Editorial - Matlan - Latihan Integral Tentu

Problem 1

(∫_^)((x2-3)*d(x))=1/3*x3-3*x+C

(∫_-*1^2)((x2-3)*d(x))=(23)/3-3⋅2-((-1)3)/3+3⋅(-1)=-6

(B)

Problem 2

(-x3+2*x-1)2=x6+4*x2+1-4*x4+2*x3-4*x

(∫_-*1^1)((x6-4*x4+2*x3+4*x2-4*x+1)2*d(x))

(∣_-*1^1)(1/7*x7-4/5*x5+1/2*x4+4/3*x3-2*x2+x)

Calculatenya banyak, maka untuk disimplify, observe bahwa integralnya di 1 dan -1, artinya kalau cari selisih dari x dengan pangkat genap, sama aja 0, maka tersimplifikasi:

(∣_-*1^1)(1/7*x7-4/5*x5+4/3*x3+x)=1/7+1/7-4/5-4/5+4/3+4/3+1+1=352/105

(C)

Problem 3

5*x2-6√(,x)+2/(x2)=5*x2-6*x(1/2)+2*x(-2)

(∫_1^4)((5*x2-6*x(1/2)+2*x(-2))*d(x))=(∣_1^4)(5/3*x3-4*x(3/2)-2*x(-1))

5/3*(43-13)-4*(4(3/2)-1(3/2))-2*(4(-1)-1(-1))=157/2=78+1/2

(D)

Problem 4

5-x=u

d(u)/d(x)=-1

d(x)=-d(u)

(∫_1^4)(-ƒ(u)*d(u))=-(∫_4^1)(ƒ(u)*d(u))=-1⋅-(∫_1^4)(ƒ(u)*d(u))=6

(A)

Problem 5

(∫_1^a)((2*x+3)*d(x))=(∣_1^a)(x2+3*x)

a2+3*a-1-3=6

a2+3*a-10=0

(a+5)*(a-2)=0

∴a=-5,2

(B) ?

Problem 6

(∫_0^4)((3*x2+p*x-3)*d(x))=(∣_0^4)(x3+p/2*x2-3*x)

43-03+p/2*(42-02)-3*(4-0)=52+8*p=68

8*p=68-52=16

∴p=2

(C)

Problem 7

(∫_9^16)((2+x)/(2*x(1/2))*d(x))=(∫_9^16)(x(-1/2)+1/2*x(1/2))=(∣_9^16)(2*x(1/2)+1/3*x(3/2))

2⋅(16(1/2)-9(1/2))+1/3⋅(16(3/2)-9(3/2))=43/3

(E)

Problem 8

2 saja, karena sifat perkalian distributif dan komutatif. okay i'll prove it.

(B)

Subproblem 8.1

LHS:

(∫_a^b)(ƒ(x)*g(x)*d(x))=ƒ(b)*g(b)-ƒ(a)*g(a)

RHS:

((∫_a^b)(ƒ(x)*d(x)))*((∫_a^b)(g(x)*d(x)))=(ƒ(b)-ƒ(a))*(g(b)-g(a))

Berbeda.

Subproblem 8.2

LHS:

(∫_a^b)((ƒ(x)+g(x))*d(x))=ƒ(b)-ƒ(a)+g(b)-g(a)

RHS:

(∫_a^b)(ƒ(x)*d(x))+(∫_a^b)(g(x)*d(x))=ƒ(b)-ƒ(a)+g(b)-g(a)

Subproblem 8.3

Pasti ada yang berbeda:

√(,ƒ(b))-√(,ƒ(a))≠√(,ƒ(b)-ƒ(a))

Problem 9

1 pasti benar karena g(a) adalah konstanta.

2 anggap saja benar idk

3 benar karena g(a) konstanta

4 anggap saja benar idk

(E)

Problem 10

(∫_^)((a*x+b)*d(x))=(a*x2)/2+b*x+C

ƒ(1)-ƒ(0)=a/2*(12-02)+b*(1-0)=a/2+b=1

ƒ(2)-ƒ(1)=a/2*(22-12)+b*(2-1)=3/2*a+b=5

∴a=4

b=1-2=-1

∴a+b=4-1=3

(C)

Problem 11

(∫_1^3)(ƒ(x)*d(x))=F(3)-F(1)=3

3*(∫_-*1^3)(g(x)*d(x))=3*(G(3)-G(1))=-6

G(3)-G(1)=-2

(∫_-*1^3)(2*ƒ(x)-g(x))=2*(F(3)-F(1))-(G(3)-G(1))=2⋅3+2=8

(E)

Problem 12

(∫_-*5^2)(ƒ(x)*d(x))=F(2)-F(-5)=-17

(∫_5^2)(ƒ(x)*d(x))=F(2)-F(5)=-4

∴F(5)-F(-5)=-13

(B)

Problem 13

Karena fungsi ganjil, (∫_-*a^a)(ƒ(x)*d(x))=0

(∫_-*2^1)(ƒ(x)*d(x))=(∫_-*2^-*1)(ƒ(x)*d(x))+(∫_-*1^1)(ƒ(x)*d(x))=4

(∫_-*2^-1)(ƒ(x)*d(x))=4

(D)

Problem 14

F(a)-F(b)=5

F(a)-F(c)=0

F(b)-F(c)=-5

(D)

Problem 15

ƒ(x) genap.

(∫_-*3^3)(ƒ(x)*d(x))=6

(∫_0^3)(ƒ(x)*d(x))+(∫_-*3^0)(ƒ(x)*d(x))=6

2*(∫_0^3)(ƒ(x)*d(x))=6

(∫_0^3)(ƒ(x)*d(x))=3

∴(∫_0^2)(ƒ(x)*d(x))=(∫_0^3)(ƒ(x))*d(x)-(∫_2^3)(ƒ(x))*d(x)=3-1=2

(B)

I UNDERSTAND IT NOW.

Problem 16

(∫_1^10)(ƒ(x)*d(x))=12=(∫_4^13)(ƒ(x)*d(x))=(∫_7^16)(ƒ(x)*d(x))

(∫_-*4^-*2)(ƒ(x)*d(x))=-10=(∫_2^4)(ƒ(x)*d(x))=(∫_5^7)(ƒ(x)*d(x))

(∫_5^16)(ƒ(x)*d(x))=(∫_5^7)(ƒ(x)*d(x))+(∫_7^16)(ƒ(x)*d(x))=12-10=2

∴(∫_16^5)(ƒ(x))*d(x)=-2

(B)

Problem 17

(∫_1^5)(ƒ(x)*d(x))=(∫_6^10)(ƒ(x)*d(x))=(∫_11^15)(ƒ(x)*d(x))=3

(∫_-*5^-*4)(ƒ(x)*d(x))=(∫_5^6)(ƒ(x)*d(x))=(∫_10^11)(ƒ(x)*d(x))=2

(∫_5^15)(ƒ(x)*d(x))=2+3+2+3=10

(D)

Problem 18

F(5)-F(1)=2

F(5)-F(3)=-3

∴F(3)-F(1)=5=(∫_1^3)(ƒ(x)*d(x))

(∫_^)(ƒ(-x)*d(x))=(∫_^)((ƒ(x)-3)*d(x))=F(x)-3*x+C

-x=u

d(u)/d(x)=-1

d(x)=-d(u)

(∫_-*3^-*1)(ƒ(x)*d(x))=(∫_3^1)(ƒ(-u)*(-d(u)))=(∫_1^3)(ƒ(-u)*d(u))=F(3)-F(1)-3*(3-1)=5-6=-1

(B)

Problem 19

(∫_1^4)(1/√(,x)*ƒ(√(,x))*d(x))=?

(∫_1^2)(ƒ(x))*d(x)=F(2)-F(1)=√(,2)

u=√(,x)

d(u)/d(x)=1/(2√(,x))

d(x)=2*d(u)⋅√(,x)

(∫_1^4)(1/√(,x)*ƒ(u)⋅2*d(u)⋅√(,x))=(∫_1^4)(2*ƒ(u))=2*(∫_1^4)(ƒ(√(,x)))=2*(F(√(,4))-F(√(,1)))=2⋅√(,2)=2√(,2)

(D)

Problem 20

(∫_-*2^2)(ƒ(x)*(x3+1)*d(x))=4

(∫_-*2^2)(ƒ(x)*(x3)*d(x))+(∫_-*2^2)(ƒ(x)*d(x))=4

ƒ(x)*(x3) adalah fungsi ganjil, maka (∫_-*2^2)(ƒ(x)*(x3)*d(x))=0 karena asimetris.

(∫_-*2^2)(ƒ(x)*d(x))=4

(∫_0^2)(ƒ(x)*d(x))=2

(∫_1^2)(ƒ(x)*d(x))=2-3=-1

∴(∫_-*2^-*1)(ƒ(x)*d(x))=-1

(B)

Problem 21

ƒ(x)=(x2)/2+C+(1-0)/2+(4-1)/2=2+C+(x2)/2

ƒ(2)=C+4=4

∴C=0

∴ƒ(0)=2

(B)

Problem 22

C=(∫_0^2)(ƒ(x)*d(x))

Misalkan sebagai C karena (ƒ^′)(x) menghilangkannya.

ƒ(x)=4*x3+3*x2+2*x+C

(ƒ^′)(x)=12*x2+6*x+2

(ƒ^′*′)(x)=24*x+6

(∫_0^2)(ƒ(x)*d(x))=(∫_0^2)((4*x3+3*x2+2*x+C)*d(x))

C=(∣_0^2)(x4+x3+x2+C*x)=24+23+22+2*C=28+2*C

∴C=(∫_0^2)(ƒ(x))=-28

ƒ(x)=4*x3+3*x2+2*x-28

ƒ(2)=4⋅8+3⋅4+2⋅2-28=20

(∫_0^2)(24*x+6+20)=(∫_0^2)(24*x+26)=(∣_0^2)(12*x2+26*x)=12⋅4+26⋅2=100

(E)

Problem 23

(∫_-*1^1)(ƒ(x)*d(x))=C=(∫_-*1^1)((x3+3*x2-5*x+C)*d(x))

C=(∣_-*1^1)(1/4*x4+x3-5/2*x2+C*x)=1+1+C+C=2+2*C

∴C=-2

ƒ(x)=x3+3*x2-5*x-2

ƒ(1)=1+3-5-2=-3

(A)

Problem 24

okay.

G(x)=A/3*x3+B/2*x2+C*x

A=(∫_0^2)(g(x)*d(x))=g(2)-g(0)=8/3*A+4/2*B+2*C

B=(∫_0^1)(g(x)*d(x))=1/3*A+1/2*B+C

C=(∫_0^3)(g(x)*d(x))+2=27/3*A+9/2*B+3*C+2

Consolidate.

5/3*A+2*B+2*C=0

5*A+6*B+6*C=0

1/3*A-1/2*B+C=0

2*A-3*B+6*C=0

9*A+9/2*B+2*C+2=0

18*A+9*B+4*C+4=0

Solve...

1 and 2

3*A+9*B=0

2 and 3

4*A-6*B+12*C=0

54*A+27*B+12*C+12=0

50*A+33*B=-12

yes I'll show all my work

150*A+450*B=0

150*A+99*B=-36

351*B=36

B=4/39

Sub.

150*A+450⋅4/39=0

A=-4/13

Sub.

2⋅(-4/13)-3⋅(4/39)+6*C=0

C=2/13

∴g(5)=25⋅(-4/13)+5⋅4/39+2/13=-274/39

(A)

Problem 25

(∫_1^a)(3*x√(,-2*x2+9)*d(x))=13

u=-2*x2+9

d(u)/d(x)=-4*x

d(x)=d(u)/(-4*x)

(∫_^)(3*x√(,-2*x2+9)*d(x))=(∫_^)((3*x√(,u)*d(u))/(-4*x))=-1/2*u(3/2)+C

apply u to √(,(log_2)(5*y+1)) and 0

√(,(log_2)(5*y+1))

-2*(√(,(log_2)(5*y+1)))2+9=-2*((log_2)(5*y+1))+9

from now on p=(log_2)(5*y+1)

0

9

(∣_9^-*2*p+9)(-1/2*u(3/2))=13

(∣_9^-*2*p+9)(u(3/2))=-26

(-2*p+9)(3/2)-9(3/2)=-26

(-2*p+9)(3/2)=1

Root both sides by 3/2

-2*p+9=1

p=4

Sub.

(log_2)(5*y+1)=4

5*y+1=24

5*y=16-1

∴y=3

LET'S GO.

(C)

Problem 26

(∣_0^π)(-1/2*cos(2*x)+sin(x))

(-1/2*cos(2*π)+sin(π))-(-1/2*cos(0)+sin(0))

(-1/2⋅1+0)-(-1/2⋅1+0)=0

(C)