### Mathematics Assignments | PDF | 8 to 12

PDF Files of Mathematics Assignments From VIII Standard to XII Standard PDF of mathematics Assignments for the students from VIII standard to XII standard.These assignments are strictly according to the CBSE and DAV Board Final question Papers

### 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

$1.)\: \int_{a}^{b}f(x)dx=\int_{a}^{b}f(t)dt$
⇒ Integration is independent of change of variable.

$2.)\: \int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$

$3.)\: \int_{a}^{a}f(x)dx=0$

$4.)\: \int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx$ ,  where a < c < b

$5.)\: \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx$

$6.)\: \int_{0}^{a}f(x)dx=2\int_{0}^{\frac{a}{2}}f(x)dx\: if\: \: \: f(a-x)=f(x)$

(OR)

$\int_{0}^{a}f(x)dx=0$, if f(a-x) = -f(x)

$7.)\: \int_{0}^{2a}f(x)dx= \int_{0}^{a}f(x)dx+\int_{0}^{a}f(2a-x)dx$

$8.)\: \int_{0}^{2a}f(x)dx= \int_{0}^{2a}f(2a-x)dx$

(OR)

$\int_{0}^{2a}f(x)dx= 2\int_{0}^{a}f(x)dx$, if  f (2a-x) = f(x)

(OR)

$\int_{0}^{2a}f(x)dx= 0$,  if f(2a-x) = -f(x)

$9.)\: \int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx$ ,  If f(x) is an even function

(OR)
$\int_{-a}^{a}f(x)dx=0$ , 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.
$10.)\: \int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$

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
$\int_{0}^{\frac{\pi }{2}}log\: sinxdx=\int_{0}^{\frac{\pi }{2}}log\: cosxdx=-\frac{\pi }{2}log\: 2$
$\int_{0}^{\frac{\pi }{2}}log\: cosecxdx=\int_{0}^{\frac{\pi }{2}}log\: secxdx=\frac{\pi }{2}log\: 2$
$\int_{0}^{\frac{\pi }{2}}log\: tanxdx=\int_{0}^{\frac{\pi }{2}}log\: cotxdx=0$