Berikut ini adalah kumpulan soal mengenai integral tentu tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
No. 11
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. 12
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. 13
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. 14
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. 15
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. 16
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}
0 Response to "Exercise Zone : Integral Tentu [2]"
Post a Comment