Some properties of definite integrals: We use integrals to calculate areas of irregular shapes in a two-dimensional space. We use definite integrals in industries in real life, and calculus is used in engineering. Engineers use integrals to determine the shape and height of a building that needs to be constructed or the length of power cable required to connect the two substations. 

In Mathematics, we start with a function defined on an interval and find its derivative. But there are several situations, practical and theoretical, where it is necessary to reverse this process, i.e. the derivative of a function is known, and we are required to know the function. Here, we use integrals to find the function.

Types of Integrals

Now let us learn everything about integrals and their types here:

1. Indefinite Integrals

The formula that gives the antiderivatives is called the indefinite integral of the function, and the process of evaluating the integral is called integration.

If \(F\left( x \right)\) is the primitive of \(f\left( x \right)\), then \(F\left( x \right) + c\) is also a primitive of \(f\left( x \right)\), where \(c\) is a constant.
i.e.,  \(\int {f\left( x \right)} \,dx = F\left( x \right) + c\)

2. Definite Integrals

Unlike the indefinite integral, the definite integral has a unique value. It is formulated as
\(\int\limits_a^b {f\left( x \right)\,dx = } F\left( b \right) – F\left( a \right)\)
Where,
 1. \(F\) is primitive of \(f\)
2. \(a\) is a lower limit of \(f\left( x \right)\), and \(b\) is an upper limit of \(f\left( x \right)\), 
3. \(F\left( a \right)\) and \(F\left( b \right)\) are the values of primitive at the lower limit and upper limit respectively.

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Definite Integral Formula

If a continuous function \(f\left( x \right)\) is defined on the closed interval \(\left[ {a,\,b} \right]\) and \(F\left( x \right)\) is an indefinite integral of \(f\left( x \right)\) and \(x=a\) and \(x=b\) be the two given values of \(x\), then \(\left[ {F\left( b \right) – F\left( a \right)} \right]\) is called the definite integral of \(f\left( x \right)\) between the limits \(a\) and \(b\).

Notation: It is denoted by \(\int\limits_a^b {f\left( x \right)\,dx} \) and read as integral of \(f\left( x \right)\) from \(a\) to \(b\).
Here \(‘a’\) is called the lower limit and \(‘b’\) the upper limit of integration.

Therefore, \(\int\limits_a^b {f\left( x \right)\,dx}  = \left[ {F\left( x \right)} \right]_a^b = F\left( b \right) – F\left( a \right)\) where \(\int {f\left( x \right)} \,dx = F\left( x \right)\)

Note:

1. After integration, we first substitute \(x = {\text{upper}}\,{\text{limit}}\) and then subtract \(x = {\text{lower}}\,{\text{limit}}\) from it.
2. An arbitrary constant \(‘c’\) will get eliminated while applying the limits. Therefore, while evaluating definite integrals, arbitrary constant \(‘c’\) is not added.

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Properties of Definite Integral with Examples

As the value of a definite integral \(\int\limits_a^b {f\left( x \right)\,dx} \) is independent of the variable of integration \(x,x\) can be treated as a dummy variable (or symbol) i.e., it can be replaced by any other variable (or symbol) having the same range.

Properties

1. \(\int\limits_a^b {f\left( x \right)\,dx}  = \int\limits_a^b {f\left( t \right)\,dt} \)
   The value of the definite integral does not change with the change of argument (variable of integration) provided the limit of integration remains same.
   Proof:
   If \(\int {f\left( x \right)} \,dx = F\left( x \right) + c\) then \(\int {f\left( t \right)} \,dx = F\left( t \right) + c\) and
   \(\int\limits_a^b {f\left( x \right)\,dx}  = \left[ {F\left( x \right) + c} \right]_a^b\)
   \( = F\left( b \right) – F\left( a \right)\)
   \(\therefore \,\int\limits_a^b {f\left( t \right)\,dt}  = \left[ {F\left( t \right) + c} \right]_a^b\)
   \( = F\left( b \right) – F\left( a \right) = \int\limits_a^b {f\left( x \right)\,dx} \)
   Thus, \(\int\limits_a^b {f\left( x \right)\,dx}  = \int\limits_a^b {f\left( t \right)\,dt}\) or \(\int\limits_a^b {f\left( y \right)\,dy} \)

Example:

Show that \(\int\limits_0^a {\frac{{dx}}{{{x^2} + {a^2}}}}  = \,\int\limits_0^a {\frac{{dt}}{{{t^2} + {a^2}}}} \)
Sol: \(\int\limits_0^a {\frac{{dx}}{{{x^2} + {a^2}}}}  = \left[ {\left( {\frac{1}{a}} \right)\,{{\tan }^{ – 1}}\left( {\frac{x}{a}} \right)} \right]_0^a\)
\( = \left( {\frac{1}{a}} \right)\left[ {{{\tan }^{ – 1}}\left( 1 \right) – {{\tan }^{ – 1}}\left( 0 \right)} \right] = \frac{\pi }{{4a}}\)
Also \(\int\limits_0^a {\frac{{dt}}{{{t^2} + {a^2}}}}  = \left[ {\left( {\frac{1}{a}} \right)\,{{\tan }^{ – 1}}\left( {\frac{t}{a}} \right)} \right]_0^a\)
\( = \left( {\frac{1}{a}} \right)\left[ {{{\tan }^{ – 1}}\left( 1 \right) – {{\tan }^{ – 1}}\left( 0 \right)} \right] = \frac{\pi }{{4a}}\)
Hence, \(\int\limits_0^a {\frac{{dx}}{{{x^2} + {a^2}}}}  = \,\int\limits_0^a {\frac{{dt}}{{{t^2} + {a^2}}}} \, = \,\frac{\pi }{{4a}}\,\,\)

2. \(\int\limits_a^a {f\left( x \right)\,dx}  = 0\)
Proof: \(\int\limits_a^a {f\left( x \right)\,dx}  = F\left( a \right) – F\left( a \right)\)
\(\therefore \,\int\limits_a^a {f\left( x \right)\,dx}  = 0\,\)

3. \(\int\limits_a^b {f\left( x \right)\,dx}  = \,\, – \int\limits_b^a {f\left( x \right)\,dx} \)
Proof: \(\int\limits_a^b {f\left( x \right)\,dx}  = F\left( b \right) – F\left( a \right)\)
\( =  – \left( {F\left( a \right) – F\left( b \right)} \right)\)
\(\therefore \,\,\int\limits_a^b {f\left( x \right)\,dx}  = \, – \int\limits_b^a {f\left( x \right)\,dx} \)
Hence, the sign of definite integral changes when the order of limits is changed.

4. Evaluation of definite integral when \(f\left( x \right)\) is discontinuous or \(f\left( x \right)\) involves modulus and greatest integer function.
\(\,\int\limits_a^b {f\left( x \right)\,dx}  = \,\int\limits_a^c {f\left( x \right)\,dx} \, + \,\,\int\limits_c^b {f\left( x \right)\,dx} \)
where, \(c\) may lie inside or outside the interval \(\left[ {a,\,b} \right]\)
Note: This property is to be used when  \(f\) is piecewise continuous in \(\left[ {a,\,b} \right]\)
Proof:
 Let \(F’\left( x \right) = f\left( x \right)\)
\({\text{RHS = }}\int\limits_a^c {f\left( x \right)\,dx} \, + \,\,\int\limits_c^b {f\left( x \right)\,dx}  = F\left( c \right) – F\left( a \right) + F\left( b \right) – F\left( c \right)\)
\( = F\left( b \right) – F\left( a \right) = \int\limits_a^b {f\left( x \right)\,dx}  = {\text{LHS}}\)

Note:

1. For \(a < {c_1} < {c_2} < …. < {c_n} < b,\,{c_1},\,{c_2},\,{c_3},….{c_n}\) are points of discontinuities of \(f\) and all points lie in \(\left[ {a,\,b} \right]\) or outside the closed interval \(\left[ {a,\,b} \right]\) then
   \(\int\limits_a^b {f\left( x \right)\,dx}  = \,\int\limits_a^{{c_1}} {f\left( x \right)\,dx} \, + \,\,\int\limits_{{c_1}}^{{c_2}} {f\left( x \right)\,dx} \, + …. + \int\limits_{{c_n}}^b {f\left( x \right)\,dx} \)
2. This property will be used when the integration is discontinuous at a finite number of points or the integrand involves a modulus function, the greatest integer function, or inverse trigonometric functions (not in all cases).
3. A function having a finite number of points of discontinuities on \(\left[ {a,\,b} \right]\) is integrable on \(\left[ {a,\,b} \right]\)

Example: Evaluate \(\int\limits_0^e {\left| {x\,{\text{ln}}\,x} \right|\,dx} \)
Sol: \({\text{ln}}\,x < 0\) for \(0<x<1\) and \({\text{ln}}\,x > 0\) for \(1<x<e\)
\( \Rightarrow \int_0^e {\left| {x\,{\text{ln}}\,x} \right|\,dx\, = \,\int_0^{1\,} {\left| {x\,{\text{ln}}\,x} \right|} } \,dx + \int_1^{e\,} {\left| {x\,{\text{ln}}\,x} \right|} \,dx\)
\( =  – \left[ {\frac{{{x^2}}}{2}{\text{ln}}\,x – \frac{{{x^2}}}{4}} \right]_0^1\, + \left[ {\frac{{{x^2}}}{2}{\text{ln}}\,x – \frac{{{x^2}}}{4}} \right]_1^e\)
\( = \frac{1}{4} + \frac{{{e^2}}}{2} – \frac{{{e^2}}}{4} + \frac{1}{4}\)
\(\therefore \int_0^e {\left| {x\,{\text{ln}}\,x} \right|\,dx\, = \,} \frac{1}{2} + \frac{{{e^2}}}{4}\)

5. a. \(I = \int_0^a {f\left( x \right)dx = } \int_0^a {f\left( {a – x} \right)dx} \)
b. \(I = \int_0^a {f\left( x \right)dx = } \,\left( {\frac{1}{2}} \right)\int_0^a {\left( {f\left( x \right) + f\left( {a – x} \right)} \right)dx}\)
Proof:  Let \(x = a – t \Rightarrow dx =  – dt\)
\(\int_0^a {f\left( x \right)dx = } \,\int_a^0 {f\left( {a – t} \right)\left( { – dt} \right)}  = \int_0^a {f\left( {a – t} \right)\,dt} \)
Now \(I = \int_0^a {f\left( x \right)dx\,\,\,} \) and \(I = \int_0^a {f\left( a-x \right)dx\,\,\,} \) on adding, we get
\(\therefore \,I = \left( {\frac{1}{2}} \right)\int_0^a {\left( {f\left( x \right) + f\left( {a – x} \right)} \right)dx\,\,\,} \)

6. a. \(I = \int_a^b {f\left( x \right)dx\, = } \int_a^b {f\left( {a + b – x} \right)dx\,} \)
b. \(I = \int_a^b {f\left( x \right)dx\, = \left( {\frac{1}{2}} \right)} \int_a^b {\left( {f\left( x \right) + f\left( {a + b – x} \right)} \right)dx\,}\)
Proof: Consider \(I = \int_a^b {f\left( {a + b – x} \right)dx\,} \)
Put \(a+b-x\) then \(dx=-dt\)
\(x = a \Rightarrow t = b\), and \(x = b \Rightarrow t = a\)
\( \Rightarrow \int_a^b {f\left( {a + b – x} \right)dx\, = \,\int_b^a {f\left( t \right)\,\left( { – dt} \right)} \,} \)
\( = \, – \int_b^a {f\left( t \right)\,dt} \)
\( = \, \int_a^b {f\left( t \right)\,dt} \)
\(\therefore \,\int_a^b {f\left( {a + b – x} \right)dx\, = \,\int_a^b {f\left( x \right)\,dx} \,} \)

7. \(\int_{ – a}^a {f\left( x \right)dx = } \int_0^a {\left[ {f\left( x \right) + f\left( { – x} \right)} \right]dx} = \left\{ \begin{gathered} 0\,,\,{\text{if}}\,f\left( { x} \right)\, {\text{is}}\,{\text{odd}}\, \hfill \\ 2\int_0^a {f\left( x \right)\,dx,\,\,{\text{if}}\,f\left( { x} \right)\, {\text{is}}\,{\text{even}}}  \hfill \\\end{gathered}  \right.\)
Proof: \(I = \int_{ – a}^a {f\left( x \right)dx = \int_{ – a}^0 {f\left( x \right)dx + \int_0^a {f\left( x \right)\,dx} } } \)
Replacing \(x\) by \(-x\) in the \({1^{{\text{st}}}}\) integral, we get
\(I = \int_{ – a}^0 {f\left( { – x} \right)\,d( – x)}  + \int_0^a {f\left( x \right)} \,dx\)
\(I = \int_{ 0}^a {f\left( { – x} \right)\,dx}  + \int_0^a {f\left( x \right)} \,dx\)
\( = \left\{ \begin{gathered} 0\,,\,\,\,\,{\text{if}}\,f\left( {- x} \right)\, =  – f(x)\, \hfill \\ 2\int_0^a {f\left( x \right)\,dx,\,\,\,\,\,{\text{if}}\,f\left( { – x} \right)\, = f(x)}  \hfill \\\end{gathered}  \right.\)
\(\therefore I = \left\{ \begin{gathered} 0\,,\,{\text{if}}\,f\left( { x} \right)\, {\text{is}}\,{\text{odd}}\, \hfill \\ 2\int_0^a {f\left( x \right)\,dx,\,\,{\text{if}}\,f\left( { x} \right)\, {\text{is}}\,{\text{even}}}  \hfill \\\end{gathered}  \right.\)

8. \(\int_0^{2a} {f\left( x \right)\,dx = \int_0^a {\left[ {f\left( x \right) + f\left( {2a – x} \right)} \right]dx = } }   \left\{ \begin{gathered} 0\,,\,\,\,\,{\text{if}}\,f\left( {2a – x} \right)\, =  – f(x)\, \hfill \\ 2\int_0^a {f\left( x \right)\,dx,\,\,\,\,\,{\text{if}}\,f\left( {2a – x} \right)\, = f(x)}  \hfill \\\end{gathered}  \right.\)
Proof: \(I = \int_0^{2a} {f\left( x \right)dx = } \int_0^a {f\left( x \right)\,dx + \int_a^{2a} {f\left( x \right)dx} } \)
\( = \int_0^a {f\left( x \right)dx + \int_0^a {f\left( {2a – x} \right)dx} } \)
\( = \int_0^a {\left( {f\left( x \right)dx + f\left( {2a – x} \right)} \right)} dx\)
\(\therefore I = \left\{ \begin{gathered} 0\,,\,\,\,\,{\text{if}}\,f\left( {2a – x} \right)\, =  – f\left( x \right)\, \hfill \\ 2\int_0^a {f\left( x \right)\,dx,\,\,\,\,\,{\text{if}}\,f\left( {2a – x} \right)\, = f\left( x \right)}  \hfill \\\end{gathered}  \right.\)

9. If \(f\left( x \right)\) is discontinuous at \(x=a\) then \(\int_0^{2a} {f\left( x \right)dx = \int_0^a {\left\{ {f\left( {a – x} \right) + f\left( {a + x} \right)} \right\}} } dx\)
Proof: \(\int_0^{2a} {f\left( x \right)dx = \int_0^a {f\left( x \right)} } dx + \int_a^{2a} {f\left( x \right)\,dx} \)
\( = \int_0^a {f\left( {a – x} \right)\,dx + \int_0^a {f\left( {a + x} \right)dx} } \)
\(\therefore \,\int_0^{2a} {f\left( x \right)dx = \int_0^a {\left[ {f\left( {a – x} \right) + f\left( {a + x} \right)} \right]\,dx} } \)

10.\(\int_a^b {f\left( x \right)\,dx = \left\{ \begin{gathered} \int_{a \pm c}^{b \pm c} {f\left( {x \mp c} \right)} \,dx \hfill \\ k\int_{\frac{a}{k}}^{\frac{b}{k}} {f\left( {kx} \right)\,dx}  \hfill \\ \left( {\frac{1}{k}} \right)\int_{ak}^{bk} {f\left( {\frac{x}{k}} \right)\,dx,\,k \ne 0}  \hfill \\\end{gathered}  \right.} \)

11. Change of limits of the integral to the limits \(0\) and \(1\).
\(\int_a^b {f\left( x \right)dx} = \left( {b – a} \right)\int_0^1 {f\left( {a + \left( {b – a} \right)x} \right)dx} \)
Proof:
Substitute \(t = \frac{{x – a}}{{b – a}} \Rightarrow x = a + \left( {b – a} \right)t\)
\(dx = \left( {b – a} \right)\,dt\)
\(\therefore \,\int_a^b {f\left( x \right)dx = \,\int_0^1 {\left( {b – a} \right)f\left( {a + \left( {b – a} \right)t} \right)\,dt} } \)
This formula is useful when we want to change the limits of integration from \(0\) to \(1\).

What is a Periodic Function?

A function \(f\left( x \right)\) is periodic, if there exists a real positive and finite constant \(T\) independent of \(x\) such that \(f\left( {x + T} \right) = f\left( x \right),\forall x \in {D_f}\) provided, \(\left( {x + T} \right) \in {D_f}\left( {{\text{domain}}\,{\text{of}}\,f} \right)\). Since \(…. = f\left( {x – 2T} \right) = f\left( {x – T} \right) = f\left( x \right) = f\left( {x + T} \right) = f\left( {x + 2T} \right)…..\)
Therefore, we conclude \(f\left( x \right) = f\left( {x + nT} \right)\forall n \in Z\), for which \(x + nT \in {D_f}\).

The least positive value of such \(T\) (if exists), is called the period/principal period or fundamental period of \(f\left( x \right)\).

e.g., \(f\left( x \right) = \sin \,x,\,f\left( x \right) = \tan \,x\) are periodic functions with periods \(2\pi \) and \(\pi \), respectively.

Note:

1. \(\left[ {\sin \,x} \right]\) and \(\left\{ {\sin \,x} \right\}\) are periodic functions with the period \(2\pi \) (where \(\left[ . \right]\) and \(\left\{ . \right\}\) be respectively denotes G.I.F. and F.P.F.)
2. \(\left\{ x \right\}\) is a periodic function with period \(‘1’\).
3. \(\left| {\sin \,x} \right|,\,\left| {\cos \,x} \right|\) are periodic functions with the period \(\pi \).

Non-Periodic Functions: The functions which are not periodic are called non-periodic functions.

Properties of Definite Integral Periodic Function

Let \(f\left( x \right)\) be a periodic function with the period \(T\). Then \(f\left( {x + nT} \right) = f\left( x \right),n \in N\).

1. \(\int_0^{nT} {f\left( x \right)\,dx = n\int_0^T {f\left( x \right)\,dx} } \)
Proof:
\(\int_0^{nT} {f\left( x \right)\,dx = \int_0^T {f\left( x \right)\,dx} }  + \int_T^{2T} {f\left( x \right)\,dx}  + …. + \int_{\left( {n – 1} \right)T}^{nT} {f\left( x \right)\,dx} \)
\( = \int_0^T {f\left( x \right)\,dx + \int_0^T {f\left( {x + T} \right)\,dx} }  + …. + \int_0^T {f\left( {x+\left( {n – 1} \right)T} \right)\,dx} \)
\(\therefore \,\int_0^{nT} {f\left( x \right)\,dx = n\int_0^T {f\left( x \right)\,dx} } \)

2. \(\int_{mT}^{nT} {f\left( x \right)\,dx = \left( {n – m} \right)\,\int_0^T {f\left( x \right)\,dx\,} } \) where \(m,\,n \in Z\)
Proof:
\(\int_{mT}^{nT} {f\left( x \right)\,dx = \,\int_{mT – mT}^{nT – mT} {f\left( {x + mT} \right)\,dx\,} } \)
\( = \int_0^{\left( {n – m} \right)T} {f\left( x \right)\,dx} \)
\(\therefore \,\int_{mT}^{nT} {f\left( x \right)\,} dx = \left( {n – m} \right)\,\int_0^T {f\left( x \right)\,dx\,} \)

3. \(\int_{a + nT}^{b + nT} {f\left( x \right)\,} dx = \int_a^b {f\left( x \right)\,} dx\,\left( {n \in N} \right)\)
Proof:
Apply translation properties, we get
\(\int_{a + nT}^{b + nT} {f\left( x \right)\,} dx = \int_{a + nT – nT}^{b + nT – nT} {f\left( {x + nT} \right)\,} dx\)
\( = \int_a^b {f\left( {x + nT} \right)\,} dx\)
\(\therefore \,\int_{a + nT}^{b + nT} {f\left( x \right)\,} dx = \int_a^b {f\left( x \right)\,} dx\)

4. \(\int_a^{a + T} {f\left( x \right)\,} dx = \int_0^T {f\left( x \right)\,} dx\), which is independent of \(‘a’\).
Proof:
\(\int_a^{a + T} {f\left( x \right)\,} dx = \int_a^T {f\left( x \right)\,} dx + \int_T^{a + T} {f\left( x \right)\,} dx\)
\( = \,\int_a^T {f\left( x \right)\,} dx + \int_{T – T}^{a + T – T} {f\left( {x + T} \right)\,} dx = \int_a^T {f\left( x \right)\,} dx + \int_0^a {f\left( x \right)\,dx} \)
\(\therefore \,\int_a^{a + T} {f\left( x \right)\,} dx = \,\int_0^T {f\left( x \right)\,dx\,} \)
This is independent of \(‘a’\).

5. \(\int_a^{a + nT} {f\left( x \right)\,} dx = \,n\int_0^T {f\left( x \right)\,dx\,} ,\,n \in N\)
Proof:
\(\int_a^{a + nT} {f\left( x \right)\,} dx = \,\int_a^0 {f\left( x \right)\,dx\,}  + \int_0^{nT} {f\left( x \right)\,dx\,}  + \int_{nT}^{a + nT} {f\left( x \right)\,} dx\)
\( = \int_a^0 {f\left( x \right)\,} dx + n\int_0^T {f\left( x \right)\,} dx + \int_0^a {f\left( x \right)\,} dx\)
\(\therefore \int_a^{a + nT} {f\left( x \right)\,} dx = n\int_0^T {f\left( x \right)\,} dx\)

Periodic Function when \({e^x}\) is involved

Theorem: Let \(f\left( x \right)\) be a periodic function with the period \(T\). Then \(\frac{{\int_0^{nT} {{e^x}} f\left( x \right)\,dx}}{{\int_0^T {{e^x}} f\left( x \right)\,dx}} = \frac{{1 – {e^{nT}}}}{{1 – {e^T}}}\)
Proof:
Let \(I = \int_0^{nT} {{e^x}} f\left( x \right)\,dx\)
Then \(I = \int_0^{T} {{e^x}} f\left( x \right)\,dx + \int_T^{2T} {{e^x}} f\left( x \right)\,dx + …. + \int_{\left( {n – 1} \right)T}^{nT} {{e^x}} f\left( x \right)\,dx\)
\( = \int_0^T {{e^x}} f\left( x \right)\,dx + \int_0^T {{e^{x + T}}} f\left( {x + T} \right)\,dx + …. + \int_0^T {{e^{x + \left( {n – 1} \right)T}}} f\left( {x + \left( {n – 1} \right)T} \right)\,dx\)
\( = \int_0^T {{e^x}} f\left( x \right)\,dx + {e^T}\int_0^T {{e^x}} f\left( x \right)\,dx + {e^{2T}}\int_0^T {{e^x}} f\left( x \right)\,dx + …. + {e^{\left( {n – 1} \right)T}}\int_0^T {{e^x}} f\left( x \right)\,dx\)
\( = \left[ {1 + {e^T} + {e^{2T}} + …. + {e^{\left( {n – 1} \right)T}}} \right]\int_0^T {{e^x}} f\left( x \right)\,dx\)
\( = \left[ {\frac{{1 – {e^{nT}}}}{{1 – {e^T}}}} \right]\int_0^T {{e^x}} f\left( x \right)\,dx\)
\(\therefore \frac{{\int_0^{nT} {{e^x}} f\left( x \right)\,dx}}{{\int_0^T {{e^x}} f\left( x \right)\,dx}} = \frac{{1 – {e^{nT}}}}{{1 – {e^T}}}\)

Solved Examples on Some Properties of Definite Integrals

Q.1. Evaluate \(\sum\nolimits_{m = 0}^{2013} {\int_0^1 x {{\left( {1 – x} \right)}^m}\,dx} \)
Sol: \(I = \int_0^1 x {\left( {1 – x} \right)^m}\,dx = \int_0^1 {{{\left( {1 – x} \right)}}} {x^m}\,dx\)
\( = \int_0^1 {\left( {{x^m} – {x^{m + 1}}} \right)} \,dx\)
\( = \frac{1}{{m + 1}} – \frac{1}{{m + 2}}\)
\({\sum\limits_{m = 0}^{2013} {\left( {\frac{1}{{m + 1}} – \frac{1}{{m + 2}}} \right)} }\)
\( = 1 – \frac{1}{{2015}}\)
\( = \frac{{2014}}{{2015}}\)

Q.2. Compute the integral \(\int_{-\sqrt 2 }^{\sqrt 2 } {\frac{{2{x^7} + 3{x^6} – 10{x^5} – 7{x^3} – 12{x^2} + x + 1}}{{{x^2} + 2}}} dx\)
Sol: Break the integrand as the sum of two functions, one being even and the other being odd.
\(I = \int_{ – \sqrt 2 }^{\sqrt 2 } {\frac{{3{x^6} – 12{x^2} + 1}}{{{x^2} + 2}}} dx + \int_{ – \sqrt 2 }^{\sqrt 2 } {\frac{{2{x^7} – 10{x^5} – 7{x^3} + x}}{{{x^2} + 2}}} dx\)
It is obvious that the integral of \({2^{{\text{nd}}}}\) integrand is an odd function and \({1^{{\text{st}}}}\) is an even function.
\(\therefore I = 2\int_0^{\sqrt 2 } {\frac{{3{x^6} – 12{x^2} + 1}}{{{x^2} + 2}}} dx + 0\,\)
\( = 2\int_0^{\sqrt 2 } {\frac{{3{x^2}\left( {{x^4} – 4} \right) + 1}}{{{x^2} + 2}}} dx\)
\( = 2\int_0^{\sqrt 2 } {3{x^2}\left( {{x^2} – 2} \right)} \,dx + \,2\,\int_0^{\sqrt 2 } {\frac{1}{{{x^2} + 2}}} dx\)
\( = \left[ {\left( {\frac{{6{x^5}}}{5} – 4{x^3}} \right)} \right]_0^{\sqrt 2 } + \frac{2}{{\sqrt 2 }}\left[ {{{\tan }^{ – 1}}\frac{x}{{\sqrt 2 }}} \right]_0^{\sqrt 2 }\)
\( = \frac{\pi }{{2\sqrt 2 }} – \frac{{16}}{5}\sqrt 2 \)

Q.3. Evaluate \(\int_0^\pi  {\frac{{x\,dx}}{{1 + {{\cos }^2}\,x}}} \)
Sol:
\(I = \int_0^\pi  {\frac{{x\,dx}}{{1 + {{\cos }^2}\,x}}} \)
\( = \int_0^\pi  {\frac{{\left( {\pi  – x} \right)\,dx}}{{1 + {{\cos }^2}\,\left( {\pi  – x} \right)}}} \)
\( = \int_0^\pi  {\frac{{\left( {\pi  – x} \right)\,dx}}{{1 + {{\cos }^2}\,x}}} \)
On adding both, we get
\(2I = \int_0^\pi  {\frac{{\pi \,dx}}{{1 + {{\cos }^2}\,x}}} \)
\( \Rightarrow I = \frac{\pi }{2}\int_0^\pi  {\frac{{\,dx}}{{1 + {{\cos }^2}\,x}}} \)
\( = \frac{{2\pi }}{2}\mathop \smallint \nolimits_0^{\frac{\pi }{2}} \frac{{dx}}{{1 + {{\cos }^2}x}}\)
\( = \pi \int_0^{\frac{\pi }{2}} {\frac{{{{\sec }^2}\,x\,dx}}{{2 + {{\tan }^2}\,x}}} \)
Put \(\tan \,x = t\)
\(I = \pi \int_0^\infty  {\frac{{dt}}{{{t^2} + 2}} = \frac{\pi }{{\sqrt 2 }}} {\tan ^{ – 1}}\left( {\frac{t}{{\sqrt 2 }}} \right)_0^\infty \)
\(\therefore \,I = \frac{{{\pi ^2}}}{{2\sqrt 2 }}\)

Q.4. Let \(I = \int_0^\pi  {\frac{{f\left( x \right)\sin \,2x\,\sin \,\left( {\frac{\pi }{2}\,\cos \,x} \right)}}{{2x – \pi }}} dx\). If \(f\left( x \right) = {x^2}\) then \(I=\)
Sol: If \(f\left( x \right) = {x^2}\) then \(f\left( {\frac{\pi }{2} + x} \right) – f\left( {\frac{\pi }{2} – x} \right)\)
\( = {\left( {\frac{\pi }{2} + x} \right)^2} – {\left( {\frac{\pi }{2} – x} \right)^2} = 4 \cdot \frac{\pi }{2} \cdot x\)
\( \Rightarrow I = \int_0^{\frac{\pi }{2}} {\frac{{\sin \,2x\,\sin \,\left( {\frac{\pi }{2}\,\sin \,x} \right)}}{{2x}} \times 2\pi \,x\,dx} \)
\( = \pi \int_0^{\frac{\pi }{2}} {\sin \,2x\,\sin \,\left( {\frac{\pi }{2}\,\sin \,x} \right)\,dx}  = \pi  \cdot \frac{8}{{{\pi ^2}}}\)
\(\therefore \,I = \frac{8}{\pi }\)

Q.5. If \(\int_0^1 {\frac{{\sin \,x}}{{1 + x}}\,dx = \alpha } \) then what is \(\int_{4\pi – 2}^{4\pi } {\frac{{\sin \frac{x}{2}}}{{4\pi + 2 – x}}dx} \)
Sol: Let \(I = \int_{4\pi  – 2}^{4\pi } {\frac{{\sin \,\frac{x}{2}}}{{4\pi  + 2 – x}}\,dx} \)
Let \(t = \frac{{x – a}}{{b – a}} = \frac{{x – 4\pi }}{{ – 2}}\)
\( \Rightarrow dt =  – \frac{1}{2}dx\) and \(4\pi  – x = 2t\)
\(I = \int_0^1 {\frac{{\sin \,\left( {2\pi  – t} \right)}}{{2t + 2}}\left( { – 2dt} \right)} \)
\( = \int_0^1 {\frac{{\sin \,t}}{{1 + t}}\,dt} \)
\( = \alpha \)

Summary

If \(f\) is defined on an interval \(I\) and \(F\) is a function such that \(F’\left( x \right) = f\left( x \right)\) for all \(x \in I\), then \(F\) is said to be the primitive or antiderivative of \(f\). If \(f\left( x \right)\) is continuous and is defined on the closed interval \(\left[ {a,\,b} \right]\) and \(F\left( x \right)\) is an indefinite integral of \(f\left( x \right)\) and \(x=a\) and \(x=b\) be the two given values of \(x\), then \(\int_a^b {f\left( x \right)dx = \left[ {F\left( x \right)} \right]} _a^b = F\left( b \right) – F\left( a \right)\) where \(\int {f\left( x \right)\,dx = F\left( x \right)} \). If \(f\left( x \right) = f\left( {x + nT} \right)\forall n \in Z\) , for which \(\left( {x + nT} \right)\, \in {D_f}\).Then, \(f\left( x \right)\) is said to be a periodic function. The least positive value of such \(T\), (if exists), is called the principal period or fundamental period of \(f\left( x \right)\).

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Frequently Asked Questions (FAQs)

The frequently asked questions on some properties of definite integrals are given below:

Q.1. What is meant by definite integral?
Ans: If \(f\left( x \right)\) is continuous and is defined on the closed interval \(\left[ {a,\,b} \right]\) and \(F\left( x \right)\) is an indefinite integral of \(f\left( x \right)\) and \(x=a\) and \(x=b\) be the two given values of \(x\), then \(\left[ {F\left( b \right) – F\left( a \right)} \right]\) is called the definite integral of \(f\left( x \right)\) between the limits \(a\) and \(b\).
Therefore, \(\int_a^b {f\left( x \right)\,dx = \left[ {F\left( x \right)} \right]} _a^b = F\left( b \right) – F\left( a \right)\) where \(\int {f\left( x \right)\,dx = F\left( x \right)} \)

Q.2. What are the rules for definite integrals?
Ans: Rules of definite integrals are as follows:
1. \(\int_a^b {f\left( x \right)\,dx = } \int_a^b {f\left( t \right)} \,dt\)
2. \(\int_a^a {f\left( x \right)\,dx = } 0\)
3. \(\int_a^b {f\left( x \right)\,dx = }  – \int_b^a {f\left( x \right)\,dx} \)
4. \(\int_a^b {f\left( x \right)\,dx = } \int_a^c {f\left( x \right)\,dx}  + \int_c^b {f\left( x \right)\,dx} \) where \(c\) may lie inside or outside the interval \(\left[ {a,\,b} \right]\).
5. a. \(I = \int_0^a {f\left( x \right)dx = } \int_0^a {f\left( {a – x} \right)dx} \)
b. \(I = \int_0^a {f\left( x \right)dx} = \left( {\frac{1}{2}} \right)\int_0^a {\left( {f\left( x \right) + f\left( {a – x} \right)} \right)dx} \)
6. a. \(I = \int_a^b {f\left( x \right)dx} = \int_a^b {f\left( {a + b – x} \right)dx} \)
b. \(I = \int_a^b {f\left( x \right)dx} = \left( {\frac{1}{2}} \right)\int_a^b {\left( {f\left( x \right) + f\left( {a + b – x} \right)} \right)dx} \)
7. \(\int_{ – a}^a {f\left( x \right)dx = } \int_0^a {\left[ {f\left( x \right) + f\left( { – x} \right)} \right]dx} = \left\{ \begin{gathered} 0\,,\,{\text{if}}\,f\left( { x} \right)\, {\text{is}}\,{\text{odd}}\, \hfill \\ 2\int_0^a {f\left( x \right)\,dx,\,\,{\text{if}}\,f\left( { x} \right)\, {\text{is}}\,{\text{even}}}  \hfill \\\end{gathered}  \right.\)
8. \(\int_0^{2a} {f\left( x \right)\,dx = \int_0^a {\left[ {f\left( x \right) + f\left( {2a – x} \right)} \right]dx = } }   \left\{ \begin{gathered} 0\,,\,\,\,\,{\text{if}}\,f\left( {2a – x} \right)\, =  – f(x)\, \hfill \\ 2\int_0^a {f\left( x \right)\,dx,\,\,\,\,\,{\text{if}}\,f\left( {2a – x} \right)\, = f(x)}  \hfill \\\end{gathered}  \right.\)
9. If \(f\left( x \right)\) is discontinuous at \(x=a\) then \(\int_0^{2a} {f\left( x \right)dx = \int_0^a {\left\{ {f\left( {a – x} \right) + f\left( {a + x} \right)} \right\}} } dx\)
10. \(\int_a^b {f\left( x \right)\,dx = \left\{ \begin{gathered} \int_{a \pm c}^{b \pm c} {f\left( {x \mp c} \right)} \,dx \hfill \\ k\int_{\frac{a}{k}}^{\frac{b}{k}} {f\left( {kx} \right)\,dx}  \hfill \\ \left( {\frac{1}{k}} \right)\int_{ak}^{bk} {f\left( {\frac{x}{k}} \right)\,dx,\,k \ne 0}  \hfill \\\end{gathered}  \right.} \)
11. \(\int_a^b {f\left( x \right)dx = \left( {b – a} \right)\,\int_0^1 {f\left( {a + \left( {b – a} \right)x} \right)\,dx} } \)

Q.3. What is the application of definite integrals?
Ans: Definite integrals can be used to determine the mass of an object if its density function is known. Work can also be calculated from integrating a force function, or when counting the force of gravity, as in a pumping problem.

Q.4.  How do you integrate upper and lower limits?
Ans: The definite integral has a unique value. It is formulated as
\(\int_a^b {f\left( x \right)dx = } F\left( b \right) – F\left( a \right)\)
Where,
i)  \(F\) is primitive of \(f\).
ii)  \(a\) is a lower limit of \(f\left( x \right)\), and \(b\) is an upper limit of \(f\left( x \right)\).
iii) \(F\left( a \right)\) and \(F\left( b \right)\) are the values of primitive at the lower limit and upper limit, respectively.

Q.5. What is the value of a definite integral?
Ans: The value of a definite integral of \(f\left( x \right)\) on \(\left[ {a,\,b} \right]\) is the difference between the primitive or antiderivative of \(f\left( x \right)\) evaluated at the upper limit and the lower limit of integration.

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