Definite and Indefinite Integration

Indefinite Integration

Integration means summation. Integration originated during the course of finding the area of a plane figure. Integration is the reverse of differentiation. It is also called as antiderivative. In this section, aspirants will learn the list of important formulas, how to use integral properties to solve integration problems, integration methods and many more.

The process of finding an indefinite integral of a given function is called integration of the function. Integrating a function f(x) means finding a function Φ(x)suchthatddx(Φ(x))=f(x))\Phi (x) \: such \: that \frac{d}{dx}\left ( \Phi (x) \right ) = f(x)) .

Standard formulas for Indefinite Integration

  1. xndx=xn+1n+1+c\int{{{x}^{n}}dx=\frac{{{x}^{n+1}}}{n+1}}+c
  2. axdx=axna+c\int{{{a}^{x}}dx}=\frac{{{a}^{x}}}{\ell n\,a}+c
  3. exdx=ex+c\int{{{e}^{x}}}dx={{e}^{x}}+c
  4. 1xdx=nx+c\int{\frac{1}{x}}dx=\ell n\,x+c
  5. sinxdx=cosx+c\int{\sin x\,dx=-\cos x+c}
  6. cosxdx=sinx+c\int{\cos x\,dx=\sin x+c}
  7. sec2xdx=tanx+c\int{{{\sec }^{2}}x\,\,dx=\tan x+c}
  8. cosec2xdx=cotx+c\int{\cos e{{c}^{2}}x\,\,dx=-\cot x\,+c}
  9. secxtanxdx=secx+c\int{\sec x\tan xdx=\sec x+c}
  10. cosecxcotxdx=cosecx+c\int{\cos ec\,\,x\,\,}\cot x\,dx=-\cos ec\,x+c
  11. 11x2dx=={sin1x+ccos1x+c\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. 11+x2dx=={tan1x+ccot1x+c\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. 1xx21={sec1x+ccosec1x+c\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. kf(x)dx=kf(x)k\int{f\left( x \right)dx=k\int{f\left( x \right)}}
  2. (f(x)+g(x))dx=f(x)dx+g(x)dx\int{\left( f\left( x \right)+g\left( x \right) \right)dx=\int{f\left( x \right)dx+\int{g\left( x \right)dx}}}

Below are some Illustrations based on integration properties:

Illustration 1: Solve (1+x)3x23dx\int{\frac{{{\left( 1+x \right)}^{3}}}{\sqrt[3]{{{x}^{2}}}}}dx

Solution:  (1+x)3x23dx=1+3x+3x2+x3x2/3dx\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 =x2/3+3x1/3+3x4/3+x7/3dx=\int{{{x}^{2/3}}+3{{x}^{1/3}}+3{{x}^{4/3}}}+{{x}^{7/3}}dx =3x1/3+94x4/3+97x7/3+310x10/3+c=3{{x}^{1/3}}+\frac{9}{4}{{x}^{4/3}}+\frac{9}{7}{{x}^{7/3}}+\frac{3}{10}{{x}^{10/3}}+c

Illustration 2: Solve  sec2xcosec2xdx\int{{\sec }^{2}}x\cos e{{c}^{2}}x\,\,dx

Solution: sec2xcosec2xdx=sin2x+cos2xsin2xcos2xdx\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} =sec2x+cosec2xdx=\int{{{\sec }^{2}}x+\cos e{{c}^{2}}x\,\,dx} =tanxcotx+c=\tan x-\cot x+c

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

and so on…

we follow as below

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

Problem 1. 2xsin(x2)dx\int{2x\,\,\sin \left( {{x}^{2}} \right)dx}

Solution: Put x2=t{{x}^{2}}=t

the 2xdx=dt2x\,\,dx=dt sintdt=cost+c\int{\sin t\,\,dt=-\cos t+c} =cosx2+c=-\cos {{x}^{2}}+c

Problem 2. sin3xcos5xdx\int{{{\sin }^{3}}x{{\cos }^{5}}x\,\,dx}

Solution: =sin2xcos5xsinxdx=\int{{{\sin }^{2}}x{{\cos }^{5}}x\sin x\,\,dx} =(1cos2x)cos5xsinxdx=\int{\left( 1-{{\cos }^{2}}x \right)}{{\cos }^{5}}x\,\,\sin x\,\,dx

put cosx=t\cos x=t

sinxdx=dt\,\,\,-\sin x\,dx=dt =(1t2)t5(dt)=\int{\left( 1-{{t}^{2}} \right)}{{t}^{5}}\left( -dt \right) =t5t7dt=\int{{{t}^{5}}-{{t}^{7}}}dt =(t66t88)=-\left( \frac{{{t}^{6}}}{6}-\frac{{{t}^{8}}}{8} \right) =cos6x6+cos8x8+c=-\frac{{{\cos }^{6}}x}{6}+\frac{{{\cos }^{8}}x}{8}+c

Note: Type I

sinmxcosnxdx\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. a2x2x=asinθ\sqrt{{{a}^{2}}-{{x}^{2}}}\rightarrow{{}}x=a\sin \theta
  2. x2+a2x=atanθ\sqrt{{{x}^{2}}+{{a}^{2}}}\rightarrow{{}}x=a\tan \theta
  3. x2a2x=asecθ\sqrt{{{x}^{2}}-{{a}^{2}}}\rightarrow{{}}x=a\sec \theta
  4. xabxx=acos2θ+bsin2θ\sqrt{\frac{x-a}{b-x}}\rightarrow{{}}x=a{{\cos }^{2}}\theta +b{{\sin }^{2}}\theta
  5. xaxbx=asec2θb\sqrt{\frac{x-a}{x-b}}\rightarrow{{}}x=a{{\sec }^{2}}\theta -b

Important formulae set for Indefinite Integration

  1. dxa2x2=sin1(xa)+c\int{\frac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}={{\sin }^{-1}}\left( \frac{x}{a} \right)+c
  2. dxx2a2=nx+x2a2\int{\frac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}=\ell n\left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|}
  3. dxx2+a2=nx+x2+a2\int{\frac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}=\ell n\left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|}
  4. 1a2+x2dx=1atan1(xa).\int{\frac{1}{{{a}^{2}}+{{x}^{2}}}}dx=\frac{1}{a}{{\tan }^{-1}}\left( \frac{x}{a} \right).
  5. 1a2x2dx=12ana+xax.\int{\frac{1}{{{a}^{2}}-{{x}^{2}}}}dx=\frac{1}{2a}\ell n\left| \frac{a+x}{a-x} \right|.
  6. 1x2a2dx=12anxax+a.\int{\frac{1}{{{x}^{2}}-{{a}^{2}}}}dx=\frac{1}{2a}\ell n\left| \frac{x-a}{x+a} \right|.
  7. a2x2dx=x2a2x2+a22sin1(xa).\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. a2+x2dx=x2x2+a2+a22nx+x2+a2\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. x2a2dx=x2x2a2+a22nx+x2a2\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|

Problem Type 1:

dxx2+2x+2=dx(x+1)2+1\int{\frac{dx}{\sqrt{{{x}^{2}}+2x+2}}}=\int{\frac{dx}{\sqrt{{{\left( x+1 \right)}^{2}}+1}}} =log(x+1)+(x+1)2+1.=\log \left| \left( x+1 \right)+\sqrt{{{\left( x+1 \right)}^{2}}+1} \right|.

Problem Type 2:

5x2+2x+3dx\int{\sqrt{5{{x}^{2}}+2x+3}}\,\,dx

here

5x2+2x+3=5[(x+15)2+(1425)2]5{{x}^{2}}+2x+3=5\left[ {{\left( x+\frac{1}{5} \right)}^{2}}+{{\left( \sqrt{\frac{14}{25}} \right)}^{2}} \right] 5x2+2x+3dx=5(x+15)2+(1425)2\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}}}} =5(x+15)2(x+15)2+(1415)2+1425.2log(x+15)+(x+15)2+1415=\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

px+qax2+bx+cdx&px+qax2+bx+cdx&(px+q)ax2+bx+cdx\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+qpx+q as m(ax2+bx+c)1+nm{{\left( a{{x}^{2}}+bx+c \right)}^{1}}+n then this gets changed to standard integral.

Problem Type 3:

I=x+1x2+3x+4dxI=\int{\frac{x+1}{{{x}^{2}}+3x+4}}dx x+1=a(x2+3x+4)1+bx+1=a{{\left( {{x}^{2}}+3x+4 \right)}^{1}}+b =2ax+(3a+b)=2ax+\left( 3a+b \right)

a=12b=12a=\frac{1}{2}\,\,\,\,b=\frac{-1}{2}

I=12(x2+3x+4)1x2+3x+412(1x2+3x+4)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) =12log(x2+3x+4)121(x+32)2+74=\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}}} =12log(x2+3x+4)12174tan1(x+32)74=\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}}} =12logx2+3x+417tan1(2x+37)=\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=f(x+1x)(11x2)dxI=\int{f\left( x+\frac{1}{x} \right)}\left( 1-\frac{1}{{{x}^{2}}} \right)dx

Put x+1x=tx+\frac{1}{x}=t

2. I=f(x1x)(1+1x2)dxI=\int{f\left( x-\frac{1}{x} \right)}\left( 1+\frac{1}{{{x}^{2}}} \right)dx

Put (x1x)=t\left( x-\frac{1}{x} \right)=t

Problem Type 4:

1+x21+x4dx=(1+1x2)(1x2+x2)dx\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 =(1+1x2)dx(x1x)2+2=\int{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)dx}{{{\left( x-\frac{1}{x} \right)}^{2}}+2}} =d(x1x)(x1x)2+2=\int{\frac{d\left( x-\frac{1}{x} \right)}{{{\left( x-\frac{1}{x} \right)}^{2}}+2}} =12tan1((x1x)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.

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

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

Otherwise, it is an improper fraction.

Then f(x)g(x)=h(x)+d(x)g(x)\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(x)g\left( x \right) is expressed as product of non-repeating linear factors.

g(x)=(xa1)(xa2).(xan)g\left( x \right)=\left( x-{{a}_{1}} \right)\left( x-{{a}_{2}} \right)….\left( x-{{a}_{n}} \right) then

fg=A1xa1+A2xa2+.Anxan.\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(x)=(xa)k(xa1)(xa2)g\left( x \right)={{\left( x-a \right)}^{k}}\left( x-{{a}_{1}} \right)\left( x-{{a}_{2}} \right)… fg=A1(xa)+A2(xa)2+..Ak(xa)n+\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(x)g\left( x \right) has a quadratic term then fg=Ax+Bax2+bx+c\frac{f}{g}=\frac{Ax+B}{a{{x}^{2}}+bx+c}

Where A & B are constants determined by comparing coefficients.

dx(x+1)(x2)=ax+1+bx2dx\int{\frac{dx}{\left( x+1 \right)\left( x-2 \right)}}=\int{\frac{a}{x+1}+\frac{b}{x-2}}\,dx =13x+1+12x2dx=\int{\frac{-\frac{1}{3}}{x+1}}+\int{\frac{\frac{1}{2}}{x-2}}\,dx =13nx+1+12nx2=-\frac{1}{3}\ell n\left| x+1 \right|+\frac{1}{2}\ell n\left| x-2 \right| 2x21(x1)(x+1)2dx=Ax1+B(x+1)+C(x+1)2\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}}}} 2x21=A(x+1)2+B(x+1)(x+1)+C(x1)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=1A=12x=1\Rightarrow A=\frac{1}{2} x=1C=12x=-1\Rightarrow \,\,C=\frac{-1}{2} B=32B=\frac{3}{2} I=12(x1)dx+32dx(x+1)+(12)dx(x2)2I=\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}}}} =12nx1+32(nx+1)+12(x+2)+c=\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 dx(x+2)(x2+1)=Ax+2+Bx+Cx2+1\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=2A=15x=-2\Rightarrow \,\,A=\frac{1}{5} by x2{{x}^{2}} coeff. B=12B=\frac{-1}{2} C=25C=\frac{2}{5} I=15dxx+215xx2+1dxI=\frac{1}{5}\left| \frac{dx}{x+2} \right.-\frac{1}{5}\int{\frac{x}{{{x}^{2}}+1}}\,dx I=15nx+2110n(x2+1)+25tan1(x)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=sinmxcosnxdxI=\int{{{\sin }^{m}}x{{\cos }^{n}}xdx}

1. If m –odd put cosx=t\cos x=t

2. If n odd put sinx=t\sin x=t

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

4. If both even then use reduction method

Let us understand with the help of an example:

cos3xsin6xdx=1t2t6dt\int{\frac{{{\cos }^{3}}x}{{{\sin }^{6}}x}dx=\int{\frac{1-{{t}^{2}}}{{{t}^{6}}}dt}}

Where t=sinxt=\sin x =t6t4dt=\int{{{t}^{-6}}-{{t}^{-4}}dt} =15sin5x+13sin3x+c=-\frac{1}{5si{{n}^{5}}x}+\frac{1}{3{{\sin }^{3}}x}+c

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

Put t=tan(x2)t=\tan \left( \frac{x}{2} \right)

Illustration

dx2+sinx\int{\frac{dx}{2+\sin x}} t=tan(x2)\Rightarrow t=\tan \left( \frac{x}{2} \right) dx=2dt1+t2dx=\frac{2dt}{1+{{t}^{2}}} =2dt1+t22+2t1+t2=\int{\frac{\frac{2dt}{1+{{t}^{2}}}}{2+\frac{2t}{1+{{t}^{2}}}}} dtt2+t+1\Rightarrow \int{\frac{dt}{{{t}^{2}}+t+1}} =23tan1(2t+13)=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2t+1}{\sqrt{3}} \right) =23tan1(2tanx2+13)+c=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2\tan \frac{x}{2}+1}{\sqrt{3}} \right)+c

Type II

dxacos2x+bsin2x=dxa+bsin2x\int{\frac{dx}{a{{\cos }^{2}}x+b{{\sin }^{2}}x}}=\int{\frac{dx}{a+b{{\sin }^{2}}x}} 1a+bcos2xdx,1(asinx+bcosx)2dx\int{\frac{1}{a+b{{\cos }^{2}}x}}\,dx,\int{\frac{1}{{{\left( a\sin x+b\cos x \right)}^{2}}}}\,dx or 1a+bsin2x+cos2xdx\int{\frac{1}{a+b{{\sin }^{2}}x+{{\cos }^{2}}x}}\,dx

Rule:

Divide both numerator & denominator by cos2x{{\cos }^{2}}x

Illustration:

134sin2xdx=1cos2x3cos2x4sin2xcos2x\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}}} =sec2x3sec2x4tan2xdx=\int{\frac{{{\sec }^{2}}x}{3{{\sec }^{2}}x-4{{\tan }^{2}}x}}\,dx =dt3(1+t2)4t2=\int{\frac{dt}{3\left( 1+{{t}^{2}} \right)-4{{t}^{2}}}} (Since, by tanx=tsec2xdx=dt\tan x=t\,\,\,{{\sec }^{2}}x\,dx=dt)

=dt3t2=\int{\frac{dt}{3-{{t}^{2}}}} =1(3)2t2=\int{\frac{1}{{{\left( \sqrt{3} \right)}^{2}}-{{t}^{2}}}} =123n3+t3t=\frac{1}{2\sqrt{3}}\ell n\left| \frac{\sqrt{3}+t}{\sqrt{3}-t} \right| =123log3+tanx3tanx=\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=ban=h=\frac{b-a}{n}=. partition of interval putting (n – 1) parts in between

abf(x)dx=limh0hn=0n1f(a+rh)\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

=limh0h[f(a)+f(a+h)+f(a+2h)+..+f(a+(n1)h)]=\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=banh=b-a

Integrals using limits

Evaluate as Limit of a Sum

I=abxdxf(x)=xh=banI=\int\limits_{a}^{b}{x\,\,dx\Rightarrow f\left( x \right)=x\,\,\,\,h=\frac{b-a}{n}} f(a)=af\left( a \right)=a f(a+h)=a+hf\left( a+h \right)=a+h I=h[a+(a+h)+.(a+(n1)h)]I=h\left[ a+\left( a+h \right)+….\left( a+\left( n-1 \right)h \right) \right] look for nh

=h[(na)+h+2h+3h.(n1)h]=h\left[ \left( na \right)+h+2h+3h….\left( n-1 \right)h \right] =hna+h2n(n1)2=hna+\frac{{{h}^{2}}n\left( n-1 \right)}{2} =(ba)a+h2n22(ba)22=b2a22=\left( b-a \right)a+\frac{{{h}^{2}}{{n}^{2}}}{2}-\frac{{{\left( b-a \right)}^{2}}}{2}=\frac{{{b}^{2}}-{{a}^{2}}}{2} r=1nn=n(n+1)2\sum\limits_{r=1}^{n}{n}=\frac{n\left( n+1 \right)}{2} r=1nn2=n(n+1)(2n+1)6\sum\limits_{r=1}^{n}{{{n}^{2}}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} r=1nn3=[n(n+1)2]2\sum\limits_{r=1}^{n}{{{n}^{3}}}={{\left[ \frac{n\left( n+1 \right)}{2} \right]}^{2}}

GP … a+ar+ar2.arn1=a(1rn)1ra+ar+a{{r}^{2}}….a{{r}^{n-1}}=\frac{a\left( 1-{{r}^{n}} \right)}{1-r} sinα+sin(α+β)+sin(α+2β).sin(α+(n1)β)\sin \alpha +\sin \left( \alpha +\beta \right)+\sin \left( \alpha +2\beta \right)….\sin \left( \alpha +\left( n-1 \right)\beta \right) =sin(nβ2)sin(β2)sin[α+α+(n1)β2]=\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] =sin(ndiff2)sin(diff2)sin(1st+last2)=\frac{\sin \left( n\frac{diff}{2} \right)}{\sin \left( \frac{diff}{2} \right)}\sin \left( \frac{1st+last}{2} \right) =sin(ndiff2)sin(diff2)cos(1st+last2)=\frac{\sin \left( \frac{n\,diff}{2} \right)}{\sin \left( \frac{diff}{2} \right)}\cos \left( \frac{1st+last}{2} \right) 1122+132142+152162=π2121-\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+122+132+142+152.=π2121+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}….=\frac{{{\pi }^{2}}}{12} abexdx=limnh[ea+ea+h+ea+2hea+(n1)h]\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] =hea[1+en+e2h+.]=h{{e}^{a}}\left[ 1+{{e}^{n}}+{{e}^{2h}}+…. \right] =ea[enh1]h(en1)={{e}^{a}}\left[ {{e}^{nh}}-1 \right]\frac{h}{\left( {{e}^{n}}-1 \right)} =ebea={{e}^{b}}-{{e}^{a}} absinxdx=limnh[sina+sin(a+n)+sin(a+(n1)h)]\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]