Quick answer: By the fundamental theorem of calculus , if F ′ ( x ) = f ( x ) F'(x) = f(x) F ′ ( x ) = f ( x ) ∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x)\,dx = F(b) - F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) King's property : ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x \int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x \int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x ∫ 0 π 2 sin x sin x + cos x d x = π 4 \int_0^{\frac{\pi}{2}}\frac{\sin x}{\sin x + \cos x}\,dx = \frac{\pi}{4} ∫ 0 2 π s i n x + c o s x s i n x d x = 4 π
This uses the techniques from indefinite integrals and integration by parts .
Fundamental theorem of calculus
Part Statement Evaluation ∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x)\,dx = F(b) - F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) F ′ = f F' = f F ′ = f Derivative of an integral d d x ∫ a x f ( t ) d t = f ( x ) \frac{d}{dx}\int_a^x f(t)\,dt = f(x) d x d ∫ a x f ( t ) d t = f ( x )
No constant of integration is needed for definite integrals.
Substitution with limits
When you substitute t = g ( x ) t = g(x) t = g ( x ) g ( a ) g(a) g ( a ) g ( b ) g(b) g ( b )
Basic properties
Property Statement Dummy variable ∫ a b f ( x ) d x = ∫ a b f ( t ) d t \int_a^b f(x)\,dx = \int_a^b f(t)\,dt ∫ a b f ( x ) d x = ∫ a b f ( t ) d t Reversing limits ∫ a b f d x = − ∫ b a f d x \int_a^b f\,dx = -\int_b^a f\,dx ∫ a b f d x = − ∫ b a f d x Equal limits ∫ a a f d x = 0 \int_a^a f\,dx = 0 ∫ a a f d x = 0 Splitting ∫ a b f d x = ∫ a c f d x + ∫ c b f d x \int_a^b f\,dx = \int_a^c f\,dx + \int_c^b f\,dx ∫ a b f d x = ∫ a c f d x + ∫ c b f d x King's property ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x \int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x Special case ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x \int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x Doubling ∫ 0 2 a f d x = 2 ∫ 0 a f d x \int_0^{2a} f\,dx = 2\int_0^a f\,dx ∫ 0 2 a f d x = 2 ∫ 0 a f d x f ( 2 a − x ) = f ( x ) f(2a - x) = f(x) f ( 2 a − x ) = f ( x ) f ( 2 a − x ) = − f ( x ) f(2a - x) = -f(x) f ( 2 a − x ) = − f ( x )
Splitting is essential for ∣ x ∣ \lvert x\rvert ∣ x ∣ [ x ] [x] [ x ]
Standard results from King's property
Integral Value ∫ 0 π 2 sin n x sin n x + cos n x d x \int_0^{\frac{\pi}{2}}\frac{\sin^n x}{\sin^n x + \cos^n x}\,dx ∫ 0 2 π s i n n x + c o s n x s i n n x d x π 4 \frac{\pi}{4} 4 π ∫ 0 π 2 ln sin x d x \int_0^{\frac{\pi}{2}}\ln\sin x\,dx ∫ 0 2 π ln sin x d x − π 2 ln 2 -\frac{\pi}{2}\ln 2 − 2 π ln 2 ∫ a b f ( x ) f ( x ) + f ( a + b − x ) d x \int_a^b\frac{f(x)}{f(x) + f(a + b - x)}\,dx ∫ a b f ( x ) + f ( a + b − x ) f ( x ) d x b − a 2 \frac{b - a}{2} 2 b − a ∫ 0 π x f ( sin x ) d x \int_0^{\pi}x\,f(\sin x)\,dx ∫ 0 π x f ( sin x ) d x π 2 ∫ 0 π f ( sin x ) d x \frac{\pi}{2}\int_0^{\pi}f(\sin x)\,dx 2 π ∫ 0 π f ( sin x ) d x
Worked examples
Example 1: direct evaluation
∫ 1 2 ( 3 x 2 + 1 ) d x = [ x 3 + x ] 1 2 = 10 − 2 = 8 \int_1^2(3x^2 + 1)\,dx = \left[x^3 + x\right]_1^2 = 10 - 2 = 8 ∫ 1 2 ( 3 x 2 + 1 ) d x = [ x 3 + x ] 1 2 = 10 − 2 = 8
Example 2: substitution with limits
Evaluate ∫ 0 1 2 x e x 2 d x \int_0^1 2x\,e^{x^2}\,dx ∫ 0 1 2 x e x 2 d x
Put t = x 2 t = x^2 t = x 2 ∫ 0 1 e t d t = e − 1 \int_0^1 e^t\,dt = e - 1 ∫ 0 1 e t d t = e − 1
Example 3: King's property
Evaluate I = ∫ 0 π 2 sin x sin x + cos x d x I = \int_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx I = ∫ 0 2 π s i n x + c o s x s i n x d x
Step 1: Replacing x by π 2 − x \frac{\pi}{2} - x 2 π − x I = ∫ 0 π 2 cos x cos x + sin x d x I = \int_0^{\frac{\pi}{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx I = ∫ 0 2 π c o s x + s i n x c o s x d x
Step 2: Adding, 2 I = ∫ 0 π 2 1 d x = π 2 2I = \int_0^{\frac{\pi}{2}}1\,dx = \frac{\pi}{2} 2 I = ∫ 0 2 π 1 d x = 2 π I = π 4 I = \frac{\pi}{4} I = 4 π
Example 4: splitting at a corner
Evaluate ∫ − 1 2 ∣ x ∣ d x \int_{-1}^{2}\lvert x\rvert\,dx ∫ − 1 2 ∣ x ∣ d x
∫ − 1 0 ( − x ) d x + ∫ 0 2 x d x = 1 2 + 2 = 5 2 \int_{-1}^0(-x)\,dx + \int_0^2 x\,dx = \frac{1}{2} + 2 = \frac{5}{2} ∫ − 1 0 ( − x ) d x + ∫ 0 2 x d x = 2 1 + 2 = 2 5
Example 5: greatest integer
Evaluate ∫ 0 3 [ x ] d x \int_0^3[x]\,dx ∫ 0 3 [ x ] d x
0 ⋅ 1 + 1 ⋅ 1 + 2 ⋅ 1 = 3 0 \cdot 1 + 1 \cdot 1 + 2 \cdot 1 = 3 0 ⋅ 1 + 1 ⋅ 1 + 2 ⋅ 1 = 3
Example 6: derivative of an integral
Find d d x ∫ 0 x 2 sin t d t \frac{d}{dx}\int_0^{x^2}\sin t\,dt d x d ∫ 0 x 2 sin t d t
By the chain rule, sin ( x 2 ) ⋅ 2 x \sin(x^2) \cdot 2x sin ( x 2 ) ⋅ 2 x
Summary table
Idea Key result Evaluation F ( b ) − F ( a ) F(b) - F(a) F ( b ) − F ( a ) King's property Replace x by a + b − x a + b - x a + b − x Splitting Break at corners and jumps Symmetric fraction Gives b − a 2 \frac{b - a}{2} 2 b − a d d x ∫ a g ( x ) f ( t ) d t \frac{d}{dx}\int_a^{g(x)}f(t)\,dt d x d ∫ a g ( x ) f ( t ) d t f ( g ( x ) ) g ′ ( x ) f(g(x))\,g'(x) f ( g ( x )) g ′ ( x )
Exam tips
If the integrand looks hard and the limits are 0 to π 2 \frac{\pi}{2} 2 π
Change limits when substituting; it saves the back-substitution step.
Sketch ∣ f ( x ) ∣ \lvert f(x)\rvert ∣ f ( x )∣ [ x ] [x] [ x ]
Next, study definite integrals of even and odd functions , or see the NIMCET Calculus chapter .
Common mistakes
Forgetting to change the limits after substitution.
Adding a constant C to a definite integral.
Not splitting the interval at points where ∣ f ∣ \lvert f\rvert ∣ f ∣
Using King's property with the wrong replacement (a − x a - x a − x a + b − x a + b - x a + b − x
Frequently Asked Questions
What is King's property of definite integrals?
∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx.
What is the fundamental theorem of calculus?
If F′(x) = f(x), then ∫ₐᵇ f(x) dx = F(b) − F(a). Also, d/dx ∫ₐˣ f(t) dt = f(x).
What is the value of ∫ from 0 to π/2 of sin x/(sin x + cos x) dx?
It equals π/4, by King's property.
Why do we split a definite integral?
To handle functions like |x| or [x] whose formula changes inside the interval.
Do definite integrals need a constant of integration?
No. The constant cancels when F(b) − F(a) is computed.
