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CBSE Assignments class 09 Mathematics

  Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX 

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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