Berikut ini adalah kumpulan soal mengenai integral tentu tipe standar. Jika ingin bertanya soal, silahkan gabung ke grup
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No. 1
Jika
\displaystyle\intop_0^3f(x)\ dx=-5 dan
\displaystyle\intop_0^7f(x)\ dx=8 maka
\displaystyle\intop_3^7f(x)\ dx= ....
\(\begin{aligned}
\displaystyle\intop_0^7f(x)\ dx&=\displaystyle\intop_0^3f(x)\ dx+\displaystyle\intop_3^7f(x)\ dx\\
8&=-5+\displaystyle\intop_3^7f(x)\ dx\\
\displaystyle\intop_3^7f(x)\ dx&=8+5\\
&=\boxed{\boxed{13}}
\end{aligned}\)
No. 2
Jika
\displaystyle\intop_0^4f(x)\ dx=5 dan
\displaystyle\intop_4^{10}f(x)\ dx=10 maka
2\displaystyle\intop_0^{10}f(x)\ dx= ....
\(\begin{aligned}
\displaystyle\intop_0^{10}f(x)\ dx&=\displaystyle\intop_0^4f(x)\ dx+\displaystyle\intop_4^{10}f(x)\ dx\\
&=5+10\\
&=15\\
2\displaystyle\intop_0^{10}f(x)\ dx&=2(15)\\
&=\boxed{\boxed{30}}
\end{aligned}\)
No. 3
Jika
\displaystyle\intop_0^22f(x)\ dx=4 dan
\displaystyle\intop_4^23f(x)\ dx=12 maka nilai
\displaystyle\intop_0^4f(x)\ dx=
\(\begin{aligned}
\displaystyle\intop_0^22f(x)\ dx&=4\\
2\displaystyle\intop_0^2f(x)\ dx&=4\\
\displaystyle\intop_0^2f(x)\ dx&=2
\end{aligned}\)
\(\begin{aligned}
\displaystyle\intop_4^23f(x)\ dx&=12\\
3\displaystyle\intop_4^2f(x)\ dx&=12\\
\displaystyle\intop_4^2f(x)\ dx&=4\\
\displaystyle\intop_2^4f(x)\ dx&=-4
\end{aligned}\)
\(\begin{aligned}
\displaystyle\intop_0^4f(x)\ dx&=\displaystyle\intop_0^2f(x)\ dx+\displaystyle\intop_2^4f(x)\ dx\\
&=2+(-4)\\
&=\boxed{\boxed{-2}}
\end{aligned}\)
No. 4
Nilai dari
\displaystyle\intop_1^4\left(2\sqrt{x}-\dfrac3{\sqrt{x}}-1\right)dx adalah
\(\begin{aligned}
\displaystyle\intop_1^4\left(2\sqrt{x}-\dfrac3{\sqrt{x}}-1\right)dx&=\displaystyle\intop_1^4\left(2x^{\frac12}-\dfrac3{x^{\frac12}}-1\right)dx\\
&=\displaystyle\intop_1^4\left(2x^{\frac12}-\dfrac3x^{-\frac12}-1\right)dx\\
&=\left[2\cdot\dfrac23x^{\frac32}-3\cdot2x^{\frac12}-x\right]_1^4\\
&=\left[\dfrac43x\sqrt{x}-6\sqrt{x}-x\right]_1^4\\
&=\left[\dfrac43(4)\sqrt4-6\sqrt4-4\right]-\left[\dfrac43(1)\sqrt1-6\sqrt1-1\right]\\
&=\left[\dfrac{16}3(2)-6(2)-4\right]-\left[\dfrac43(1)-6(1)-1\right]\\
&=\left[\dfrac{32}3-12-4\right]-\left[\dfrac43-6-1\right]\\
&=-\dfrac{16}3-\left(-\dfrac{17}3\right)\\
&=-\dfrac{16}3+\dfrac{17}3\\
&=\boxed{\boxed{\dfrac13}}
\end{aligned}\)
No. 5
Nilai dari
\displaystyle\intop_{-1}^1\left(x^2+2x-3\right)\ dx-\displaystyle\intop_{-1}^1\left(x^2-3x+4\right)\ dx=
\(\begin{aligned}
\displaystyle\intop_{-1}^1\left(x^2+2x-3\right)\ dx-\displaystyle\intop_{-1}^1\left(x^2-3x+4\right)\ dx&=\displaystyle\intop_{-1}^1\left(x^2+2x-3-\left(x^2-3x+4\right)\right)\ dx\\
&=\displaystyle\intop_{-1}^1\left(x^2+2x-3-x^2+3x-4\right)\ dx\\
&=\displaystyle\intop_{-1}^1\left(5x-7\right)\ dx\\
&=\left[\dfrac52x^2-7x\right]_{-1}^1\\[10pt]
&=\left[\dfrac52(1)^2-7(1)\right]-\left[\dfrac52(-1)^2-7(-1)\right]\\[10pt]
&=\left[\dfrac52-7\right]-\left[\dfrac52+7\right]\\[10pt]
&=\dfrac52-7-\dfrac52-7\\
&=\boxed{\boxed{-14}}
\end{aligned}\)
No. 6
\displaystyle\intop_{-1}^2x\left(x^2+1\right)\ dx=
Misal u=x^2+1
\(\begin{aligned}
du&=2x\ dx\\
x\ dx&=\dfrac12du
\end{aligned}\)
x=-1\rightarrow u=(-1)^2+1=2,
x=2\rightarrow u=2^2+1=5,
\(\begin{aligned}
\displaystyle\intop_{-1}^2x\left(x^2+1\right)\ dx&=\dfrac12\displaystyle\intop_2^5u\ du\\
&=\dfrac12\left[\dfrac12u^2\right]_2^5\\
&=\dfrac12\left[\left(\dfrac12(5)^2\right)-\left(\dfrac12(2)^2\right)\right]\\
&=\dfrac12\left[\dfrac{25}2-\dfrac42\right]\\
&=\dfrac12\left[\dfrac{21}2\right]\\
&=\boxed{\boxed{\dfrac{21}4}}
\end{aligned}\)
No. 7
Jika
\displaystyle\intop_1^p3x\left(x+\dfrac23\right)\ dx=78 maka nilai
3p=
\(\eqalign{
\displaystyle\intop_1^p3x\left(x+\dfrac23\right)\ dx&=78\\
\displaystyle\intop_1^p\left(3x^2+2x\right)\ dx&=78\\
\left[x^3+x^2\right]_1^p&=78\\
\left[p^3+p^2\right]-\left[1^3+1^2\right]&=78\\
p^3+p^2-2&=78\\
p^3+p^2-80&=0\\
(p-4)\left(p^2+5p+20\right)&=0\\
p&=4\\
3p&=3(4)\\
&=\boxed{\boxed{12}}}\)
No. 8
Nilai dari
\displaystyle\intop_2^3\left(x^2+10x+25\right)\ dx adalah
\(\begin{aligned}
\displaystyle\intop_2^3\left(x^2+10x+25\right)\ dx&=\left[\dfrac13x^3+5x^2+25x\right]_2^3\\
&=\left[\dfrac13(3)^3+5(3)^2+25(3)\right]-\left[\dfrac13(2)^3+5(2)^2+25(2)\right]\\
&=\left[\dfrac13(27)+5(9)+75\right]-\left[\dfrac13(8)+5(4)+50\right]\\
&=\left[9+45+75\right]-\left[\dfrac83+20+50\right]\\
&=129-\dfrac83-70\\
&=59-2\dfrac23\\
&=\boxed{\boxed{56\dfrac13}}
\end{aligned}\)
No. 9
\displaystyle\intop_{\frac12}^1\left(\sin\pi x+\sqrt[3]{2x-1}\right)\ dx=
- \dfrac{3\pi+4}{4\pi}
- \dfrac{3\pi+8}{8\pi}
- \dfrac{3\pi-4}{4\pi}
- \dfrac{3\pi-8}{8\pi}
- \dfrac34+\pi
\(\begin{aligned}
\displaystyle\intop_{\frac12}^1\left(\sin\pi x+\sqrt[3]{2x-1}\right)\ dx&=\displaystyle\intop_{\frac12}^1\left(\sin\pi x+(2x-1)^{\frac13}\right)\ dx\\
&=\left[-\dfrac1{\pi}\cos\pi x+\dfrac12\cdot\dfrac34(2x-1)^{\frac43}\right]_{\frac12}^1\\
&=\left[-\dfrac1{\pi}\cos\pi x+\dfrac38(2x-1)\sqrt[3]{2x-1}\right]_{\frac12}^1\\
&=\left[-\dfrac1{\pi}\cos\pi(1)+\dfrac38(2(1)-1)\sqrt[3]{2(1)-1}\right]-\left[-\dfrac1{\pi}\cos\pi\left(\dfrac12\right)+\dfrac38\left(2\left(\dfrac12\right)-1\right)\sqrt[3]{2\left(\dfrac12\right)-1}\right]\\
&=\left[-\dfrac1{\pi}\cos\pi+\dfrac38(1)\sqrt[3]{1}\right]-\left[-\dfrac1{\pi}\cos\dfrac{\pi}2+\dfrac38\left(0\right)\sqrt[3]{0}\right]\\
&=\left[-\dfrac1{\pi}(-1)+\dfrac38\right]-\left[-\dfrac1{\pi}(0)+0\right]\\
&=\left[\dfrac1{\pi}+\dfrac38\right]-0\\
&=\dfrac{8+3\pi}{8\pi}\\
&=\boxed{\boxed{\dfrac{3\pi+8}{8\pi}}}
\end{aligned}\)
No. 10
\displaystyle\intop_2^6\sqrt{x}\ dx= ....
\(\begin{aligned}
\displaystyle\intop_9^16\sqrt{x}\ dx&=\displaystyle\intop_9^16x^{\frac12}\ dx\\
&=\left.\dfrac23x^{\frac32}\right|_9^{16}\\
&=\left.\dfrac23x\sqrt{x}\right|_9^{16}\\
&=\dfrac23(16)\sqrt{16}-\dfrac23(9)\sqrt{9}\\[8pt]
&=\dfrac{32}3(4)-6(3)\\[8pt]
&=\dfrac{128}3-18\\
&=\boxed{\boxed{\dfrac{74}3}}
\end{aligned}\)
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