Indefinite Integration

Integration means summation. Integration originated during the course of finding the area of a plane figure. Integration is reverse of differentiation it is also called as antiderivative.

Standard formulas for Indefinite Integration

  1. \(\int{{{x}^{n}}dx=\frac{{{x}^{n+1}}}{n+1}}+c\)
  2. \(\int{{{a}^{x}}dx}=\frac{{{a}^{x}}}{\ell n\,a}+c\)
  3. \(\int{{{e}^{x}}}dx={{e}^{x}}+c\)
  4. \(\int{\frac{1}{x}}dx=\ell n\,x+c\)
  5. \(\int{\sin x\,dx=-\cos x+c}\)
  6. \(\int{\cos x\,dx=\sin x+c}\)
  7. \(\int{{{\sec }^{2}}x\,\,dx=\tan x+c}\)
  8. \(\int{\cos e{{c}^{2}}x\,\,dx=-\cot x\,+c}\)
  9. \(\int{\sec x\tan xdx=\sec x+c}\)
  10. \(\int{\cos ec\,\,x\,\,}\cot x\,dx=-\cos ec\,x+c\)
  11. \(\int{\frac{1}{\sqrt{1-{{x}^{2}}}}dx\begin{matrix} = \\ = \\ \end{matrix}}\left\{ \begin{matrix} {{\sin }^{-1}}x+c \\ -{{\cos }^{-1}}x+c \\ \end{matrix} \right.\)
  12. \(\int{\frac{1}{1+{{x}^{2}}}}dx\begin{matrix} = \\ = \\ \end{matrix}\left\{ \begin{matrix} {{\tan }^{-1}}x+c \\ -{{\cot }^{-1}}x+c \\ \end{matrix} \right.\).
  13. \(\frac{1}{x\sqrt{{{x}^{2}}-1}}=\left\{ \begin{matrix} {{\sec }^{-1}}x+c \\ -\cos e{{c}^{-1}}x+c \\ \end{matrix} \right.\)

⇒ Also Read – Introduction to Integration

Properties of Indefinite Integration

  1. \(k\int{f\left( x \right)dx=k\int{f\left( x \right)}}\)
  2. \(\int{\left( f\left( x \right)+g\left( x \right) \right)dx=\int{f\left( x \right)dx+\int{g\left( x \right)dx}}}\)

Illustration:

Q. \(\int{\frac{{{\left( 1+x \right)}^{3}}}{\sqrt[3]{{{x}^{2}}}}}dx=\int{\frac{1+3x+3{{x}^{2}}+{{x}^{3}}}{{{x}^{2/3}}}}dx\) \(=\int{{{x}^{2/3}}+3{{x}^{1/3}}+3{{x}^{4/3}}}+{{x}^{7/3}}dx\) \(=3{{x}^{1/3}}+\frac{9}{4}{{x}^{4/3}}+\frac{9}{7}{{x}^{7/3}}+\frac{3}{10}{{x}^{10/3}}+c\)

Q. \(\int{{{\sec }^{2}}x\cos e{{c}^{2}}x\,\,dx=\int{\frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\sin }^{2}}x\,{{\cos }^{2}}x}}dx}\) \(=\int{{{\sec }^{2}}x+\cos e{{c}^{2}}x\,\,dx}\) \(=\tan x-\cot x+c\)

Standard integral form \(f\left( ax+b \right)\) \(\int{{{\left( ax+b \right)}^{n}}dx=\frac{{{\left( ax+b \right)}^{n+1}}}{\left( n+1 \right)a}}\) \(\int{\frac{1}{ax+b}}=dx=\frac{\ell n\left| ax+b \right|}{a}+c\)

and so on…

we follow as below

If \(\int{f\left( x \right)}=F\left( x \right)+c\) then \(\int{f\left( ax+b \right)}=\frac{F\left( ax+b \right)}{a}+c\)

Methods of Integration

  1. Substitution
  2. Integration by partial fractions
  3. By parts
  4. Euler substitution
  5. Reduction method

Substitution method

[Integration by change of variable]

A new variable is introduced such that direct integration formula can be applied.

Problems on Indefinite Integration

1. \(\int{2x\,\,\sin \left( {{x}^{2}} \right)dx}\)

Put \({{x}^{2}}=t\)

the \(2x\,\,dx=dt\) \(\int{\sin t\,\,dt=-\cos t+c}\) \(=-\cos {{x}^{2}}+c\)

2. \(\int{{{\sin }^{3}}x{{\cos }^{5}}x\,\,dx}\) \(=\int{{{\sin }^{2}}x{{\cos }^{5}}x\sin x\,\,dx}\) \(=\int{\left( 1-{{\cos }^{2}}x \right)}{{\cos }^{5}}x\,\,\sin x\,\,dx\)

put \(\cos x=t\)

∴ \(\,\,\,-\sin x\,dx=dt\) \(=\int{\left( 1-{{t}^{2}} \right)}{{t}^{5}}\left( -dt \right)\) \(=\int{{{t}^{5}}-{{t}^{7}}}dt\) \(=-\left( \frac{{{t}^{6}}}{6}-\frac{{{t}^{8}}}{8} \right)\) \(=-\frac{{{\cos }^{6}}x}{6}+\frac{{{\cos }^{8}}x}{8}+c\)

Note: Type I

\(\int{{{\sin }^{m}}x{{\cos }^{n}}x\,\,dx}\)

Rule:

  1. If m, n one of them odd then substitute for even power.
  2. If both odd then substitute either of them.
  3. If both even use trigonometric identities.

Type II: Substitution of Trigonometric functions.

  1. \(\sqrt{{{a}^{2}}-{{x}^{2}}}\rightarrow{{}}x=a\sin \theta\)
  2. \(\sqrt{{{x}^{2}}+{{a}^{2}}}\rightarrow{{}}x=a\tan \theta\)
  3. \(\sqrt{{{x}^{2}}-{{a}^{2}}}\rightarrow{{}}x=a\sec \theta\)
  4. \(\sqrt{\frac{x-a}{b-x}}\rightarrow{{}}x=a{{\cos }^{2}}\theta +b{{\sin }^{2}}\theta\)
  5. \(\sqrt{\frac{x-a}{x-b}}\rightarrow{{}}x=a{{\sec }^{2}}\theta -b\)

Important formulae set for Indefinite Integration

  1. \(\int{\frac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}={{\sin }^{-1}}\left( \frac{x}{a} \right)+c\)
  2. \(\int{\frac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}=\ell n\left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|}\)
  3. \(\int{\frac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}=\ell n\left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|}\)
  4. \(\int{\frac{1}{{{a}^{2}}+{{x}^{2}}}}dx=\frac{1}{a}{{\tan }^{-1}}\left( \frac{x}{a} \right).\)
  5. \(\int{\frac{1}{{{a}^{2}}-{{x}^{2}}}}dx=\frac{1}{2a}\ell n\left| \frac{a+x}{a-x} \right|.\)
  6. \(\int{\frac{1}{{{x}^{2}}-{{a}^{2}}}}dx=\frac{1}{2a}\ell n\left| \frac{x-a}{x+a} \right|.\)
  7. \(\int{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx=\frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \frac{x}{a} \right).\)
  8. \(\int{\sqrt{{{a}^{2}}+{{x}^{2}}}}dx=\frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\frac{{{a}^{2}}}{2}\ell n\left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|\)
  9. \(\int{\sqrt{{{x}^{2}}-{{a}^{2}}}}dx=\frac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}+\frac{{{a}^{2}}}{2}\ell n\left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|\)

Illustration:

Q. \(\int{\frac{dx}{\sqrt{{{x}^{2}}+2x+2}}}=\int{\frac{dx}{\sqrt{{{\left( x+1 \right)}^{2}}+1}}}\) \(=\log \left| \left( x+1 \right)+\sqrt{{{\left( x+1 \right)}^{2}}+1} \right|.\)

Q. \(\int{\sqrt{5{{x}^{2}}+2x+3}}\,\,dx\)

here

\(5{{x}^{2}}+2x+3=5\left[ {{\left( x+\frac{1}{5} \right)}^{2}}+{{\left( \sqrt{\frac{14}{25}} \right)}^{2}} \right]\) \(\int{\sqrt{5{{x}^{2}}+2x+3}\,dx}=\sqrt{5}\int{\sqrt{{{\left( x+\frac{1}{5} \right)}^{2}}+{{\left( \sqrt{\frac{14}{25}} \right)}^{2}}}}\) \(=\sqrt{5}\frac{\left( x+\frac{1}{5} \right)}{2}\sqrt{{{\left( x+\frac{1}{5} \right)}^{2}}+{{\left( \sqrt{\frac{14}{15}} \right)}^{2}}}+\frac{14}{25.2}\log \left| \left( x+\frac{1}{5} \right)+\sqrt{{{\left( x+\frac{1}{5} \right)}^{2}}+\frac{14}{15}} \right|\)

Important forms

\(\int{\frac{px+q}{a{{x}^{2}}+bx+c}}dx\And \int{\frac{px+q}{\sqrt{a{{x}^{2}}+bx+c}}}\,dx\And \int{\left( px+q \right)}\sqrt{a{{x}^{2}}+bx+c}\,dx\)

We express \(px+q\) as \(m{{\left( a{{x}^{2}}+bx+c \right)}^{1}}+n\) then this gets changed to standard integral.

Illustration:

\(I=\int{\frac{x+1}{{{x}^{2}}+3x+4}}dx\) \(x+1=a{{\left( {{x}^{2}}+3x+4 \right)}^{1}}+b\) \(=2ax+\left( 3a+b \right)\)

∴ \(a=\frac{1}{2}\,\,\,\,b=\frac{-1}{2}\)

⇒ \(I=\int{\frac{1}{2}}\frac{{{\left( {{x}^{2}}+3x+4 \right)}^{1}}}{{{x}^{2}}+3x+4}-\frac{1}{2}\left( \frac{1}{{{x}^{2}}+3x+4} \right)\) \(=\frac{1}{2}\log \left( {{x}^{2}}+3x+4 \right)-\frac{1}{2}\int{\frac{1}{{{\left( x+\frac{3}{2} \right)}^{2}}+\frac{7}{4}}}\) \(=\frac{1}{2}\log \left( {{x}^{2}}+3x+4 \right)-\frac{1}{2}\frac{1}{\sqrt{\frac{7}{4}}}{{\tan }^{-1}}\frac{\left( x+\frac{3}{2} \right)}{\sqrt{\frac{7}{4}}}\) \(=\frac{1}{2}\log \left| {{x}^{2}}+3x+4 \right|-\frac{1}{\sqrt{7}}{{\tan }^{-1}}\left( \frac{2x+3}{\sqrt{7}} \right)\)

Biquadratic Substitutions

1. \(I=\int{f\left( x+\frac{1}{x} \right)}\left( 1-\frac{1}{{{x}^{2}}} \right)dx\)

Put \(x+\frac{1}{x}=t\)

2. \(I=\int{f\left( x-\frac{1}{x} \right)}\left( 1+\frac{1}{{{x}^{2}}} \right)dx\)

Put \(\left( x-\frac{1}{x} \right)=t\)

Illustration:

\(\int{\frac{1+{{x}^{2}}}{1+{{x}^{4}}}}\,dx=\int{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)}{\left( \frac{1}{{{x}^{2}}}+{{x}^{2}} \right)}}\,dx\) \(=\int{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)dx}{{{\left( x-\frac{1}{x} \right)}^{2}}+2}}\) \(=\int{\frac{d\left( x-\frac{1}{x} \right)}{{{\left( x-\frac{1}{x} \right)}^{2}}+2}}\) \(=\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{\left( x-\frac{1}{x} \right)}{\sqrt{2}} \right)\)

Integration by Partial Fraction

Integrals of rational functions can be evaluated by splitting them into partial fractions.

\(\frac{f\left( x \right)}{g\left( x \right)}\) where \(f\And g\) are polynomial is called as rational function.

If the degree of \(f<\), the degree of g it is called proper fraction.

Otherwise, it is an improper fraction.

Then \(\frac{f\left( x \right)}{g\left( x \right)}=h\left( x \right)+\frac{d\left( x \right)}{g\left( x \right)}\) degree of d < degree of g

Cases:

1. When \(g\left( x \right)\) is expressed as product of non-repeating linear factors.

\(g\left( x \right)=\left( x-{{a}_{1}} \right)\left( x-{{a}_{2}} \right)….\left( x-{{a}_{n}} \right)\) then

\(\frac{f}{g}=\frac{{{A}_{1}}}{x-{{a}_{1}}}+\frac{{{A}_{2}}}{x-{{a}_{2}}}+….\frac{{{A}_{n}}}{x-{{a}_{n}}}.\)

2. Some factors of g are repeating then \(g\left( x \right)={{\left( x-a \right)}^{k}}\left( x-{{a}_{1}} \right)\left( x-{{a}_{2}} \right)…\) \(\frac{f}{g}=\frac{{{A}_{1}}}{\left( x-a \right)}+\frac{{{A}_{2}}}{{{\left( x-a \right)}^{2}}}+…..\frac{{{A}_{k}}}{{{\left( x-a \right)}^{n}}}+…\)

3. \(g\left( x \right)\) has a quadratic term then \(\frac{f}{g}=\frac{Ax+B}{a{{x}^{2}}+bx+c}\)

Where A & B are constants determined by comparing coefficients.

\(\int{\frac{dx}{\left( x+1 \right)\left( x-2 \right)}}=\int{\frac{a}{x+1}+\frac{b}{x-2}}\,dx\) \(=\int{\frac{-\frac{1}{3}}{x+1}}+\int{\frac{\frac{1}{2}}{x-2}}\,dx\) \(=-\frac{1}{3}\ell n\left| x+1 \right|+\frac{1}{2}\ell n\left| x-2 \right|\) \(\int{\frac{2{{x}^{2}}-1}{\left( x-1 \right){{\left( x+1 \right)}^{2}}}dx=\int{\frac{A}{x-1}+\frac{B}{\left( x+1 \right)}}+\frac{C}{{{\left( x+1 \right)}^{2}}}}\) \(2{{x}^{2}}-1=A{{\left( x+1 \right)}^{2}}+B\left( x+1 \right)\left( x+1 \right)+C\left( x-1 \right)\)

Put \(x=1\Rightarrow A=\frac{1}{2}\) \(x=-1\Rightarrow \,\,C=\frac{-1}{2}\) \(B=\frac{3}{2}\) \(I=\int{\frac{\frac{1}{2}}{\left( x-1 \right)}}dx+\frac{3}{2}\int{\frac{dx}{\left( x+1 \right)}}+\left( \frac{-1}{2} \right)\int{\frac{dx}{{{\left( x-2 \right)}^{2}}}}\) \(=\frac{1}{2}\ell n\left| x-1 \right|+\frac{3}{2}\left( n\left| x+1 \right. \right)+\frac{1}{2\left( x+2 \right)}+c\) \(\int{\frac{dx}{\left( x+2 \right)\left( {{x}^{2}}+1 \right)}}=\frac{A}{x+2}+\frac{Bx+C}{{{x}^{2}}+1}\)

Comparing by \(x=-2\Rightarrow \,\,A=\frac{1}{5}\) by \({{x}^{2}}\) coeff. \(B=\frac{-1}{2}\) \(C=\frac{2}{5}\) \(I=\frac{1}{5}\left| \frac{dx}{x+2} \right.-\frac{1}{5}\int{\frac{x}{{{x}^{2}}+1}}\,dx\) \(I=\frac{1}{5}\ell n\left| x+2 \right|-\frac{1}{10}\ell n\left( {{x}^{2}}+1 \right)+\frac{2}{5}{{\tan }^{-1}}\left( x \right)\)

Integration of Trigonometric Functions

Type 1: \(I=\int{{{\sin }^{m}}x{{\cos }^{n}}xdx}\)

1. If m –odd put \(\cos x=t\)

2. If n odd put \(\sin x=t\)

3. If m, n rationales then put \(\tan x=t\)

4. If both even then use reduction method

\(Q\int{\frac{{{\cos }^{3}}x}{{{\sin }^{6}}x}dx=\int{\frac{1-{{t}^{2}}}{{{t}^{6}}}dt}}\)

Where \(t=\sin x\) \(=\int{{{t}^{-6}}-{{t}^{-4}}dt}\) \(=-\frac{1}{5si{{n}^{5}}x}+\frac{1}{3{{\sin }^{3}}x}+c\)

Type 2: \(\int{\frac{dx}{a\cos x+b\sin x+c}}\)

Put \(t=\tan \left( \frac{x}{2} \right)\)

Illustration

\(\int{\frac{dx}{2+\sin x}}\) \(\Rightarrow t=\tan \left( \frac{x}{2} \right)\) \(dx=\frac{2dt}{1+{{t}^{2}}}\) \(=\int{\frac{\frac{2dt}{1+{{t}^{2}}}}{2+\frac{2t}{1+{{t}^{2}}}}}\) \(\Rightarrow \int{\frac{dt}{{{t}^{2}}+t+1}}\) \(=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2t+1}{\sqrt{3}} \right)\) \(=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2\tan \frac{x}{2}+1}{\sqrt{3}} \right)+c\)

Type II

\(\int{\frac{dx}{a{{\cos }^{2}}x+b{{\sin }^{2}}x}}=\int{\frac{dx}{a+b{{\sin }^{2}}x}}\) \(\int{\frac{1}{a+b{{\cos }^{2}}x}}\,dx,\int{\frac{1}{{{\left( a\sin x+b\cos x \right)}^{2}}}}\,dx\) or \(\int{\frac{1}{a+b{{\sin }^{2}}x+{{\cos }^{2}}x}}\,dx\)

Rule:

Divide both numerator & denominator by \({{\cos }^{2}}x\)

Illustration:

\(\int{\frac{1}{3-4{{\sin }^{2}}x}}\,dx=\int{\frac{\frac{1}{{{\cos }^{2}}x}}{\frac{3}{{{\cos }^{2}}x}-\frac{4{{\sin }^{2}}x}{{{\cos }^{2}}x}}}\) \(=\int{\frac{{{\sec }^{2}}x}{3{{\sec }^{2}}x-4{{\tan }^{2}}x}}\,dx\) \(=\int{\frac{dt}{3\left( 1+{{t}^{2}} \right)-4{{t}^{2}}}}\) (Since, by \(\tan x=t\,\,\,{{\sec }^{2}}x\,dx=dt\))

\(=\int{\frac{dt}{3-{{t}^{2}}}}\) \(=\int{\frac{1}{{{\left( \sqrt{3} \right)}^{2}}-{{t}^{2}}}}\) \(=\frac{1}{2\sqrt{3}}\ell n\left| \frac{\sqrt{3}+t}{\sqrt{3}-t} \right|\) \(=\frac{1}{2\sqrt{3}}\log \left| \frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x} \right|\)

Definite Integrals

Definite Integrals

Choose n divide [a, b] in n parts of width \(h=\frac{b-a}{n}=\). partition of interval putting (n – 1) parts in between

\(\int\limits_{a}^{b}{f\left( x \right)}dx=\underset{h\to 0}{\mathop{\lim }}\,h\,\sum\limits_{n=0}^{n-1}{f\left( a+rh \right)}\)

Definition of Definite Integral as limit of a solution

\(=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( a \right)+f\left( a+h \right)+f\left( a+2h \right)+…..+f\left( a+\left( n-1 \right)h \right) \right]\) \(nh=b-a\)

Integrals using limits

Evaluate as Limit of a Sum

\(I=\int\limits_{a}^{b}{x\,\,dx\Rightarrow f\left( x \right)=x\,\,\,\,h=\frac{b-a}{n}}\) \(f\left( a \right)=a\) \(f\left( a+h \right)=a+h\) \(I=h\left[ a+\left( a+h \right)+….\left( a+\left( n-1 \right)h \right) \right]\) look for nh

\(=h\left[ \left( na \right)+h+2h+3h….\left( n-1 \right)h \right]\) \(=hna+\frac{{{h}^{2}}n\left( n-1 \right)}{2}\) \(=\left( b-a \right)a+\frac{{{h}^{2}}{{n}^{2}}}{2}-\frac{{{\left( b-a \right)}^{2}}}{2}=\frac{{{b}^{2}}-{{a}^{2}}}{2}\) \(\sum\limits_{r=1}^{n}{n}=\frac{n\left( n+1 \right)}{2}\) \(\sum\limits_{r=1}^{n}{{{n}^{2}}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\) \(\sum\limits_{r=1}^{n}{{{n}^{3}}}={{\left[ \frac{n\left( n+1 \right)}{2} \right]}^{2}}\)

GP … \(a+ar+a{{r}^{2}}….a{{r}^{n-1}}=\frac{a\left( 1-{{r}^{n}} \right)}{1-r}\) \(\sin \alpha +\sin \left( \alpha +\beta \right)+\sin \left( \alpha +2\beta \right)….\sin \left( \alpha +\left( n-1 \right)\beta \right)\) \(=\frac{\sin \left( n\frac{\beta }{2} \right)}{\sin \left( \frac{\beta }{2} \right)}\sin \left[ \frac{\alpha +\alpha +\left( n-1 \right)\beta }{2} \right]\) \(=\frac{\sin \left( n\frac{diff}{2} \right)}{\sin \left( \frac{diff}{2} \right)}\sin \left( \frac{1st+last}{2} \right)\) \(=\frac{\sin \left( \frac{n\,diff}{2} \right)}{\sin \left( \frac{diff}{2} \right)}\cos \left( \frac{1st+last}{2} \right)\) \(1-\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}-\frac{1}{{{6}^{2}}}=\frac{{{\pi }^{2}}}{12}\) \(1+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}….=\frac{{{\pi }^{2}}}{12}\) \(\int\limits_{a}^{b}{{{e}^{x}}dx=\underset{n\to \infty }{\mathop{\lim }}\,}h\left[ {{e}^{a}}+{{e}^{a+h}}+{{e}^{a+2h}}…{{e}^{a+\left( n-1 \right)h}} \right]\) \(=h{{e}^{a}}\left[ 1+{{e}^{n}}+{{e}^{2h}}+…. \right]\) \(={{e}^{a}}\left[ {{e}^{nh}}-1 \right]\frac{h}{\left( {{e}^{n}}-1 \right)}\) \(={{e}^{b}}-{{e}^{a}}\) \(\int\limits_{a}^{b}{\sin x\,dx=\underset{n\to \infty }{\mathop{\lim }}\,}h\left[ \sin a+\sin \left( a+n \right)+…\sin \left( a+\left( n-1 \right)h \right) \right]\) \(=\underset{n\to \infty }{\mathop{\lim }}\,h\frac{\sin \left( \frac{nh}{2} \right)}{\sin \left( \frac{h}{2} \right)}\left[ \sin \left( \frac{2a+\left( n-1 \right)h}{2} \right) \right]\) \(=\frac{\sin \left( \frac{b-a}{2} \right)}{\left( \frac{1}{h} \right)\sin \left( \frac{h}{2} \right)}\left[ \sin \left( a+\frac{nh}{2}-\frac{h}{2} \right) \right]=2\sin \left( \frac{b-a}{2} \right)\sin \left( \frac{a+b}{2} \right)\) \(=\cos \left( \frac{b-a-\left( a+b \right)}{2} \right)-\cos \left( \frac{\left( b-a \right)+\left( a+b \right)}{2} \right)\) \(=\cos a-\cos b\) \(\int\limits_{0}^{2}{\left( {{x}^{2}}+1 \right)dx=\frac{14}{3}}\) \(\int\limits_{0}^{2}{\left( 3{{x}^{2}}+2x+1 \right)dx=3}\) \(\int\limits_{0}^{\frac{\pi }{2}}{\cos x\,dx=1}\) \(\int\limits_{0}^{\frac{\pi }{2}}{{{\sin }^{2}}x\,dx=h\left[ \frac{1-\cos 2x}{2}+\frac{1-{{\cos }^{\left( 2n+2n \right)}}}{2}…… \right]}\) \(=\frac{1}{2}\left( b-a \right)-\frac{1}{4}\left( \sin 2b-\sin 2n \right)\)

Limits using Definite Integration

\(\int\limits_{a}^{b}{f\left( x \right)dx=\underset{h\to 0}{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{f\left( a+rh \right)}}\)

Special case a = 0 b = 1

\(\int\limits_{0}^{1}{f\left( x \right)\,dx=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{n}}\sum\limits_{r=1}^{n}{f\left( \frac{r}{n} \right)}\) \(\int\limits_{0}^{k}{f\left( x \right)\,dx=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{n}}\sum\limits_{r=1}^{kn}{f\left( \frac{r}{n} \right)}\) \(\sum{\rightarrow{Multiple\,\,of\,n}}\int{{}}\) \(\frac{1}{n}\to dx\,\,\,\,\frac{r}{n}\Rightarrow x\)

 

Q. \(\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}+…\frac{1}{nb} \right]\) \(=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=0}^{\left( b-a \right)}{\frac{1}{na+r}}\) \(=\sum\limits_{n\to \infty }{\frac{1}{n}\sum\limits_{r=0}^{\left( b-a \right)n}{\frac{1}{a+\frac{r}{n}}}}=\int\limits_{0}^{b-a}{\frac{dx}{a+x}}=\left[ \log \left( a+x \right) \right]_{a}^{b-a}=\log \left( \frac{b}{a} \right)\)

Practise This Question

A copper sphere is suspended in an evacuated chamber maintained at 300 K. The sphere is maintained at a constant temperature of 500 K by heating it electrically. A total of 210 W of electric power is needed to do it. When the surface of the copper sphere is completely blackened, 700 W is needed to maintain the same temperature of the sphere. Calculate the emissivity of copper.