Mathematics

Integral Calculus

Posted by Asok K Nadhani on

Last Updated on: 8th September 2024, 09:18 am

Integral Calculus

Integral Calculus

Integral Calculus is the branch of mathematics that deals with integrals, especially the methods of  ascertaining indefinite integrals, and applying them to the solution of differential equations

Integration

Integration is the reverse process of getting the original function f(x) from its derivative f’(x)

  • Indefinite integral :  Indefinite integral of a function f is a differentiable function F whose derivative is equal to the original function f. This can be stated symbolically as F′ = f.
  • Definite integral : Definite integral of F(x) is derived for specified interval (e.g between x=a and x=b

Standard Formula of integration

\displaystyle \int{{{{x}^{n}}dx=\frac{{{{x}^{{n+1}}}}}{{n+1}}+C\ \ \ where\ \ n\ne -1}}
\displaystyle \int{{{{e}^{x}}dx={{e}^{x}}+C}}
\displaystyle \int{{dx=x+C}}
\displaystyle \int{{\frac{1}{x}dx={{{\log }}_{e}}\ \left| {\,x\,} \right|}}+C
\displaystyle \int{{{{e}^{{ax}}}dx=\frac{{{{e}^{{ax}}}}}{a}+C}}
\displaystyle \int{{{{a}^{x}}dx=\frac{{{{a}^{x}}}}{{{{{\log }}_{e}}a}}}}+C

Simple Integration Problems & Solutions

 (i)   \displaystyle \int{{{{x}^{5}}dx=\frac{{{{x}^{{5+1}}}}}{{5+1}}+C=\frac{1}{6}{{x}^{6}}+C}}         

(ii)   \displaystyle \int{{{{x}^{{200}}}dx=\frac{{{{x}^{{200+1}}}}}{{200+1}}=\frac{{{{x}^{{201}}}}}{{201}}+C}}

(iii)   \displaystyle \int{{xdx=\frac{{{{x}^{{1+1}}}}}{{1+1}}=\frac{{{{x}^{2}}}}{2}+C}}                      

(iv)  \displaystyle \int{{1\ dx=x+c}}

(v)   \displaystyle \int{{{{x}^{{-9}}}dx=\frac{{{{x}^{{-9+1}}}}}{{-9+1}}=\frac{{{{x}^{{-8}}}}}{{-8}}=\frac{1}{{-8{{x}^{8}}}}+C}} 

(vi)   \displaystyle \int{{{{x}^{{-4/5}}}dx=\frac{{{{x}^{{-4/5+1}}}}}{{-4/5+1}}=\frac{{{{x}^{{1/5}}}}}{{1/5}}=5{{x}^{{1/5}}}+C}}

(vii)  \displaystyle \int{{{{a}^{3}}\sqrt{{{{x}^{4}}}}\ dx=\int{{{{x}^{{4/3}}}\ dx=\frac{3}{7}{{x}^{{7/3}}}+C}}}}  [Note: [(\displaystyle \frac{4}{3}) + 1 = \displaystyle \frac{7}{3}]

(viii) \displaystyle \int{{{{{\left( {x+1/x} \right)}}^{2}}dx=\int{{{{x}^{2}}\ dx+2\int{{dx+\frac{1}{{{{x}^{2}}}}\int{{dx}}}}}}}}\displaystyle =\frac{{{{x}^{3}}}}{3}+2x+\frac{{{{x}^{{-2+1}}}}}{{-2+1}}=\frac{{{{x}^{3}}}}{3}+2x-\frac{1}{x}+c    (\displaystyle \frac{1}{{{{x}^{2}}}}={{x}^{{-2}}})

(ix) \displaystyle \int{{\left[ {{{x}^{3}}+5{{x}^{2}}-6x-8} \right]dx=\int{{{{x}^{3}}dx+}}}}\int{{5{{x}^{2}}dx-}}\int{{6xdx-}}\int{{8dx}}\displaystyle =\frac{{{{x}^{4}}}}{4}+\frac{{5{{x}^{3}}}}{3}-\frac{{6{{x}^{2}}}}{2}-8x+C=\frac{{{{x}^{4}}}}{4}+\frac{5}{3}{{x}^{3}}-3{{x}^{2}}-8x+C

(x) \displaystyle \int{{\frac{{6{{x}^{7}}+8{{x}^{3}}-5{{x}^{2}}}}{{{{x}^{4}}}}dx=}}\int{{\left[ {6{{x}^{3}}+\frac{8}{x}-5{{x}^{{-2}}}} \right]dx}}=\int{{6{{x}^{3}}dx+\int{{8\frac{1}{x}dx-}}\int{{5{{x}^{{-2}}}dx}}}}\displaystyle =\frac{{6{{x}^{{3+1}}}}}{4}+8\ \log \ x-5\frac{{{{x}^{{-2+1}}}}}{{-2+1}}+C=\frac{3}{2}{{x}^{4}}+8\ \log x+\frac{5}{x}+C

Integration Common Formula

a) \displaystyle \int{{\frac{{dx}}{{{{x}^{2}}-{{a}^{2}}}}=\frac{1}{{2a}}\log \ \left| {\,\frac{{x-a}}{{x+a}}\,} \right|}}
b) \displaystyle \int{{\frac{{dx}}{{{{a}^{2}}-{{x}^{2}}}}=\frac{1}{{2a}}\log \ \left| {\,\frac{{a+x}}{{a-x}}\,} \right|}}
c) \displaystyle \int{{\frac{{dx}}{{\sqrt{{{{a}^{2}}-{{x}^{2}}}}}}={{{\sin }}^{{-1}}}\frac{x}{a}+C}}
d) \displaystyle \int{{\frac{{dx}}{{\sqrt{{{{x}^{2}}+{{a}^{2}}}}}}=\log \ \left| {\,x+\sqrt{{{{x}^{2}}+{{a}^{2}}}}\,} \right|}}+C
e) \displaystyle \int{{\frac{{dx}}{{\sqrt{{{{x}^{2}}-{{a}^{2}}}}}}=\log \ \left| {\,x+\sqrt{{{{x}^{2}}-{{a}^{2}}}}\,} \right|}}+C
f) \displaystyle \int{{\sqrt{{{{a}^{2}}-{{x}^{2}}dx}}}}\ =\frac{x}{2}\sqrt{{{{a}^{2}}-{{x}^{2}}}}+\frac{{{{a}^{2}}}}{2}{{\sin }^{{-1}}}\frac{x}{a}+C
g) \displaystyle \int{{\sqrt{{{{x}^{2}}+{{a}^{2}}}}}}\ dx\ =\ \frac{x}{2}\sqrt{{{{x}^{2}}+{{a}^{2}}}}+\frac{{{{a}^{2}}}}{2}\ \log \ \left| {\,x+\sqrt{{{{x}^{2}}+{{a}^{2}}}}\,} \right|+C

Integration Problems & Solutions

(i) \displaystyle \int{{\frac{1}{{{{x}^{2}}-25}}dx}}
\displaystyle =\int{{\frac{1}{{{{x}^{2}}-{{5}^{2}}}}=\frac{1}{{2\times 5}}\ \log }}\left| {\frac{{x-5}}{{x+5}}} \right|+C
\displaystyle =\frac{1}{{10}}\log \left| {\frac{{x-5}}{{x+5}}} \right|+C

(ii) \displaystyle \int{{\frac{1}{{7-{{x}^{2}}}}dx\ =}}
\displaystyle \int{{\frac{1}{{{{{\left( {\sqrt{7}} \right)}}^{2}}-{{x}^{2}}}}dx\ =\frac{1}{{2\sqrt{7}}}\log \frac{{\sqrt{{7+x}}}}{{\sqrt{{7-x}}}}+C}}

(iii) \displaystyle \int{{\frac{{{{e}^{x}}}}{{{{e}^{{2x}}}-4}}dx\ =}}\int{{\frac{{dz}}{{{{z}^{2}}-{{2}^{2}}}}\left( {let\ z={{e}^{x}},\ \ so\ dz={{e}^{x}}dx} \right)\ =\frac{1}{4}\log \frac{{(z-2)}}{{(z+2)}}=\frac{1}{4}}}\ \log \left[ {\frac{{{{e}^{x}}-2}}{{{{e}^{x}}+2}}} \right]+C

(iv) \displaystyle \int{{\frac{1}{{x+\sqrt{{{{x}^{2}}-1}}}}dx\ =}}\int{{\frac{{x-\sqrt{{{{x}^{2}}-1}}}}{{\left( {x+\sqrt{{{{x}^{2}}-1}}} \right)\left( {x-\sqrt{{{{x}^{2}}-1}}} \right)}}dx\ =\ }}
\displaystyle \int{{\left( {x-\sqrt{{{{x}^{2}}-1}}} \right)\ dx}}=
\displaystyle \frac{{{{x}^{2}}}}{2}-\frac{x}{2}\sqrt{{{{x}^{2}}-1}}+\frac{1}{2}\ \log \left( {x+\sqrt{{{{x}^{2}}-1}}} \right)+C

Integration Methods

When the function is not in standard integrable form, it may be evaluated by 1. Integration by Substitution, 2. Integration by Parts, 3. Integration through Partial Fractions

Integration by Substitution

By suitable substitution, the variable x in ∫f(x) dx is altered into another variable t so that the integrand f(x) is changed into F(t) which is some standard integral or algebraic sum of standard integrals.

Integration by Substitution – Problems and solutions

i) \displaystyle \int{{\frac{1}{{3x-5}}dx}}
Step 1 : Let 3x – 5 = t, so, \displaystyle \frac{{dt}}{{dx}}=\frac{d}{{dx}}\left( {3x-5} \right)
Step 2 : So, \displaystyle \frac{{dt}}{{dx}}=\frac{d}{{dx}}\left( {3x-5} \right)
Step 3 : so, \displaystyle I=\int{{\frac{1}{{3x-5}}dx}}
Step 4 : \displaystyle =\frac{1}{3}\ \int{{\frac{3}{{3x-5}}dx}}=\frac{1}{3}\int{{\frac{{dt}}{t}=\frac{1}{3}}}\,\,\log \,\left| {\,t\,} \right|+C
Step 5 : substituting value of t\displaystyle =\frac{1}{3}\log \ \left| {\,\,3x-5\,} \right|+C

ii) \displaystyle \int{{\frac{{{{x}^{3}}}}{{{{{\left( {{{x}^{2}}+1} \right)}}^{4}}}}dx}}
Step 1 : Let x2 + 1 = t,    so, 2x dx = dt, or x dx = \displaystyle \frac{{dt}}{2}
Step 2 : so, \displaystyle \int{{\frac{{{{x}^{3}}}}{{{{{\left( {{{x}^{2}}+1} \right)}}^{4}}}}dx}}=\displaystyle \int{{\frac{{{{x}^{2}}.\,x}}{{{{t}^{4}}}}\,dx}}
Step 3 : \displaystyle =\frac{1}{2}\int{{\frac{{t-1}}{{{{t}^{4}}}}}}\,dt
Step 4 : \displaystyle =\frac{1}{2}\int{{\frac{{dt}}{{{{t}^{3}}}}}}-\frac{1}{2}\,\,\int{{\frac{{dt}}{{{{t}^{4}}}}}}
Step 5 : \displaystyle =\frac{1}{2}\times \frac{{{{t}^{{-3+1}}}}}{{\left( {-3+1} \right)}}-\frac{1}{2}\times \frac{{{{t}^{{-4+1}}}}}{{\left( {-4+1} \right)}}
Step 6 : \displaystyle =\frac{1}{2}.\frac{{{{t}^{{-2}}}}}{{-2}}-\frac{1}{2}\times \frac{{{{t}^{{-3}}}}}{{-3}}
Step 7 : \displaystyle =\frac{1}{6}{{t}^{{-3}}}-\frac{1}{4}{{t}^{{-2}}}
Step 8 : \displaystyle =\frac{1}{6}{{\left( {{{x}^{2}}+1} \right)}^{{-3}}}-\frac{1}{4}{{\left( {{{x}^{2}}+1} \right)}^{{-2}}}+c

Integration by Parts

Integration by Parts is the process of breaking the function into product of two functions and the integration of the product of two functions 

For Example, if u and v are two functions of x, then
\displaystyle \int{{\left( {uv} \right)dx=u.}}\int{{v\,dx-\int{{\left( {\frac{{du}}{{dx}}.\int{{v\,dx}}} \right)\,dx}}}}

So, integral of the product of two functions = (first function ´integral of the second) – (integral of (differential of first ´ integral of the second function).

Integration by Parts – Problems & Solutions

(i) \displaystyle \int{{x{{e}^{x}}\,\,dx}}

Step 1 : \displaystyle =x\int{{{{e}^{x}}dx-}}\displaystyle \int{{}}{\displaystyle \frac{d}{{dx}}(x)\displaystyle \int{{}}ex dx}dx
Step 2 : \displaystyle =x{{e}^{x}}-\int{{1.{{e}^{x}}dx=x{{e}^{x}}-\,{{e}^{x}}+c}}

(ii) \displaystyle \int{{\log \,\,x\,\,dx}}

Step 1 : \displaystyle =\log \,x\,\int{{x\,\,dx-}}\int{{\frac{d}{{dx}}\,\left( {\log \,x} \right)\int{{x\,\,dx}}}}
Step 2 :  \displaystyle =x\,\,\log \,x-\int{{\frac{1}{x}\,.\,\frac{{{{x}^{2}}}}{2}}}\,dx
Step 3 : \displaystyle =x\,\log \,x-\frac{1}{2}\int{{xdx}}
Step 4 :  \displaystyle =x\,\log \,x-\frac{1}{2}\,.\,\frac{{{{x}^{2}}}}{2}+c
Step 5 : \displaystyle =x\,\,\log \,x-\frac{{{{x}^{2}}}}{4}+c

Integration of Partial Fractions

Integrals of type∫ [p(x) ¸ g(x)] dx can be integrated by resolving the integrand into partial fractions

Steps :1.Check degree of p(x) and g(x). 2. If degree of p(x) > degree of g(x), then divide p(x) by g(x) till its degree is less, i.e. put in the form [p(x) ¸ g(x)] = r(x) + [f(x) ¸ g(x)], where degree of f (x) <degree of g(x).

3. When the denominator contains non – repeated linear factors, i.e g(x) = (x – \displaystyle \alpha 1) (x – \displaystyle \alpha 2) .. (x – \displaystyle \alpha n), write f (x) and g (x) as :

[f (x) ¸g (x)] = [(A1)¸( x – \displaystyle \alpha 1)] + [(A2)¸( x – \displaystyle \alpha 2)] … [(An)¸( x – \displaystyle \alpha n)], where A 1, A 2, A 3, ………… A n are constants to be determined by comparing the coefficients of various powers of x on both sides after taking LCM

Integration of Partial Fractions – Problems and Solutions

Ex.: Evaluate \displaystyle \int{{\frac{{\left( {3x+2} \right)dx}}{{\left( {x-2} \right)\left( {x-3} \right)}}}}. Here degree of the numerator is lower than that of the denominator; the denominator contains non−repeated linear factor.

Step 1: So  we may break the expression and write as
\displaystyle \frac{{\left( {3x+2} \right)}}{{\left( {x-2} \right)\left( {x-3} \right)}}=\frac{A}{{\left( {x-2} \right)}}+\frac{B}{{\left( {x-3} \right)}}

Step 2  : so 3x + 2 = A (x − 3) + B (x − 2)

Step 3 : putting, x = 2, we get (3 × 2) + 2 = A × (2 − 3) + B × (2 − 2), or A = − 8

Step 4 : putting, x = 3, we get (3 × 3) + 2 = A × (3 − 3)+ B × (3 − 2), or B = 11

Putting the values of A & B, we get

Step 5 : \displaystyle \int{{\frac{{\left( {3x+2} \right)dx}}{{\left( {x-2} \right)\left( {x-3} \right)}}}}=-\,8\int{{\frac{{dx}}{{x-2}}}}\,+\,11\int{{\frac{{dx}}{{x-3}}}}

Step 6 : = – 8log (x − 2) + 11 log (x − 3) + c

Ex.2 : Evaluate \displaystyle \int{{\frac{{dx}}{{x\left( {{{x}^{3}}+1} \right)}}}}

Step 1 : Let z = x3, so, dz = 3x2dx.

Rewriting the expression in terms of z, we get   

Step 2 :  \displaystyle \int{{\frac{{dx}}{{x\left( {{{x}^{3}}+1} \right)}}=\,\,\frac{1}{3}\int{{\frac{{dz}}{{z\left( {z+1} \right)}}}}}}

Step 3 : \displaystyle =\frac{1}{3}\int{{\,\left[ {\frac{1}{z}-\frac{1}{{z+1}}} \right]}}\,dz

Step 4 : =\displaystyle \frac{1}{3}{logz-log(z-1)}\displaystyle =\frac{1}{3}\log \frac{z}{{z-1}}

Step 5 : \displaystyle =\frac{1}{3}\,\,\log \,\frac{{{{x}^{3}}}}{{{{x}^{3}}-1}}\,\,+c

Definite Integration

Definite integration is the process of evaluating the difference in the values of the integral of a function f(x),  between the two assigned values of an independent variable. It is expressed as:
\displaystyle \int\limits_{a}^{b}{{f\left( x \right)dx\,=\,g\left( b \right)-g\left( a \right)}}

Here, ‘a’ and ‘b’ are the lower and upper limits of integration respectively. The arbitrary constant C which is invariably attached to the model of an indefinite integral, disappears here.

Definite Integrals – Important Properties

(i)   \displaystyle \int\limits_{a}^{b}{{f\left( x \right)}}\,dx=\int\limits_{a}^{b}{{f\left( t \right)}}\,dt

(ii) \displaystyle \int\limits_{a}^{b}{{f\left( x \right)}}\,dx=-\int\limits_{b}^{a}{{f\left( x \right)}}\,dx

(iii) \displaystyle \int\limits_{a}^{b}{{f\left( x \right)}}\,dx=\int\limits_{b}^{c}{{f\left( x \right)}}\,dx+\int\limits_{c}^{b}{{f\left( x \right)}}\,dx,wher,a<c<b

(iv) \displaystyle \int\limits_{0}^{a}{{f\left( x \right)}}\,dx=\int\limits_{0}^{a}{{f\left( {a-x} \right)}}\,dx

(v) When \displaystyle f\left( x \right)=f\left( {a+x} \right),\int\limits_{0}^{{na}}{{f\left( x \right)dx\,=\,\,n}}\int\limits_{0}^{a}{{f\left( x \right)dx}}

(vi) \displaystyle \int\limits_{{-a}}^{a}{{f\left( x \right)dx\,=\,}}2\int\limits_{0}^{a}{{f\left( x \right)dx}}, if f(–x) = f(x), or = 0, if f(– x) = – f(x)

Definite Integration – Problems and Solutions

(i) \displaystyle \int\limits_{1}^{2}{{xdx=\left[ {\frac{{{{x}^{2}}}}{2}} \right]}}_{1}^{2}=\frac{{{{2}^{2}}}}{2}-\frac{{{{1}^{2}}}}{2}=\frac{4}{2}-\frac{1}{2}=2-\frac{1}{2}=\frac{3}{2}

(ii)  \displaystyle \int\limits_{0}^{1}{{{{x}^{4}}dx=\left[ {\frac{{{{x}^{5}}}}{5}} \right]}}_{0}^{1}=\,\frac{1}{5}-\,\frac{0}{5}\,=\,\frac{1}{5}\,-\,0\,=\,\frac{1}{2}

(iii) \displaystyle \int\limits_{4}^{9}{{\sqrt{{xdx}}=\left[ {\frac{{{{x}^{{\frac{1}{2}+1}}}}}{{^{{\frac{1}{2}+1}}}}} \right]_{4}^{9}}}\displaystyle =\left[ {\frac{2}{3}\,\,{{x}^{{3/2}}}} \right]_{4}^{9}=\,\,\frac{2}{3}\,\,{{x}^{9}}^{{^{{3/2}}}}-\frac{2}{3}\,\,{{x}^{4}}^{{^{{3/2}}}}

   = (2 × 19)/3 = \displaystyle \frac{{38}}{3}

(iv)  \displaystyle \int\limits_{3}^{5}{{\frac{{dx}}{x}}}=\left[ {\log \,x} \right]_{3}^{5}=\,\,\log \,\,5\,-\,\,\log 3\,\,=\,\log \,\frac{5}{3}

(v)  \displaystyle \int\limits_{1}^{2}{{{{e}^{{2x}}}\,dx=\left[ {\frac{{{{e}^{{2x}}}}}{2}} \right]_{1}^{2}=\,\,\frac{1}{2}\,\left( {{{e}^{{2.2}}}-{{e}^{{2.1}}}} \right)\,=\,\,\frac{1}{2}\,\left( {{{e}^{4}}-{{e}^{2}}} \right)}}

(v) \displaystyle \int\limits_{1}^{2}{{\frac{{{{x}^{2}}+2x+5}}{x}\,dx}}

Step 1 : \displaystyle =\,\int\limits_{1}^{2}{{\left( {x+2+\frac{5}{x}} \right)\,dx=\left[ {\frac{{{{x}^{2}}}}{2}+2x+5\,\,\log \,x} \right]}}_{1}^{2}

Step 2 : \displaystyle =\left( {\frac{{{{2}^{2}}}}{2}+2.2+5\,\,\log \,\,2} \right)-\left( {\frac{1}{2}+2.2+5\,\,\log \,\,1} \right)

Step 3 : \displaystyle =\left( {6+5\,\,\log \,\,2} \right)-\left( {\frac{5}{2}} \right)\left( {as\,\,\log \,\,1\,\,=0} \right) ={-\displaystyle \left( {\frac{5}{2}} \right)+5 log 2=\displaystyle \left( {\frac{7}{2}} \right)+5 log 2}

(vi) \displaystyle \int\limits_{0}^{1}{{\frac{{1-x}}{{1+x}}\,dx}}

Step 1 : First Evaluate Indefinite Integral \displaystyle \int{{\frac{{1-x}}{{1+x}}}}\,dx=\int{{\left( {\frac{2}{{1+x}}-1} \right)}}=\,\,dx=\,2\int{{\frac{{dx}}{{1+x}}-}}\int{{dx}} = 2 log (1 + x) – x

Step 2 : Evaluate the definite integral Interval values between 0 and 1. = (2 log 2 – 1) – 2 log 1 = 2 log 2 – 1

(vii) \displaystyle \int\limits_{3}^{5}{{5{{x}^{4}}dx}}

Step 1 : \displaystyle =5.\mathop{{\left[ {\frac{{{{x}^{{4+1}}}}}{{4+1}}} \right]}}_{3}^{5}

Step 2 : \displaystyle \mathop{{\left[ {{{x}^{5}}} \right]}}_{3}^{5}

Step 3 : = 55 – 35 = 3125 – 243 = 2882.  

(viii) \displaystyle \int\limits_{1}^{e}{{\,\log \,\,x\,\,dx}}

Step 1: \displaystyle \left[ {x\,\,\log \,\,x-x} \right]_{1}^{e}

Step 2: = (e log e – e) – (log 1– 1) = e(log e –1 ) – (log 1 – 1) = e(1 – 1) – (0 –1) = 0 + 1 = 1

(ix) \displaystyle \int\limits_{2}^{4}{{{{{\left( {3x-2} \right)}}^{2}}}}dx

Step 1 : \displaystyle =\int\limits_{2}^{4}{{\left( {9{{x}^{2}}-12x+4} \right)dx}}

Step 2 : [{(9x3)/3}-{(12x2)/2}+4x]24= (3 × 43 – 6 × 42 + 4 × 4) – (3 × 23 –6 × 22 + 4 × 2) = (192 – 96 + 16) – (24 – 24 + 8) = 112 – 8 = 104

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