left[\dfrac75(1)-\dfrac{21}8(1)\right]-\left[\dfrac75(0)-\dfrac{21}8(0)\right]\\ &=\left[\dfrac75-\dfrac{21}8\right]-\left[0-0\right]\\ &=-\dfrac{49}{40}\\ &=\boxed{\boxed{-\dfrac{147}{120}}} \end{aligned}\)
CARA 2: PARSIAL
| u | dv |
|---|---|
| 3x | \(\sqrt[7]{1+x}=(1+x)^{\frac17}\) |
| 3 | \(\dfrac78(1+x)^{\frac87}\) |
| 0 | \(\dfrac78\cdot\dfrac7{15}(1+x)^{\frac{15}7}=\dfrac{49}{120}(1+x)^{\frac{15}7}\) |
\(\begin{aligned} \displaystyle\intop_{-1}^03x\sqrt[7]{1+x}\ dx&=\left[3x\cdot\dfrac78(1+x)^{\frac87}-3\cdot\dfrac{49}{120}(1+x)^{\frac{15}7}\right]_{-1}^0\\ &=\left[\dfrac{21}8x(1+x)^{\frac87}-\cdot\dfrac{147}{120}(1+x)^{\frac{15}7}\right]_{-1}^0\\ &=\left[\dfrac{21}8(0)(1+0)^{\frac87}-\cdot\dfrac{147}{120}(1+0)^{\frac{15}7}\right]-\left[\dfrac{21}8(-1)(1+(-1))^{\frac87}-\cdot\dfrac{147}{120}(1+(-1))^{\frac{15}7}\right]\\ &=\left[0-\cdot\dfrac{147}{120}(1)^{\frac{15}7}\right]-\left[-\dfrac{21}8(0)^{\frac87}-\cdot\dfrac{147}{120}(0)^{\frac{15}7}\right]\\ &=\left[-\cdot\dfrac{147}{120}(1)\right]-0\\ &=\boxed{\boxed{-\dfrac{147}{120}}} \end{aligned}\)
Jadi, \(\displaystyle\intop_{-1}^03x\sqrt[7]{1+x}\ dx=-\dfrac{147}{120}\).
JAWAB: C
JAWAB: C
No.
Hasil dari \(\displaystyle\intop_0^1\dfrac{10(18x+6)\ dx}{\sqrt{9x^2+6x+1}}=\)- 40
- 50
- 60
- 70
- 80
ALTERNATIF PENYELESAIAN
Misal u = 9x2 + 6x + 1
du = (18x + 6) dx
x = 0 ⟶ u = 9(0)2 + 6(0) + 1 = 1
x = 1 ⟶ u = 9(1)2 + 6(1) + 1 = 16
\(\begin{aligned} \displaystyle\intop_0^1\dfrac{10(18x+6)\ dx}{\sqrt{9x^2+6x+1}}&=\displaystyle\intop_1^{16}\dfrac{10\ du}{\sqrt{u}}\\ &=\displaystyle\intop_1^{16}\dfrac{10\ du}{u^{\frac12}}\\ &=\displaystyle\intop_1^{16}10u^{-\frac12}\ du\\ &=\left[10\cdot2u^{\frac12}\right]_1^{16}\\ &=\left[20\sqrt{u}\right]_1^{16}\\ &=20\sqrt{16}-20\sqrt{1}\\ &=20(4)-20(1)\\ &=80-20\\ &=\boxed{\boxed{60}} \end{aligned}\)
du = (18x + 6) dx
x = 0 ⟶ u = 9(0)2 + 6(0) + 1 = 1
x = 1 ⟶ u = 9(1)2 + 6(1) + 1 = 16
\(\begin{aligned} \displaystyle\intop_0^1\dfrac{10(18x+6)\ dx}{\sqrt{9x^2+6x+1}}&=\displaystyle\intop_1^{16}\dfrac{10\ du}{\sqrt{u}}\\ &=\displaystyle\intop_1^{16}\dfrac{10\ du}{u^{\frac12}}\\ &=\displaystyle\intop_1^{16}10u^{-\frac12}\ du\\ &=\left[10\cdot2u^{\frac12}\right]_1^{16}\\ &=\left[20\sqrt{u}\right]_1^{16}\\ &=20\sqrt{16}-20\sqrt{1}\\ &=20(4)-20(1)\\ &=80-20\\ &=\boxed{\boxed{60}} \end{aligned}\)
Jadi, \(\displaystyle\intop_0^1\dfrac{10(18x+6)\ dx}{\sqrt{9x^2+6x+1}}=60\).
JAWAB: C
JAWAB: C
No.
\(\displaystyle\intop_0^3x\sqrt{1+x}\ dx=\) ....ALTERNATIF PENYELESAIAN
Misal
u = 1 + x ⟶ x = u − 1
du = dx
x = 0 ⟶ u = 1 + 0 = 1
x = 3 ⟶ u = 1 + 3 = 4
\(\begin{aligned} \displaystyle\intop_0^3x\sqrt{1+x}\ dx&=\displaystyle\intop_1^4(u-1)\sqrt{u}\ du\\ &=\displaystyle\intop_1^4(u-1)u^{\frac12}\ du\\ &=\displaystyle\intop_1^4\left(u^{\frac32}-u^{\frac12}\right)\ du\\ &=\left[\dfrac25u^{\frac52}-\dfrac23u^{\frac32}\right]_1^4\\ &=\left(\dfrac25(4)^{\frac52}-\dfrac23(4)^{\frac32}\right)-\left(\dfrac25(1)^{\frac52}-\dfrac23(1)^{\frac32}\right)\\ &=\left(\dfrac25(32)-\dfrac23(8)\right)-\left(\dfrac25(1)-\dfrac23(1)\right)\\ &=\left(\dfrac{64}5-\dfrac{16}3\right)-\left(\dfrac25-\dfrac23\right)\\ &=\left(\dfrac{192}{15}-\dfrac{80}{15}\right)-\left(\dfrac6{15}-\dfrac{10}{15}\right)\\ &=\left(\dfrac{112}{15}\right)-\left(-\dfrac4{15}\right)\\ &=\dfrac{112}{15}+\dfrac4{15}\\ &=\boxed{\boxed{\dfrac{116}{15}}} \end{aligned}\)
u = 1 + x ⟶ x = u − 1
du = dx
x = 0 ⟶ u = 1 + 0 = 1
x = 3 ⟶ u = 1 + 3 = 4
\(\begin{aligned} \displaystyle\intop_0^3x\sqrt{1+x}\ dx&=\displaystyle\intop_1^4(u-1)\sqrt{u}\ du\\ &=\displaystyle\intop_1^4(u-1)u^{\frac12}\ du\\ &=\displaystyle\intop_1^4\left(u^{\frac32}-u^{\frac12}\right)\ du\\ &=\left[\dfrac25u^{\frac52}-\dfrac23u^{\frac32}\right]_1^4\\ &=\left(\dfrac25(4)^{\frac52}-\dfrac23(4)^{\frac32}\right)-\left(\dfrac25(1)^{\frac52}-\dfrac23(1)^{\frac32}\right)\\ &=\left(\dfrac25(32)-\dfrac23(8)\right)-\left(\dfrac25(1)-\dfrac23(1)\right)\\ &=\left(\dfrac{64}5-\dfrac{16}3\right)-\left(\dfrac25-\dfrac23\right)\\ &=\left(\dfrac{192}{15}-\dfrac{80}{15}\right)-\left(\dfrac6{15}-\dfrac{10}{15}\right)\\ &=\left(\dfrac{112}{15}\right)-\left(-\dfrac4{15}\right)\\ &=\dfrac{112}{15}+\dfrac4{15}\\ &=\boxed{\boxed{\dfrac{116}{15}}} \end{aligned}\)
Jadi, \(\displaystyle\intop_0^3x\sqrt{1+x}\ dx=\dfrac{116}{15}\).
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