Exercise Zone : Integral Tentu (Definite Integral)

Berikut ini adalah kumpulan soal mengenai integral tentu tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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No. 1

Jika $\displaystyle\intop_0^3f(x)\ dx=-5$ dan $\displaystyle\intop_0^7f(x)\ dx=8$ maka $\displaystyle\intop_3^7f(x)\ dx=$ ....

1. $3$
2. $13$
   
3. $-8$
   

4. $5$
5. $12$
   

$∫07f(x) dx=∫03f(x) dx+∫37f(x) dx8=−5+∫37f(x) dx∫37f(x) dx=8+5=13$

Jadi, $\displaystyle\intop_3^7f(x)\ dx=13$.
JAWAB: B

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No. 2

Jika $\displaystyle\intop_0^4f(x)\ dx=5$ dan $\displaystyle\intop_4^{10}f(x)\ dx=10$ maka $2\displaystyle\intop_0^{10}f(x)\ dx=$ ....

1. $30$
2. $32$
   
3. $34$
   

4. $36$
5. $38$
   

$∫010f(x) dx=∫04f(x) dx+∫410f(x) dx=5+10=152∫010f(x) dx=2(15)=30$

Jadi, $2\displaystyle\intop_0^{10}f(x)\ dx=30$.
JAWAB: A

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No. 3

Jika $\displaystyle\intop_0^22f(x)\ dx=4$ dan $\displaystyle\intop_4^23f(x)\ dx=12$ maka nilai $\displaystyle\intop_0^4f(x)\ dx=$

1. $-2$
2. $-3$
   
3. $-4$
   

4. $-5$
5. $-6$
   

$∫022f(x) dx=42∫02f(x) dx=4∫02f(x) dx=2$

$∫423f(x) dx=123∫42f(x) dx=12∫42f(x) dx=4∫24f(x) dx=−4$

$∫04f(x) dx=∫02f(x) dx+∫24f(x) dx=2+(−4)=−2$

Jadi, $\displaystyle\intop_0^4f(x)\ dx=-2$.
JAWAB: A

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No. 4

Nilai dari $\displaystyle\intop_1^4\left(2\sqrt{x}-\dfrac3{\sqrt{x}}-1\right)dx$ adalah

1. $\dfrac13$
2. $\dfrac23$
   
3. $1$
   

4. $\dfrac12$
5. $3$
   

$∫14(2x−3x−1)dx=∫14(2x12−3x12−1)dx=∫14(2x12−3x−12−1)dx=[2⋅23x32−3⋅2x12−x]14=[43xx−6x−x]14=[43(4)4−64−4]−[43(1)1−61−1]=[163(2)−6(2)−4]−[43(1)−6(1)−1]=[323−12−4]−[43−6−1]=−163−(−173)=−163+173=13$

Jadi, $\displaystyle\intop_1^4\left(2\sqrt{x}-\dfrac3{\sqrt{x}}-1\right)dx=\dfrac13$.
JAWAB: A

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No. 5

Nilai dari $\displaystyle\intop_{-1}^1\left(x^2+2x-3\right)\ dx-\displaystyle\intop_{-1}^1\left(x^2-3x+4\right)\ dx=$

1. $0$
2. $7$
   
3. $14$
   

4. $-7$
5. $-14$
   

$∫−11(x2+2x−3) dx−∫−11(x2−3x+4) dx=∫−11(x2+2x−3−(x2−3x+4)) dx=∫−11(x2+2x−3−x2+3x−4) dx=∫−11(5x−7) dx=[52x2−7x]−11=[52(1)2−7(1)]−[52(−1)2−7(−1)]=[52−7]−[52+7]=52−7−52−7=−14$

Jadi, $\displaystyle\intop_{-1}^1\left(x^2+2x-3\right)\ dx-\displaystyle\intop_{-1}^1\left(x^2-3x+4\right)\ dx=-14$.
JAWAB: E

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No. 6

$\displaystyle\intop_{-1}^2x\left(x^2+1\right)\ dx=$

Misal $u=x^2+1$
$du=2x dxx dx=12du$

$x=-1\rightarrow u=(-1)^2+1=2$,
$x=2\rightarrow u=2^2+1=5$,

$∫−12x(x2+1) dx=12∫25u du=12[12u2]25=12[(12(5)2)−(12(2)2)]=12[252−42]=12[212]=214$

Jadi, $\displaystyle\intop_{-1}^2x\left(x^2+1\right)\ dx=\dfrac{21}4$.
JAWAB: C

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No. 7

Jika $\displaystyle\intop_1^p3x\left(x+\dfrac23\right)\ dx=78$ maka nilai $3p=$

1. $18$
2. $12$
   
3. $15$
   

4. $9$
5. $6$
   

$∫1p3x(x+23) dx=78∫1p(3x2+2x) dx=78[x3+x2]1p=78[p3+p2]−[13+12]=78p3+p2−2=78p3+p2−80=0(p−4)(p2+5p+20)=0p=43p=3(4)=12$

Jadi, $3p=12$.
JAWAB: B

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No. 8

Nilai dari $\displaystyle\intop_2^3\left(x^2+10x+25\right)\ dx$ adalah

1. $56\dfrac13$
2. $56\dfrac23$
3. $57\dfrac13$

4. $57\dfrac23$
5. $58\dfrac13$

$∫23(x2+10x+25) dx=[13x3+5x2+25x]23=[13(3)3+5(3)2+25(3)]−[13(2)3+5(2)2+25(2)]=[13(27)+5(9)+75]−[13(8)+5(4)+50]=[9+45+75]−[83+20+50]=129−83−70=59−223=5613$

Jadi, $\displaystyle\intop_2^3\left(x^2+10x+25\right)\ dx=56\dfrac13$.
JAWAB: A

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No. 9

$\displaystyle\intop_{\frac12}^1\left(\sin\pi x+\sqrt[3]{2x-1}\right)\ dx=$

1. $\dfrac{3\pi+4}{4\pi}$
2. $\dfrac{3\pi+8}{8\pi}$
3. $\dfrac{3\pi-4}{4\pi}$

4. $\dfrac{3\pi-8}{8\pi}$
5. $\dfrac34+\pi$

$∫121(sin⁡πx+2x−13) dx=∫121(sin⁡πx+(2x−1)13) dx=[−1πcos⁡πx+12⋅34(2x−1)43]121=[−1πcos⁡πx+38(2x−1)2x−13]121=[−1πcos⁡π(1)+38(2(1)−1)2(1)−13]−[−1πcos⁡π(12)+38(2(12)−1)2(12)−13]=[−1πcos⁡π+38(1)13]−[−1πcos⁡π2+38(0)03]=[−1π(−1)+38]−[−1π(0)+0]=[1π+38]−0=8+3π8π=3π+88π$

Jadi, $\displaystyle\intop_{\frac12}^1\left(\sin\pi x+\sqrt[3]{2x-1}\right)\ dx=\dfrac{3\pi+8}{8\pi}$.
JAWAB: B

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No. 10

$\displaystyle\intop_2^6\sqrt{x}\ dx=$ ....

$∫916x dx=∫916x12 dx=23x32|916=23xx|916=23(16)16−23(9)9=323(4)−6(3)=1283−18=743$

Jadi, $\displaystyle\intop_2^6\sqrt{x}\ dx=\dfrac{74}3$.

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