Definite Integral
See how to solve a definite integral.
13 examples and their solutions.
Definite Integral: Meaning
Definition
∫abf(x) dx
(when y = f(x) is above the x-axis.)
It's the sum of the area of the thinly sliced rectangle.
Riemann Sum
Definite Integral: How to Solve
Formula
∫abf(x) dx = [F(x)]ab
= F(b) - F(a)
a: Lower limit = F(b) - F(a)
b: Upper limit
How to solve:
1. Find F(x): Indefinite integral of f(x)
2. Put b and a into F(x).
The result of a definite integral is a value.
(F(b) - F(a) doesn't have a variable.)
[F(x)]ab = F(b) + C - [F(a) + C]
= F(b) - F(a)
→ You don't have to write +C
when solving a definite integral.
Example
∫-12x3 dx
Solution ∫-12x3 dx
= [14x4]-12 - [1]
= 14⋅24 - 14⋅(-1)4
= 14⋅16 - 14⋅1
= 14(16 - 1)
= 154
= [14x4]-12 - [1]
= 14⋅24 - 14⋅(-1)4
= 14⋅16 - 14⋅1
= 14(16 - 1)
= 154
[1]
∫ x3 dx = [1/4]x4
Indefinite Integral
Indefinite Integral
Close
Example
∫13(6x2 - 2x + 5) dx
Solution ∫13(6x2 - 2x + 5) dx
= [6⋅13x3 -2⋅12x2 + 5x]13
= [2x3 - x2 + 5x]13
= 2⋅33 - 32 + 5⋅3 - [2⋅13 - 12 + 5⋅1]
= 2⋅27 - 9 + 15 - [2⋅1 - 1 + 5]
= 54 + 6 - [2 + 4]
= 54
= [6⋅13x3 -2⋅12x2 + 5x]13
= [2x3 - x2 + 5x]13
= 2⋅33 - 32 + 5⋅3 - [2⋅13 - 12 + 5⋅1]
= 2⋅27 - 9 + 15 - [2⋅1 - 1 + 5]
= 54 + 6 - [2 + 4]
= 54
Close
Example
∫01dx3x + 1
Solution ∫01dx3x + 1
= [13 ln |3x + 1|]01 - [1]
= 13 ln |3⋅1 + 1| - 13 ln |3⋅0 + 1|
= 13 ln |3 + 1| - 13 ln |1|
= 13 ln |4| - 0 - [2]
= 13 ln 4
= 13 ln 22
= 23 ln 2 - [2]
= [13 ln |3x + 1|]01 - [1]
= 13 ln |3⋅1 + 1| - 13 ln |3⋅0 + 1|
= 13 ln |3 + 1| - 13 ln |1|
= 13 ln |4| - 0 - [2]
= 13 ln 4
= 13 ln 22
= 23 ln 2 - [2]
[1]
∫ 1/(3x + 1) dx = [1/3] ln |3x + 1|
Indefinite Integral
Indefinite Integral
[2]
Close
Definite Integral: Property
Variable in the Definite Integral
∫abf(x) dx = ∫abf(y) dy
The result of a definite integral is a constant.So the variable in the definite integral
doesn't affect the definite integral value.
(If the lower/upper limit is a variable,
then it's a different story.)
Upper Limit = Lower Limit
∫aaf(x) dx = 0
Example
∫1xf(t) dt = x2 + px
p = ?
Solution p = ?
∫1xf(t) dt = x2 + px
x = 1
∫11f(t) dt = 12 + p⋅1
0 = 1 + p
p + 1 = 0
p = -1
x = 1
∫11f(t) dt = 12 + p⋅1
0 = 1 + p
p + 1 = 0
p = -1
[1]
Put x = 1 into the given equation.
Close
Same Limits
∫abf(x) dx ± ∫abg(x) dx
= ∫ab(f(x) ± g(x)) dx
= ∫ab(f(x) ± g(x)) dx
Example
∫13(x2 + x - 1) dx + ∫132y2 dy - ∫13z dz
Solution ∫13(x2 + x - 1) dx + ∫132y2 dy - ∫13z dz
= ∫13(x2 + x - 1) dx + ∫132x2 dx - ∫13x dx
= ∫13(x2 + x - 1 + 2x2 - x) dx
= ∫13(3x2 - 1) dx
= [3⋅13x3 - x]13 - [1]
= [x3 - x]13
= 33 - 3 - [13 - 1]
= 27 - 3 - [1 - 1]
= 24
= ∫13(x2 + x - 1) dx + ∫132x2 dx - ∫13x dx
= ∫13(x2 + x - 1 + 2x2 - x) dx
= ∫13(3x2 - 1) dx
= [3⋅13x3 - x]13 - [1]
= [x3 - x]13
= 33 - 3 - [13 - 1]
= 27 - 3 - [1 - 1]
= 24
Close
Connected Limits
∫acf(x) dx + ∫cbf(x) dx
= ∫abf(x) dx
f(x) should be the same. = ∫abf(x) dx
Example
∫01(2x + 1) dx + ∫14(2x + 1) dx
Solution ∫01(2x + 1) dx + ∫14(2x + 1) dx
= ∫04(2x + 1) dx
= [2⋅12x2 + x]04
= [x2 + x]04
= 42 + 4 - [02 + 0]
= 16 + 4
= 20
= ∫04(2x + 1) dx
= [2⋅12x2 + x]04
= [x2 + x]04
= 42 + 4 - [02 + 0]
= 16 + 4
= 20
Close
Switching Upper Limit and Lower Limit
∫abf(x) dx = -∫baf(x) dx
Example
∫085x4 dx - ∫285x4 dx
Solution ∫085x4 dx - ∫285x4 dx
= ∫085x4 dx - [-∫825x4 dx]
= ∫085x4 dx + ∫825x4 dx
= ∫025x4 dx
= [5⋅15x5]02
= [x5]02
= 25 - 05
= 32 - 0
= 32
= ∫085x4 dx - [-∫825x4 dx]
= ∫085x4 dx + ∫825x4 dx
= ∫025x4 dx
= [5⋅15x5]02
= [x5]02
= 25 - 05
= 32 - 0
= 32
Close
Definite Integral of an Odd/Even Function
Odd Function
Even Function
Property
∫-aa(Odd function) dx = 0
∫-aa(Even function) dx = 2⋅∫0a(Even function) dx
Example
∫-11(5x4 + 8x3 - 6x2 - 2x + 7) dx
Solution ∫-11(5x4 + 8x3 - 6x2 - 2x + 7) dx
= 2∫01(5x4 - 6x2 + 7) dx
= 2[5⋅15x5 - 6⋅13x3 + 7x]01 - [1]
= 2[x5 - 2x3 + 7x]01
= 2[15 - 2⋅13 + 7⋅1 - [05 - 2⋅03 + 7⋅0]]
= 2[1 - 2⋅1 + 7 - [0 - 0 + 0]]
= 2[1 - 2 + 7]
= 2[-1 + 7]
= 2⋅6
= 12
= 2∫01(5x4 - 6x2 + 7) dx
= 2[5⋅15x5 - 6⋅13x3 + 7x]01 - [1]
= 2[x5 - 2x3 + 7x]01
= 2[15 - 2⋅13 + 7⋅1 - [05 - 2⋅03 + 7⋅0]]
= 2[1 - 2⋅1 + 7 - [0 - 0 + 0]]
= 2[1 - 2 + 7]
= 2[-1 + 7]
= 2⋅6
= 12
Close
Example
∫-π3π3(sin θ + cos θ) dθ
Solution ∫-π3π3(sin θ + cos θ) dθ
= 2∫0π3cos θ dθ
= 2[sin θ]0π3 - [1]
= 2[sin π3 - [sin 0]]
= 2[√32 - 0] - [2]
= 2⋅√32
= √3
= 2∫0π3cos θ dθ
= 2[sin θ]0π3 - [1]
= 2[sin π3 - [sin 0]]
= 2[√32 - 0] - [2]
= 2⋅√32
= √3
[1]
∫ cos θ dθ = sin θ
Indefinite Integral
Indefinite Integral
[2]
sin π/3 = sin 60°
sin 60°
SOH: Sine, Opposite side (√3), Hypotenuse (2)
sin 0 = 0
Trigonometry (Right Triangle)
Trigonometric Function Value
sin 60°
SOH: Sine, Opposite side (√3), Hypotenuse (2)
sin 0 = 0
Trigonometry (Right Triangle)
Trigonometric Function Value
Close
Integral by Substitution
Example
∫0π2esin x cos x dx
Solution ∫0π2esin x cos x dx - [1]
sin x = t
cos x dx = dt - [2]
x = 0 → t = sin 0
= 0
x = π2 → t = sin π2
= 1 - [3]
= ∫01et dt
= [et]01 - [4]
= e1 - e0
= e - 1
sin x = t
cos x dx = dt - [2]
x = 0 → t = sin 0
= 0
x = π2 → t = sin π2
= 1 - [3]
= ∫01et dt
= [et]01 - [4]
= e1 - e0
= e - 1
[3]
sin x = tx = 0 → t = sin 0
x = π/2 → t = sin π/2
x = π/2 → t = sin π/2
[4]
∫ et dt = et
Indefinite Integral
Indefinite Integral
Close
Example
∫ee4dxx ln x
Solution ∫ee4dxx ln x
ln x = t
1x dx = dt - [1]
x = e → t = ln e
= 1 - [2]
x = e4 → t = ln e4
= 4 ln e
= 4⋅1
= 4 - [3]
= ∫141t dt
= [ln |t|]14 - [4]
= ln |4| - ln |1|
= ln 4 - 0
= ln 22
= 2 ln 2 - [2]
ln x = t
1x dx = dt - [1]
x = e → t = ln e
= 1 - [2]
x = e4 → t = ln e4
= 4 ln e
= 4⋅1
= 4 - [3]
= ∫141t dt
= [ln |t|]14 - [4]
= ln |4| - ln |1|
= ln 4 - 0
= ln 22
= 2 ln 2 - [2]
[2]
[3]
ln x = tx = e → t = ln e
x = e4 → t = ln e4
x = e4 → t = ln e4
[4]
∫ 1/t dt = ln |t|
Indefinite Integral
Indefinite Integral
Close
Integral by Parts
Formula
∫abuv' dx = [uv]ab - ∫abu'v dx
Integral by Parts: Indefinite Integral How to Determine u and v'
ln x, loga x
x, x2, x3, ...
sin x, cos x
ex, ax u↕v'
Upper function → u x, x2, x3, ...
sin x, cos x
ex, ax u↕v'
(= The integral is complex.)
Lower function → v'
(= The integral is simple.)
Example
∫0π3x sin x dx
Solution u v'
∫0π3x sin x dx
u = x v = -cos x
u' = 1 v' = sin x - [1]
= [x⋅(-cos x)]0π3 - ∫0π31⋅(-cos x) dx - [2]
= [-x cos x]0π3 + ∫0π3cos x dx
= -π3 cos π3 - [-0 cos 0] + [sin x]0π3
= -π3⋅12 - 0 + sin π3 - sin 0 - [3]
= -π6 + √32 - 0 - [4]
= √32 - π6
∫0π3x sin x dx
u = x v = -cos x
u' = 1 v' = sin x - [1]
= [x⋅(-cos x)]0π3 - ∫0π31⋅(-cos x) dx - [2]
= [-x cos x]0π3 + ∫0π3cos x dx
= -π3 cos π3 - [-0 cos 0] + [sin x]0π3
= -π3⋅12 - 0 + sin π3 - sin 0 - [3]
= -π6 + √32 - 0 - [4]
= √32 - π6
[1]
Write u = x.
→ u' = x
Derivative Rules
Write v' = sin x next to u' = 1.
→ Write v = -cos x next to u = x.
Indefinite Integral
→ u' = x
Derivative Rules
Write v' = sin x next to u' = 1.
→ Write v = -cos x next to u = x.
Indefinite Integral
[2]
uv: x⋅(-cos x)
u'v: 1⋅(-cos x)
u'v: 1⋅(-cos x)
[3]
cos π/3 = cos 60°
cos 60°
CAH: Cosine, Adjacent side (1), Hypotenuse (2)
Trigonometry (Right Triangle)
Trigonometric Function Value
cos 60°
CAH: Cosine, Adjacent side (1), Hypotenuse (2)
Trigonometry (Right Triangle)
Trigonometric Function Value
[4]
sin π/3 = sin 60°
sin 60°
SOH: Sine, Opposite side (√3), Hypotenuse (2)
sin 0 = 0
sin 60°
SOH: Sine, Opposite side (√3), Hypotenuse (2)
sin 0 = 0
Close
Example
∫1e(ln x)2 dx
Solution ∫1e(ln x)2 dx
u v'
= ∫1e(ln x)2⋅1 dx
u = (ln x)2 v = x
u' = 2(ln x)1⋅1x
= (2 ln x)⋅1x v' = 1 - [1]
= [(ln x)2⋅x]1e - ∫1e(2 ln x)⋅1x⋅x dx - [2]
= [x (ln x)2]1e - 2∫1eln x dx
= e⋅(ln e)2 - 1⋅(ln 1)2 - 2[x ln x - x]1e - [3]
= e⋅12 - 1⋅02 - 2[e ln e - e - [1 ln 1 - 1]] - [4]
= e⋅1 - 0 - 2[e⋅1 - e - [0 - 1]]
= e - 2[e - e + 1]
= e - 2⋅1
= e - 2
u v'
= ∫1e(ln x)2⋅1 dx
u = (ln x)2 v = x
u' = 2(ln x)1⋅1x
= (2 ln x)⋅1x v' = 1 - [1]
= [(ln x)2⋅x]1e - ∫1e(2 ln x)⋅1x⋅x dx - [2]
= [x (ln x)2]1e - 2∫1eln x dx
= e⋅(ln e)2 - 1⋅(ln 1)2 - 2[x ln x - x]1e - [3]
= e⋅12 - 1⋅02 - 2[e ln e - e - [1 ln 1 - 1]] - [4]
= e⋅1 - 0 - 2[e⋅1 - e - [0 - 1]]
= e - 2[e - e + 1]
= e - 2⋅1
= e - 2
[1]
Write u = (ln x)2.
→ u' = 2(ln x)(1/x)
Derivative of g(f(x))
Write v' = 1 next to u' = (2 ln x)⋅(1/x).
→ Write v = x next to u = (ln x)2.
→ u' = 2(ln x)(1/x)
Derivative of g(f(x))
Write v' = 1 next to u' = (2 ln x)⋅(1/x).
→ Write v = x next to u = (ln x)2.
[2]
uv: (ln x)2⋅x
u'v: 2(ln x)(1/x)⋅x
u'v: 2(ln x)(1/x)⋅x
[3]
∫ ln x dx = x ln x - x
Indefinite Integral
Indefinite Integral
[4]
Close