Definite integrals compute the net area under a curve between two specified points, providing a numerical value representing the total accumulation of the function over that interval.
A definite integral is denoted by ∫ a b f ( x ) d x \int_a^b f(x) dx ∫ a b f ( x ) d x which represents the
algebraic area bounded by the curve y = f(x), the ordinates x = a,
x = b and the x-axis.
Neetesh Kumar | June 02, 2024
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1. The Fundamental Theorem of Calculus:
(a) If f f f is continuous on [ a , b ] [a, b] [ a , b ] , then the function g g g defined by:
g ( x ) = ∫ a x f ( t ) d t , a ≤ x ≤ b g(x) = \int_{a}^{x} f(t) \, dt, \quad a \leq x \leq b g ( x ) = ∫ a x f ( t ) d t , a ≤ x ≤ b
is continuous on [ a , b ] [a, b] [ a , b ] and differentiable on ( a , b ) (a, b) ( a , b ) , and g ′ ( x ) = f ( x ) g'(x) = f(x) g ′ ( x ) = f ( x ) .
(b)
If f f f is continuous on [ a , b ] [a, b] [ a , b ] , then: br/>
∫ 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 ) br/>
where F F F is any antiderivative of f f f , that is, a function such that F ′ = f F' = f F ′ = f .
Note:
If ∫ a b f ( x ) d x = 0 \int_{a}^{b} f(x) \, dx = 0 ∫ a b f ( x ) d x = 0
then the equation f ( x ) = 0 f(x) = 0 f ( x ) = 0 has at least one root lying in ( a , b ) (a, b) ( a , b ) provided f f f is a continuous function in ( a , b ) (a, b) ( a , b ) .
2. Properties of Definite Integral:
( a ) \bold{(a)} ( a )
∫ a b f ( x ) d x = ∫ a b f ( t ) d t ⇒ ∫ a b f ( x ) d x \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(t) \, dt \Rightarrow \int_a^bf(x)dx ∫ a b f ( x ) d x = ∫ a b f ( t ) d t ⇒ ∫ a b f ( x ) d x the integral does not depend upon the variable of integration. It is a numerical quantity.
( b ) \bold{(b)} ( b )
∫ a b f ( x ) d x = − ∫ b a f ( x ) d x \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx ∫ a b f ( x ) d x = − ∫ b a f ( x ) d x
( c ) \bold{(c)} ( c )
∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x \int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x
where c c c may lie inside or outside the interval [ a , b ] [a, b] [ a , b ] . This property is used when f f f is piecewise continuous in ( a , b ) (a, b) ( a , b ) .
( d ) \bold{(d)} ( d )
∫ − a a f ( x ) d x = ∫ 0 a [ f ( x ) + f ( − x ) ] d x = { 0 ;if f ( x ) i s a n o d d f u n c t i o n 2 ∫ 0 a f ( x ) d x ;if f ( x ) i s a n e v e n f u n c t i o n \int_{-a}^{a} f(x)dx = \int_0^a[f(x) + f(-x)]dx = \begin{cases}
0 &\text{;if } f(x) \space is \space an \space odd \space function \\
2\int_0^a f(x)dx &\text{;if } f(x) \space is \space an \space even \space function
\end{cases} ∫ − a a f ( x ) d x = ∫ 0 a [ f ( x ) + f ( − x )] d x = { 0 2 ∫ 0 a f ( x ) d x ;if f ( x ) i s an o dd f u n c t i o n ;if f ( x ) i s an e v e n f u n c t i o n
( e ) \bold{(e)} ( e )
∫ 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 In particular, ∫ 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
( f ) \bold{(f)} ( f )
∫ 0 2 a f ( x ) d x = ∫ 0 a f ( x ) d x + ∫ 0 a f ( 2 a − x ) d x = { 2 ∫ a b f ( x ) d x ;if f ( 2 a − x ) = f ( x ) 0 ;if f ( 2 a − x ) = − f ( x ) \int_{0}^{2a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(2a - x) \, dx = \begin{cases}
2\int_a^b f(x)dx &\text{;if } f(2a-x) = f(x) \\
0 &\text{;if } f(2a-x) = -f(x)
\end{cases} ∫ 0 2 a f ( x ) d x = ∫ 0 a f ( x ) d x + ∫ 0 a f ( 2 a − x ) d x = { 2 ∫ a b f ( x ) d x 0 ;if f ( 2 a − x ) = f ( x ) ;if f ( 2 a − x ) = − f ( x )
( g ) \bold{(g)} ( g )
∫ 0 n T f ( x ) d x = n ∫ 0 T f ( x ) d x , ( n ∈ I ) ; \int_{0}^{nT} f(x) \, dx = n\int_{0}^{T} f(x)dx, (n \in I); ∫ 0 n T f ( x ) d x = n ∫ 0 T f ( x ) d x , ( n ∈ I ) ; where ‘T’ is the period of the
function i.e. f(T + x) = f(x)
Note: ∫ x T + x f ( t ) d t \int_{x}^{T + x} f(t) \, dt ∫ x T + x f ( t ) d t will be independent of x x x and equal to ∫ 0 T f ( t ) d t \int_{0}^{T} f(t) dt ∫ 0 T f ( t ) d t
( h ) \bold{(h)} ( h )
∫ a + n t b + n T f ( x ) d x = n ∫ a b f ( x ) d x \int_{a+nt}^{b+nT} f(x)dx = n \int_{a}^{b} f(x)dx ∫ a + n t b + n T f ( x ) d x = n ∫ a b f ( x ) d x where f f f is periodic with period T T T and n ∈ I n \in \mathbb{I} n ∈ I .
( i ) \bold{(i)} ( i )
∫ m a n a f ( x ) d x = ( n − m ) ∫ 0 a f ( x ) d x , ( n , m ∈ I ) \int_{ma}^{na} f(x) \, dx = (n - m) \int_{0}^{a} f(x)dx , (n,m \in I) ∫ ma na f ( x ) d x = ( n − m ) ∫ 0 a f ( x ) d x , ( n , m ∈ I ) if f ( x ) f(x) f ( x ) is periodic with period 'a'.
3. Walli's Formula:
( a ) \bold{(a)} ( a )
∫ 0 π / 2 sin n x d x = ∫ 0 π / 2 cos n x d x = ( n − 1 ) ( n − 3 ) … ( 1 or 2 ) n ( n − 2 ) … ( 1 or 2 ) k \int_{0}^{\pi/2} \sin^n x \, dx = \int_{0}^{\pi/2} \cos^n x \, dx = \frac{(n-1)(n-3)\ldots(1 \text{ or } 2)}{n(n-2)\ldots(1 \text{ or } 2)}k ∫ 0 π /2 sin n x d x = ∫ 0 π /2 cos n x d x = n ( n − 2 ) … ( 1 or 2 ) ( n − 1 ) ( n − 3 ) … ( 1 or 2 ) k
Where k = = { π 2 if n i s e v e n 1 if n i s o d d = \begin{cases}
\frac{\pi}{2} &\text{if } n \, is \, even \\
1 &\text{if } n \, is \, odd
\end{cases} = { 2 π 1 if n i s e v e n if n i s o dd
( b ) \bold{(b)} ( b )
∫ 0 π / 2 sin n x cos m x d x = [ ( m − 1 ) ( m − 3 ) … ( 1 or 2 ) ] [ ( n − 1 ) ( n − 3 ) … ( 1 or 2 ) ] ( m + n ) ( m + n − 2 ) … ( 1 or 2 ) \int_{0}^{\pi/2} \sin^n x \cos^m x \, dx = \frac{[(m-1)(m-3)\ldots(1 \text{ or } 2)][(n-1)(n-3)\ldots(1 \text{ or } 2)]}{(m+n)(m+n-2)\ldots(1 \text{ or } 2)} ∫ 0 π /2 sin n x cos m x d x = ( m + n ) ( m + n − 2 ) … ( 1 or 2 ) [( m − 1 ) ( m − 3 ) … ( 1 or 2 )] [( n − 1 ) ( n − 3 ) … ( 1 or 2 )]
Where k = = { π 2 if m , n a r e e v e n ( m , n ∈ N ) 1 if o t h e r w i s e = \begin{cases}
\frac{\pi}{2} &\text{if } m, n \, are \, even (m,n \in N) \\
1 &\text{if } otherwise
\end{cases} = { 2 π 1 if m , n a re e v e n ( m , n ∈ N ) if o t h er w i se
4. Derivative of Antiderivative Function (Newton-Leibnitz Formula):
If h ( x ) h(x) h ( x ) and g ( x ) g(x) g ( x ) are differentiable functions of x x x then,
d d x ( ∫ g ( x ) h ( x ) f ( t ) d t ) = f [ h ( x ) ] ⋅ h ′ ( x ) − f [ g ( x ) ] ⋅ g ′ ( x ) \frac{d}{dx} \left( \int_{g(x)}^{h(x)} f(t) \, dt \right) = f[h(x)] \cdot h'(x) - f[g(x)] \cdot g'(x) d x d ( ∫ g ( x ) h ( x ) f ( t ) d t ) = f [ h ( x )] ⋅ h ′ ( x ) − f [ g ( x )] ⋅ g ′ ( x )
5. Definite Integral as Limit of a Sum:
∫ a b f ( x ) d x = lim n → ∞ h [ f ( a ) + f ( a + h ) + f ( a + 2 h ) + … + f ( a + ( n − 1 ) ‾ h ) ] = lim n → ∞ h ∑ r = 1 n − 1 f ( a + r h ) \int_{a}^{b} f(x) \, dx = \lim\limits_{n \to \infty} h \left[ f(a) + f(a + h) + f(a + 2h) + \ldots + f(a + \overline{(n-1)}h) \right] = \lim\limits_{n \to \infty} h \displaystyle\sum_{r=1}^{n-1} f(a+rh) ∫ a b f ( x ) d x = n → ∞ lim h [ f ( a ) + f ( a + h ) + f ( a + 2 h ) + … + f ( a + ( n − 1 ) h ) ] = n → ∞ lim h r = 1 ∑ n − 1 f ( a + r h )
where, ( b − a ) = n h (b - a) = nh ( b − a ) = nh .
If a = 0 a = 0 a = 0 and b = 1 b = 1 b = 1 then, lim n → ∞ h ∑ r = 0 n − 1 f ( r h ) = ∫ 0 1 f ( x ) d x \lim\limits_{n \to \infty} h \displaystyle\sum_{r=0}^{n-1} f(rh) = \int_{0}^{1} f(x) \, dx n → ∞ lim h r = 0 ∑ n − 1 f ( r h ) = ∫ 0 1 f ( x ) d x where nh = 1
OR
lim n → ∞ ( 1 n ) ∑ r = 1 n − 1 f ( r n ) = ∫ 0 1 f ( x ) d x \lim\limits_{n \to \infty} (\frac{1}{n}) \displaystyle\sum_{r=1}^{n-1} f(\frac{r}{n}) = \int_{0}^{1} f(x) \, dx n → ∞ lim ( n 1 ) r = 1 ∑ n − 1 f ( n r ) = ∫ 0 1 f ( x ) d x
6. Estimation of Definite Integral:
( a ) \bold{(a)} ( a ) If f ( x ) f(x) f ( x ) is continuous in [ a , b ] [a, b] [ a , b ] and its range in this interval is [ m , M ] [m, M] [ m , M ] , then
m ( b − a ) ≤ ∫ a b f ( x ) d x ≤ M ( b − a ) m(b - a) \leq \int_{a}^{b} f(x) \, dx \leq M(b - a) m ( b − a ) ≤ ∫ a b f ( x ) d x ≤ M ( b − a )
( b ) \bold{(b)} ( b ) If f ( x ) ≤ g ( x ) f(x) \leq g(x) f ( x ) ≤ g ( x ) for a ≤ x ≤ b a \leq x \leq b a ≤ x ≤ b then
∫ a b f ( x ) d x ≤ ∫ a b g ( x ) d x \int_{a}^{b} f(x) \, dx \leq \int_{a}^{b} g(x) \, dx ∫ a b f ( x ) d x ≤ ∫ a b g ( x ) d x
( c ) \bold{(c)} ( c )
∫ a b ∣ f ( x ) ∣ d x ≥ ∣ ∫ a b f ( x ) d x ∣ \int_{a}^{b} |f(x)| \, dx \geq \left| \int_{a}^{b} f(x) \, dx \right| ∫ a b ∣ f ( x ) ∣ d x ≥ ∫ a b f ( x ) d x
( d ) \bold{(d)} ( d ) If f ( x ) ≥ 0 f(x) \geq 0 f ( x ) ≥ 0 on the interval [ a , b ] [a,b] [ a , b ] , then
∫ a b f ( x ) d x ≥ 0 \int_{a}^{b} f(x) \, dx \geq 0 ∫ a b f ( x ) d x ≥ 0
( e ) \bold{(e)} ( e ) If f ( x ) f(x) f ( x ) and g ( x ) g(x) g ( x ) are two continuous functions on [ a , b ] [a, b] [ a , b ] then
∣ ∫ a b f ( x ) g ( x ) d x ∣ ≤ ( ∫ a b f 2 ( x ) d x ) ( ∫ a b g 2 ( x ) d x ) |\int_{a}^{b} f(x)g(x) \, dx| \leq \sqrt{\left( \int_{a}^{b} f^2(x) \, dx \right) \left( \int_{a}^{b} g^2(x) \, dx \right)} ∣ ∫ a b f ( x ) g ( x ) d x ∣ ≤ ( ∫ a b f 2 ( x ) d x ) ( ∫ a b g 2 ( x ) d x )
7. Some Standard Results:
( a ) \bold{(a)} ( a )
∫ 0 π / 2 log ( sin x ) d x = ∫ 0 π / 2 log ( cos x ) d x = − π 2 log 2 \int_{0}^{\pi/2} \log(\sin x) \, dx = \int_{0}^{\pi/2} \log(\cos x) \, dx = -\frac{\pi}{2} \log 2
∫ 0 π /2 log ( sin x ) d x = ∫ 0 π /2 log ( cos x ) d x = − 2 π log 2
( b ) \bold{(b)} ( b )
∫ a b { x } d x = b − a 2 , a , b ∈ I \int_{a}^{b} \{x\} \, dx = \frac{b - a}{2}, \quad a, b \in \mathbb{I} ∫ a b { x } d x = 2 b − a , a , b ∈ I
( c ) \bold{(c)} ( c )
∫ a b 1 x d x = log ∣ b a ∣ \int_{a}^{b} \frac{1}{x} \, dx = \log \left| \frac{b}{a} \right| ∫ a b x 1 d x = log a b
( d ) \bold{(d)} ( d )
∫ a b ∣ x ∣ x d x = ∣ b ∣ − ∣ a ∣ \int_{a}^{b} \frac{|x|}{x} \, dx = |b| - |a| ∫ a b x ∣ x ∣ d x = ∣ b ∣ − ∣ a ∣
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