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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
Misal u=x^2+1
\begin{aligned}
du&=2x\ dx\\
\dfrac12du&=x\ dx
\end{aligned}
x=0\rightarrow u=0^2+1=1
x=1\rightarrow u=1^2+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}
No.
Hasil dari
{\displaystyle\intop_2^42x^3+6x^2-2x-5\ dx=} ....
\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}
No.
Nilai
{\displaystyle\intop_1^4\left(3\sqrt{x}-2\right)\ dx=} ....
\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}
No.
Nilai
{\displaystyle\intop_1^4\dfrac2{x\sqrt{x}}\ dx=} ....
\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}
No.
Hasil
{\displaystyle\intop_1^3\left(x^2+\dfrac16\right)\ dx=} ....
\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}
No.
Hasil
{\displaystyle\intop_0^2x^2\left(x+2\right)\ dx=} ....
\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}
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}
CARA 1: SUBSTITUSI
Misal
u=1+x\rightarrow x=u-1
du=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}No.
Hasil dari
\displaystyle\intop_0^1\dfrac{10(18x+6)\ dx}{\sqrt{9x^2+6x+1}}=
Misal u=9x^2+6x+1
du=(18x+6)\ dx
x=0\rightarrow u=9(0)^2+6(0)+1=1
x=1\rightarrow 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}
No.
\displaystyle\intop_0^3x\sqrt{1+x}\ dx= ....
Misal
u=1+x\to x=u-1
du=dx
x=0\to u=1+0=1
x=3\to 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}
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