Editorial - Matlan - Latihan Luas Integral

Problem 1

(∫_0^2)(x3*d(x))=(∣_0^2)(1/4*x4)=2/4-0=4

(B)

Problem 2

(∫_-2^2)((4-x2)*d(x))=(∣_-2^2)(4*x-1/3*x3)=4⋅(2-(-2))-1/3⋅(2-(-2)3)=4⋅4-1/3⋅16=32/3=10+2/3

(B)

Problem 3

x-y=2

y=-2+x

(∫_3^4)((x-2)*d(x))=(∣_3^4)(1/2*x2-2*x)=1/2⋅(4-3)-2⋅(4-3)=7/2-2=3/2=1+1/2

(D)

Problem 4

y=x*(x+1)*(x-2)=x3-x2-2*x=ƒ(x)

F(x)=1/4*x4-1/3*x3-x2+C

F(-1)=1/4⋅(-1)4-1/3⋅(-1)3-(-1)2+C=-5/12+C

F(0)=0+C

F(2)=1/4⋅2-1/3⋅2-2=-8/3+C

Titik melewati sumbu x ada pada x=-1,0,2

(∫_2^-1)((x3-x2-2*x)*d(x))=(∫_-1^0)(ƒ(x)*d(x))-(∫_0^2)(ƒ(x)*d(x))
=0-(-5/12)-(0-(-8/3))
=5/12+8/3
=37/12
=3+1/12

(D)

Problem 5

ƒ(x)=sin(x)+cos(x)

F(x)=-cos(x)+sin(x)+C=sin(x)-cos(x)+C

(∫_0^π/3)(F(x)*d(x))=(√(,3)/2-1/2)-(0-1)
=1/2+√(,3)/2
=1/2*(1+√(,3))

(E)

Problem 6

ƒ(x)=x2+2*x+4

x2+2*x+4=0 (no solutions)

F(x)=1/3*x3+x2+4*x+C

∴(∫_0^3)(ƒ(x)*d(x))=ƒ(3)-ƒ(0)=27/3+9+12=30

(C)

Problem 7

ƒ(x)=-3/(x2)=-3*x(-2)

F(x)=3*x(-1)+C

-(∫_1^3)(ƒ(x)*d(x))=-3⋅3+3⋅1=-1+3=2

(A)

Problem 8

ƒ(x)=x2-1

x2-1=0

x={-1,+1}

F(x)=1/3*x3-x+C

Also, areanya di bawah sumbu x, maka ditambah sign negatif.

-(∫_-1^0)(ƒ(x)*d(x))-(∫_0^1)(ƒ(x)*d(x))=-2*(∫_0^1)(ƒ(x)*d(x))
=-2⋅(1/3-1)=4/3=1+1/3

(B)

Problem 9

x3=0

x=0

Titik balik di (0,0)

F(x)=1/4*x4+C

-(∫_-1^0)(ƒ(x)*d(x))+(∫_0^3)(ƒ(x)*d(x))=1/4+81/4=41/2=20+1/2

(E)

Problem 10

ƒ(x)=sin(2*x)

sin(2*x)=0

x=0,π/2

so, ƒ(x) positif pada 0≤x≤π/2

F(x)=-1/2*cos(2*x)+C

(∫_0^π/2)(ƒ(x)*d(x))=-1/2⋅(-1)-(-1/2⋅1)
=1

(D)

Problem 11

ƒ(x)=3*x2-18*x+24

3*x2-18*x+24=0

x={2,+4}

F(x)=x3-9*x2+24*x+C

F(0)=0

F(2)=8-9⋅4+24⋅2+C=20+C

F(4)=64-9⋅16+24⋅4+C=16+C

(∫_0^2)(ƒ(x)*d(x))-(∫_2^4)(ƒ(x)*d(x))=20+20-16=24

(B)

Problem 12

ƒ(x)=-x2+2*x+24

-x2+2*x+24=0

F(x)=-1/3*x3+x2+24*x+C

F(-4)=1/3⋅64+16+24⋅(-4)+C=-176/3+C

F(0)=0+C

F(6)=-1/3⋅6+6+24⋅6+C=108+C

I=176/3

II=108=324/3

176/3∶324/3=176∶324=44:81

(D)

Problem 13

Titik balik di 0,2,4

ƒ(x)=x*(x-2)*(x-4)=x3-6*x2+8*x

F(x)=1/4*x4-2*x3+4*x2+C

F(2)=1/4⋅16-2⋅8+4⋅4+C=4+C

F(4)=1/4⋅256-2⋅64+4⋅16+C=0+C

(∫_0^2)(ƒ(x)*d(x))-(∫_2^4)(ƒ(x)*d(x))=(4-0)-(0-4)=8

(D)

Problem 14

x=(y+2)*(y-4)=y2-2*y-8

F(y)=1/3*y3-y2-8*y+C

F(0)=0+C

F(4)=1/3⋅64-16-8⋅4+C=-80/3+C

-(∫_0^4)(ƒ(y)*d(y))=80/3

No dice. Lalu, batasan sumbu x untuk apa?

F(-2)=-8⋅1/3-4+16+C=28/3+C

-(∫_-2^0)(ƒ(y)*d(y))=28/3

Sum.

𝛴=28/3+80/3=36

(A)

Problem 15

ƒ(x)=2*cos(2*x)

Titik balik di x=π/4

F(x)=sin(2*x)

(∫_0^π/4)(ƒ(x)*d(x))=sin(π/2)=1

(E)