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Class 12 Maths Formula Ch-7 Definite Integral

DEFINITE INTEGRAL
Class 12 Maths Formula Ch-7 Definite Integral

Definition of Definite Integral: \[If \int f(x)dx=F(x)\, then\, \int_{a}^{b}f(x)dx=F(b)-F(a)\]\[ is\, called\,the\, \, definite\, integral\, of\, f(x)\, w.r.t\, \, x\, from\, a\, to\, b\]

DEFINITE INTEGRAL AS THE LIMIT OF A SUM
\[\int_{a}^{b}f(x)dx=\lim_{x\rightarrow 0}h[f(a)+f(a+h)+f(a+2h)+.........f(a+\overline{n-1})h]\]
\[where\: nh=b-a\: or\: h=\frac{b-a}{n}\](OR)
\[\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)......f(a+\overline{n-1})h]\]
USEFUL RESULTS HELPFUL IN LIMIT OF A SUM
\[1+2+3+4+.......+(n-1)=\frac{n(n-1)}{2}\]
\[1^{2}+2^{2}+3^{2}+.......+(n-1)^{2}=\frac{n(n-1)(2n-1)}{6}\]
\[1^{3}+2^{3}+3^{3}+.......+(n-1)^{3}=\left [ \frac{n(n-1)}{2} \right ]^{2}\]
\[a+ar+ar^{2}+......+ar^{n-1}=\frac{a(r^{n}-1)}{r-1}\; if\; r> 1\]
\[\lim_{h\rightarrow 0}\frac{e^{h}-1}{h}=1\: \: or\: \: \lim_{n\rightarrow \infty }\frac{e^{\frac{1}{n}-1}}{\frac{1}{n}} =1\]
PROPERTIES OF DEFINITE INTEGRAL

 
⇒ Integration is independent of change of variable.


 

 ,  where a < c < b



                                      (OR) 

, if f(a-x) = -f(x)


 


 

                                   (OR)

, if  f (2a-x) = f(x)

                                  (OR) 

,  if f(2a-x) = -f(x)


 ,  If f(x) is an even function

                                  (OR)
 , If f(x) is an odd function

Even function:- If f(-x) = f(x) then, function f(x) is called even function.
Odd Function: If f(-x) = - f(x) then, function f(x) is called odd function.

NCERT Exercise 7.10, Q. No. 8\[Integrate\: \: \int_{1}^{2}\left ( \frac{1}{x}-\frac{1}{2x^{2}} \right )e^{2x}dx\]
Solution:
 \[I=\int_{1}^{2}\left ( \frac{1}{x}-\frac{1}{2x^{2}} \right )e^{2x}dx\]\[let\: 2x=t\Rightarrow x=\frac{t}{2}\Rightarrow dx=\frac{dt}{2}\]\[If\: x=1\: then\:\: t=2,\:\: If\: x=2, then\:\: t=4\]\[I=\int_{2}^{4}\left ( \frac{2}{t}-\frac{4}{2t^{2}} \right )e^{t}\frac{dt}{2}\\=\frac{1}{2}\times 2\int_{2}^{4}\left ( \frac{1}{t}-\frac{1}{t^{2}} \right )e^{t}dt\]\[=\left [ \frac{1}{t}e^{t} \right ]_{2}^{4}=\frac{1}{4}e^{4}-\frac{1}{2}e^{2}=\frac{e^{2}}{4}(e^{2}-1)\]
NCERT Exercise 7.10 ,Q. No. 9  Find the Integral of \[I=\int_{1/3}^{1}\frac{(x-x^{3})^{1/3}}{x^{4}}dx\]
Solution:
\[I=\int_{1/3}^{1}\frac{(x-x^{3})^{1/3}}{x^{4}}dx=\int_{1/3}^{1}\frac{x^{3\times \frac{1}{3}}\left [ \frac{x}{x^{3}}-1 \right ]^{1/3}}{x^{4}}dx\\=\int_{1/3}^{1}\frac{\left ( \frac{1}{x^{2}}-1 \right )^{1/3}}{x^{3}}dx\]\[Let\: \: \left ( \frac{1}{x^{2}}-1 \right )^{1/3}=t\Rightarrow \left ( \frac{1}{x^{2}}-1 \right )=t^{3}\]Differentiating w. r. t. x we get. \[\frac{-2}{x^{3}}dx=3t^{2}dt\Rightarrow \frac{dx}{x^{3}}=\frac{-3}{2}t^{2}dt\]\[If\: \: x=\frac{1}{3},\: then\: \: t = \left ( \frac{1}{1/9}-1 \right )^{1/3}=2\\If\: \: t=1,\: \: then\: \: t=\left ( \frac{1}{1}-1 \right )=0\]\[I=\int_{2}^{0}t\times \frac{-3}{2}t^{2}dt=\frac{-3}{2}\int_{2}^{0}t^{3}dt\]\[=\frac{-3}{2}\left [ \frac{t^{4}}{4} \right ]_{2}^{0}=\frac{-3}{8}\left [ 0^{4}-2^{4} \right ]\\=\frac{-3}{8}\times -16=6\]
NCERT Exercise 7.11,Q. No. 16 \[Integrate:\: \: I=\int_{0}^{\pi }log(1+cosx)dx\]
Solution :
\[I=\int_{0}^{\pi }log(1+cosx)dx ….. (1)\] \[=\int_{0}^{\pi }log(1+cos(\pi -x))dx\\=\int_{0}^{\pi }log(1-cos(x))dx\: \: ....\: \: (2)\]Adding equation (1) and (2) we get. \[2I=\int_{0}^{\pi }\left [ log(1+cosx)+log(1-cosx) \right ]dx\\=\int_{0}^{\pi }\left [ log(1+cosx)(1-cosx) \right ]dx\\=\int_{0}^{\pi }\left [ log(1-cos^{2}x) \right ]dx=\int_{0}^{\pi }\left [ log(sin^{2}x) \right ]dx\]\[2I=2\int_{0}^{\pi }log\: sinxdx\]\[I=\int_{0}^{\pi }log\: sinxdx\: \: ......\: \: \: (3)\] \[f(x)=logsinx,\: and\: f(\pi -x)=logsin(\pi -x)=logsinx\]\[\Rightarrow I=2\int_{0}^{\frac{\pi }{2} }logsinxdx\: \: ......\: \: (4)\] \[I=2\int_{0}^{\frac{\pi }{2} }logsin\left (\frac{\pi }{2}-x \right )dx\\I=2\int_{0}^{\frac{\pi }{2} }logcosxdx\: \: .....\: \: (5)\]Adding (4) and (5) \[2I=2\int_{0}^{\pi /2}[logsinx+logcosx]dx\\I=\int_{0}^{\pi /2}[logsinxcosx]dx=\int_{0}^{\pi /2}[logsinxcosx+log2-log2]dx\]\[=\int_{0}^{\pi /2}[log2sinxcosx-log2]dx=\int_{0}^{\pi /2}[logsin2x-log2]dx\]\[I=\int_{0}^{\pi /2}logsin2xdx-\int_{0}^{\pi /2}log2dx\\=\int_{0}^{\pi /2}logsin2xdx-\frac{\pi }{2}log2\]\[Let \: \: 2x = t, \: \: dx=\frac{dt}{2}\]\[If\: x=0, \: t=0,\: If\: \: x=\frac{\pi }{2},then\: \: t=\pi\]\[I=\frac{1}{2}\int_{0}^{\pi }logsintdt-\frac{\pi }{2}log\: 2\]Now replacing t by x we get\[I=\frac{1}{2}\int_{0}^{\pi }logsinxdx-\frac{\pi }{2}log\: 2\]\[I=\frac{1}{2}I-\frac{\pi }{2}log\: 2 \;\;............\; using(3)\\ I-\frac{1}{2}I=-\frac{\pi }{2}log\: 2\\\frac{1}{2}I=-\frac{\pi }{2}log\: 2\Rightarrow I=-\pi log2\]
STANDARD RESULTS CAN BE USED IN THE PROBLEM DIRECT
\[\int_{0}^{\frac{\pi }{2}}log\: sinxdx=\int_{0}^{\frac{\pi }{2}}log\: cosxdx=-\frac{\pi }{2}log\: 2\]
\[\int_{0}^{\frac{\pi }{2}}log\: cosecxdx=\int_{0}^{\frac{\pi }{2}}log\: secxdx=\frac{\pi }{2}log\: 2\]
\[\int_{0}^{\frac{\pi }{2}}log\: tanxdx=\int_{0}^{\frac{\pi }{2}}log\: cotxdx=0\]


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