Notes - Matlan - Integrals

Power Rule

ƒ(x)=(∫_^)((ƒ^′)(x))

(∫_^)(a*xn)*d(x)=a/(n+1)*x(n+1)+C

Ex. 1

(∫_^)(5*x3)*d(x)=5/(3+1)*x4+C=5/4*x4+C

Ex. 2

(∫_^)(x(3/5))*d(x)=1/(3/5+1)⋅x((3/5)+1)+C=5/8*x(8/5)+C=5/8*x√(5,x3)+C

Ex. 3

(∫_^)(4/(x√(,x)))*d(x)=(∫_^)(4/(x⋅x(1/2)))*d(x)=(∫_^)(4⋅x(-3/2))*d(x)

4/(-3/2+1)*x(-1/2)+C=-8*x(-1/2)+C

Ex. 4

(∫_^)(6*x2+4*x-8)*d(x)=6/(2+1)*x(2+1)+4/(1+1)*x(1+1)-8/(0+1)*x(0+1)+C

2*x3+2*x2-8*x+C

Ex. 5

(∫_^)((3*x2-2)*(4*x+1))*d(x)=(∫_^)(12*x3+3*x2-8*x-2)*d(x)

12/(3+1)*x(3+1)+3/(2+1)*x(2+1)-8/(1+1)*x(1+1)-2/(0+1)*x(0+1)+C

3*x4+x3-4*x2-2*x+C

Ex. 6

(∫_^)((4*x5-6*x3+2*x2-1)/(x2))*d(x)=(∫_^)(4*x3-6*x+2-x(-2))*d(x)

(4*x4)/4-(6*x2)/2+(2*x)/1-(x(-1))/(-1)+C

x4-3*x2+2*x+x(-1)+C

Ex. 7

(ƒ^′)(x)=(3*x-2)*(4*x+6)

ƒ(2)=6

ƒ(x)=?

(ƒ^′)(x)=12*x2+18*x-8*x-12=12*x2+10*x-12

ƒ(x)=12/(2+1)*x3+10/2*x2-12*x*C=4*x3+5*x2-12*x+C

ƒ(2)=4⋅23+5⋅22-12⋅2+C=6

4⋅8+5⋅4-24+C=6

C=6+24-20-32=-22

∴ƒ(x)=4*x3+5*x2-12*x-22

Substitution

Ex. 1

(∫_^)((2*x+3)2)*d(x)

(∫_^)(u)*d(x)

u=2*x+3

(u^′)=d(u)/d(x)=2

d(u)=2*d(x)

d(u)/2=d(x)

(∫_^)((u2*d(u))/2)=1/2*(∫_^)(u2)*d(u)=1/2⋅1/3*u3+C=1/6*u3+C=1/6*(2*x+3)3+C

Ex. 2

(∫_^)((x3)/((x2+1)3))*d(x)

u=x2+1

x2=u-1

d(u)/d(x)=2*x

d(u)/(2*x)=d(x)

(∫_^)((x3)/(u3)d(u)/(2*x))=(∫_^)((x2)/(u3)d(u)/2)=1/2*(∫_^)(x2*u(-3)*d(u))

1/2*(∫_^)((u-1)⋅u(-3)*d(u))=1/2*(∫_^)((u(-2)-u(-3))*d(u))

1/2*[-u(-1)+1/2*u(-2)]+C

-1/2*u(-1)+1/4*u(-2)+C

-(2*u)/(4*u2)+1/(4*u2)+C

(-2*u+1)/(4*u2)+C

(-2*(x2+1)+1)/(4*(x2+1)2)+C=(-2*x2-1)/(4*(x2+1)2)+C

Ex. 3

(∫_^)((x4-x2)3*(8*x3-4*x))*d(x)

u=x4-x2

d(u)/d(x)=4*x3-2*x

d(u)/(4*x3-2*x)=d(x)

(∫_^)(u3⋅(8*x3-4*x)⋅d(u)/(4*x3-2*x))

(∫_^)(u3⋅2⋅d(u))

2⋅1/4*u4+C

1/2*u4+C

1/2*(x4-x2)4+C

ln

(∫_^)(1/x*d(x))=ln|x|+C

Integration By Parts

(∫_^)(u*d(v))=u*v-(∫_^)(v*d(u))

Ex. 1

(∫_^)(6*x*(2*x-1)5*d(x))

Easily solvable via u substitution, but that's not why we're here.

u=6*x

d(u)/d(x)=6

d(u)=6*d(x)

d(v)=(2*x-1)5*d(x)

v=(∫_^)((2*x-1)5*d(x))=1/(2⋅6)*(2*x-1)6=1/12*(2*x-1)6

(∫_^)(u*d(v))=1/2*x*(2*x-1)6-(∫_^)(1/2*(2*x-1)6*d(x))

1/2*x*(2*x-1)6-1/28*(2*x-1)7+C

Integration by parts can be generalized via a tabular method.

When picking u, pick the group that if you keep deriving it, it can go to 0

Left derives, right integrates as we go downwards. Stop when u is 0

6*x⋅1/12*(2*x-1)6-6⋅1/12⋅1/7⋅1/2*(2*x-1)7+C

Alternate between plus and minus, starting from the second row.

Then, multiply u with d(v)/d(x) from the row below it.

Ex. 2

(∫_^)(12*x3*(2*x+3)4*d(x))

u=12*x3

d(v)=(2*x+3)4

6/5*x3*(2*x+3)5-3/10*x2*(2*x+3)6+3/70*x*(2*x+3)7-3/1120*(2*x+3)8+C

Definite Integrals

(∫_a^b)(ƒ(x))*d(x)=F(b)-F(a)

There are many identities related to definite integrals. The key thing to understanding them is understanding what a definite integral is: The signed sum (net area) strictly between a and b. Result is a number. But, I'll write those identities anyway.

(∫_a^b)(k*ƒ(x)*d(x))=k*(∫_a^b)(ƒ(x)*d(x))

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

(∫_a^a)(ƒ(x)*d(x))=0

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

(∫_a^c)(ƒ(x)*d(x))=(∫_a^b)(ƒ(x)*d(x))+(∫_b^c)(ƒ(x)*d(x))

Exponentials and ln

(∫_^)(ℇx*d(x))=ℇx+C

(∫_^)(1/x*d(x))=ln(x)+C

Ex. 1

(∫_^)(ℇ(x+3)*d(x))=ℇ(x+3)+C

Ex. 2

(∫_^)(ℇ(2*x-3)*d(x))=1/2*ℇ(2*x-3)+C

Ex. 3

(∫_^)(4*x*ℇ(x2-5)*d(x))

x2-5=u

d(u)/d(x)=2*x

(∫_^)(2*ℇu*d(u))=2*ℇ(x2-5)

Ex. 4

(∫_^)(1/(2*x+1)*d(x))

2*x+1

d(u)/d(x)=2

(∫_^)(1/u⋅1/2⋅d(u))=1/2⋅ln|2*x+1|+C

Ex. 5

(∫_^)((3*x-1)/(3*x+2)*d(x))=(∫_^)(((3*x+2)-3)/(3*x+2)*d(x))=(∫_^)((1-3/(3*x+2))*d(x))

x-ln|3*x+2|+C

Trigonometry

(∫_^)(cos(x)*d(x))=sin(x)+C

(∫_^)(sin(x)*d(x))=-cos(x)+C

(∫_^)(sec2(x)*d(x))=tan(x)+C

(∫_^)(csc2(x)*d(x))=-cot(x)+C

(∫_^)(sec(x)*tan(x)*d(x))=sec(x)+C

(∫_^)(csc(x)*cot(x)*d(x))=-csc(x)+C

(∫_^)(tan(x)*d(x))=ln|sec(x)|+C

(∫_^)(cot(x)*d(x))=ln|sin(x)|+C

(∫_^)(sec(x)*d(x))=ln|sec(x)+tan(x)|+C

(∫_^)(csc(x)*d(x))=-ln|csc(x)+cot(x)|+C