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Common Errors in Secondary Mathematics

Common Errors Committed  by the  Students  in Secondary Mathematics   Errors  that students often make in doing secondary mathematics  during their practice and during the examinations  and their remedial measures are well explained here stp by step.  Some Common Errors in Mathematics

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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