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Exercise Zone : Integral Tentu [2]

Written By Anonymous Monday, March 9, 2020 Add Comment Edit
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 ∫12f(x) dx=10\displaystyle\intop_1^2f(x)\ dx=10, maka nilai ∫01x⋅f(x2+1) dx\displaystyle\intop_0^1x\cdot f\left(x^2+1\right)\ dx adalah
  1. 33
  2. 44
  3. 55
  1. 66
  2. 77
Misal u=x2+1u=x^2+1
du=2x dx12du=x dxdu=2x dx12du=x dx

x=0→u=02+1=1x=0\rightarrow u=0^2+1=1
x=1→u=12+1=2x=1\rightarrow u=1^2+1=2

∫01x⋅f(x2+1) dx=12∫12f(u) du=12(10)=5∫01x⋅f(x2+1) dx=12∫12f(u) du=12(10)=5
Jadi, ∫01x⋅f(x2+1) dx=5\displaystyle\intop_0^1x\cdot f\left(x^2+1\right)\ dx=5.
JAWAB: C

No. 12

Hasil dari ∫242x3+6x2−2x−5 dx={\displaystyle\intop_2^42x^3+6x^2-2x-5\ dx=} ....
SUMBER
∫242x3+6x2−2x−5 dx=[24x4+63x3−22x2−5x]24=[12x4+2x3−x2−5x]24=(12(4)4+2(4)3−(4)2−5(4))−(12(2)4+2(2)3−(2)2−5(2))=(8+128−16−20)−(8+16−4−10)=100−20=80∫242x3+6x2−2x−5 dx=[24x4+63x3−22x2−5x]24=[12x4+2x3−x2−5x]24=(12(4)4+2(4)3−(4)2−5(4))−(12(2)4+2(2)3−(2)2−5(2))=(8+128−16−20)−(8+16−4−10)=100−20=80
Jadi, ∫242x3+6x2−2x−5 dx=80\displaystyle\intop_2^42x^3+6x^2-2x-5\ dx=80.

No. 13

Nilai ∫14(3x−2) dx={\displaystyle\intop_1^4\left(3\sqrt{x}-2\right)\ dx=} ....
  1. 22
  2. 44
  3. 66
  1. 77
  2. 88
GRUP TELEGRAM
∫14(3x−2) dx=∫14(3x12−2) dx=3⋅23x32−2x∣14=2xx−2x∣14=(2(4)4−2(4))−(2(1)1−2(1))=(8(2)−8)−(2(1)−2)=(16−8)−(2−2)=8−0=8∫14(3x−2) dx=∫14(3x12−2) dx=3⋅23x32−2x|14=2xx−2x|14=(2(4)4−2(4))−(2(1)1−2(1))=(8(2)−8)−(2(1)−2)=(16−8)−(2−2)=8−0=8
Jadi, ∫14(3x−2) dx=8\displaystyle\intop_1^4\left(3\sqrt{x}-2\right)\ dx=8.
JAWAB: E

No. 14

Nilai ∫142xx dx={\displaystyle\intop_1^4\dfrac2{x\sqrt{x}}\ dx=} ....
  1. −12-12
  2. −4-4
  3. −3-3
  1. 22
  2. 32\dfrac32
GRUP TELEGRAM
∫142xx dx=∫142x⋅x12 dx=∫142x32 dx=∫142x−32 dx=2(−21x−12)∣14=−4x12∣14=−4x∣14=(−44)−(−41)=−42+41=−2+4=2∫142xx dx=∫142x⋅x12 dx=∫142x32 dx=∫142x−32 dx=2(−21x−12)|14=−4x12|14=−4x|14=(−44)−(−41)=−42+41=−2+4=2
Jadi, ∫142xx dx=2\displaystyle\intop_1^4\dfrac2{x\sqrt{x}}\ dx=2.
JAWAB: D

No. 15

Hasil ∫13(x2+16) dx={\displaystyle\intop_1^3\left(x^2+\dfrac16\right)\ dx=} ....

Related:

  1. 9139\dfrac13
  2. 99
  3. 88
  1. 103\dfrac{10}3
  2. 33
GRUP TELEGRAM
∫13(x2+16) dx=13x3+16x∣13=(13(3)3+16(3))−(13(1)3+16(1))=(9+12)−(13+16)=9+12−36=9+12−12=9∫13(x2+16) dx=13x3+16x|13=(13(3)3+16(3))−(13(1)3+16(1))=(9+12)−(13+16)=9+12−36=9+12−12=9
Jadi, ∫13(x2+16) dx=9\displaystyle\intop_1^3\left(x^2+\dfrac16\right)\ dx=9.
JAWAB: B

No. 16

Hasil ∫02x2(x+2) dx={\displaystyle\intop_0^2x^2\left(x+2\right)\ dx=} ....
  1. 66
  2. 6136\dfrac13
  3. 6236\dfrac23
  1. 9139\dfrac13
  2. 2020
GRUP TELEGRAM
∫02x2(x+2) dx=∫02(x3+2x2) dx=14x4+23x3∣02=(14(2)4+23(2)3)−(14(0)4+23(0)3)=(4+163)−0=4+513=913∫02x2(x+2) dx=∫02(x3+2x2) dx=14x4+23x3|02=(14(2)4+23(2)3)−(14(0)4+23(0)3)=(4+163)−0=4+513=913
Jadi, ∫02x2(x+2) dx=913\displaystyle\intop_0^2x^2\left(x+2\right)\ dx=9\dfrac13.
JAWAB: D


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