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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 201. |
निम्नलिखित समाकलनों के मान ज्ञात कीजिए- `int(2x-sin2x)/(1-cos2x)dx` |
| Answer» Correct Answer - `-x cot x+c` | |
| 202. |
`int_(0)^(4)(dx)/(sqrt(x^(2)+2x+3))` का मान ज्ञात कीजिए। |
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Answer» `int_(0)^(4)(dx)/(sqrt(x^(2)+2x+3))` `=int_(0)^(4)(1)/(sqrt((x^(2)+2x+1)+2))dx` `=int_(0)^(4)(1)/(sqrt((x+1)^(2)+(sqrt2)^(2)))dx` `=[log|x+1+sqrt((x+1)^(2)+(sqrt2)^(2))|]_(0)^(4)` `=[log|x+1+sqrt(x^(2)+2x+3)|]_(0)^(4)` `=log|5+sqrt(16+8+3)|-log|1+sqrt3|` `=log|(5+3sqrt3)/(1+sqrt3)|` |
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| 203. |
`int_(2)^(4)(x^(2)+x)/(sqrt(x+1))dx` का मान ज्ञात कीजिए। |
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Answer» `int_(2)^(4)(x^(2)+x)/(sqrt(2x+1))dx=[(x^(2)+x)int(1)/(sqrt(2x+1))dx]_(2)^(4)-int_(2)^(4)(d)/(dx)(x^(2)+x)int(1)/(sqrt(2x+1))dx` `=[(x^(2)+x)sqrt(2x+1)]_(2)^(4)- int_(2)^(4)(2x+1)sqrt(2x+1)dx` `=(60-6sqrt5)-int_(2)^(4)(2x+1)^(3//2)dx` `=(60-6sqrt5)-(1)/(5)[(2x+1)^(5//2)]_(2)^(4)` `=(60-6sqrt5)-(1)/(5)[243-(5)^(5//2)]` `=(60-6sqrt5)-(1)/(5)[243-25sqrt5]` `=(57-5sqrt5)/(5)` |
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| 204. |
`int_(0)^((pi)/(4))sin 2x dx` |
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Answer» माना `I=int_(0)^(pi//4)sin 2x dx ` `=int_(0)^(pi//4)2sin x cos x dx` माना `sin x=t" "rArr" "cos x dx = dt` `x=(pi)/(4)" "rArr" "t=sin.(pi)/(4)=(1)/(sqrt2)` `x= 0 " "rArr" "t=sin 0 =0 ` `therefore" "I=2int_(0)^(1//sqrt2)tdt=2[(t^(2))/(2)]_(0)^(1//sqrt2)` `=((1)/(sqrt2))^(2)=(1)/(2)` |
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| 205. |
`int ((cos2x-cos2alpha)/(cosx-cosalpha))dx` का मान ज्ञात कीजिए । |
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Answer» `int((cos2x-cos2alpha)/(cosx-cosalpha))dx` `=int((2cos^(2)x-1)-(2cos^(2)alpha-1))/((cosx-cosalpha))dx` `=2int((cos^(2)x-cos^(2)alpha))/((cosx-cosalpha))=2int(cosx+cosalpha)dx` `=2intcosx dx+2cosalpha* intdx` `=2sinx+2xcosalpha + c` |
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| 206. |
c तथा a के किन मानो के लिए निम्नलिखिए समीकरण संतुष्ट होती है ? `int (sin2x+cos2x)dx=(1)/(sqrt(2))sin(2x-c)+a` |
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Answer» `int(sin2x+cos2x)dx=intsin2xdx+intcos2xdx` `=-(cos2x)/(2)+(sin2x)/(2)+k` `=(1)/(sqrt(2))((1)/(sqrt(2))sin2x-(1)/(sqrt(2))cos2x)+k` `=(1)/(sqrt(2))(sin2x.cos""(pi)/(4)-cos2xsin"(pi)/(4))+k` `=(1)/(sqrt(2))sin(2x-(pi)/(4))+k` प्रश्नानुसार माना दिया है `int(sin2x+cos2x)dx =(1)/(sqrt(2))sin(2x-c)+a` `implies (1)/(sqrt(2))(2x-(pi)/(4))+k =(1)/(sqrt(2))sin(2x-c)+a` `implies c=(pi)/(4)` तथा `a=k` |
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| 207. |
`(1)/(9x^(2)+6x+5)` |
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Answer» `int(1)/(9x^(2)+6x+5)dx` `=(1)/(9)int(1)/(x^(2)+(2)/(3)x+(5)/(9))dx` `=(1)/(9)int(1)/(x^(2)+(2x)/(3)+(1)/(9)+(5)/(9)-(1)/(9))dx` `=(1)/(9)int(1)/((x+(1)/(3))^(2)+((2)/(3))^(2))dx` `=(1)/(9).(1)/(2//3)tan^(-1)(x+(1)/(3))/((2)/(3))+c` `=(1)/(6)tan^(-1)((3x+1)/(2))+c` |
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| 208. |
`sqrt(4-x^(2))` |
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Answer» माना `I=intsqrt(4-x^(2))dx=intsqrt(2^(2)-x^(2))dx` `=(x)/(2)sqrt(4-x^(2))+(4)/(2)sin^(-1).(x)/(2)+C` `=(x)/(2)sqrt(4-x^(2))+2sin^(-1).(x)/(2)+C` |
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| 209. |
यदि `int(sin2x-cos2x)dx=(1)/(sqrt2)sin(2x-k)+c,` तो k का मान है -A. `(5pi)/(4)`B. `-(5pi)/(4)`C. `(pi)/(4)`D. `-(pi)/(4)` |
| Answer» Correct Answer - B | |
| 210. |
`(x)/((x+1)(x+2))` |
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Answer» `int(x)/((x+1)(x+2))dx=I" "` (माना ) माना `(x)/((x+1)(x+2))=(A)/((x+1))+(B)/((x+2))` `rArr" "(x)/((x+1)(x+2))=(A(x+2)+B(x+1))/((x+1)(x+2))` `rArr" "x=A(x+2)+B(x+1)` `{:(x=-1,"तो ",-1=A(-1+2)+0,rArr.,A=-1),(x=-2,"तो ",-2=0+B(-1),rArr.,B=2):}` `therefore" "I=int(x)/((x+1)(x+2))dx` `=int((-1))/(x+1)dx+int(2)/(x+2)dx` `" "=-log(x+1)+2log(x+2)+C` `" "=-log(x+1)+log(x+2)^(2)+C` `" "=log|((x+2)^(2))/(x+1)|+C` |
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| 211. |
निम्नलिखित के मान ज्ञात कीजिए - `intsqrt(cotx)dx` |
| Answer» `-(2)/(sqrt2)tan^(-1)((cot x-1)/(sqrt(2 cot x)))-(1)/(2sqrt2)log |(cot x+1-sqrt(2 cot x))/(cotx +1+sqrt(2 cotx))|+c` | |
| 212. |
`int_(0)^(2)(6x+3)/(x^(2)+4)dx` |
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Answer» माना `I=int_(0)^(2)(6x+3)/(x^(2)+4)dx` `=int_(0)^(2)(6x)/(x^(2)+4)dx+int_(0)^(2)(3)/(x^(2)+4)dx` `{:(माना ,x^(2)+4=t,rArr.,2xdx=dt),(,x=0,rArr.,t=0+4=4),(,x=2,rArr.,t=2^(2)+4=8):}` `therefore" "i=int_(4)^(8)(3)/(t)dt+int_(0)^(2)(3)/(x^(2)+4)dx` `=3int_(4)^(8)(1)/(t)dt+3int_(0)^(2)(1)/(x^(2)+2^(2))dx` `=3[logt]_(4)^(8)+(3)/(2)[tan^(-1).(x)/(2)]_(0)^(2)` `=3[log(8)-log(4)+(3)/(2)(tan^(-1).(2)/(2))` `=3log((8)/(4))+(3)/(2)xx(pi)/(4)` `= 3log2+(3pi)/(8)` |
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| 213. |
`int_(0)^(2)sqrt(6x+4)dx` का मान ज्ञात कीजिए। |
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Answer» `int_(0)^(2)sqrt(6x+4)dx` `=int_(0)^(2)(6x+4)^(1//2)dx=[((6x+4)^(3//2))/((3)/(2)xx6)]_(0)^(2)` `=(1)/(9)[(6xx2+4)^(3//2)-(6xx0+4)^(3//2)]` `=(1)/(9)[64-8]=(56)/(9)` |
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| 214. |
`(1)/(cos(x+a)cos(x+b))` |
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Answer» `int(1)/(cos(x+a)cos(x+b))dx=I" "` (माना ) `thereforeI=(1)/(sin(a-b))int(sin(a-b))/(cos(x+a)cos(x+b))dx` `rArr I=(1)/(sin(a-b))int(sin[(x+a)-(x+b)])/(cos(x+a)cos(x+b))dx sin(x+a)cos(x+b)` `=(1)/(sin(a-b))int(-cos(x+a)sin(x_+b))/(cos(x+a)cos(x+b))dx` `=(1)/(sin(a-b))int[(sin(x+a))/(cos(x+a))-(sin(x+b))/(cos(x+b))]dx` `=(1)/(sin(a-b))int[tan(x+a)-tan(x+b)]dx` `=(1)/(sin(a-b))[-log|cos(x+a)|+log|cos(x+b)|]+C` `=(1)/(sin(a-b))log|(cos(x+b))/(cos(x+a))|+C` |
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| 215. |
`int(1)/(sqrt(1-9x^(2)))dx` का मान ज्ञात कीजिए । |
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Answer» `int(1)/(sqrt(1-9x^(2)))dx=int(1)/(sqrt(1-(3x)^(3)))dx` `=(1)/(3)sin^(-1)(3x)+c` |
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| 216. |
`int(cos2x)/((sinx+cosx)^(2))dx` बराबर है :A. `(-1)/(sinx+cosx)+C`B. `log|sinx+cosx|+C`C. `log|sinx-cosx|+C`D. `(1)/((sinx-cosx)^(2))` |
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Answer» Correct Answer - B माना `I=int(cos2x)/((sinx+cosx)^(2))dx` `" "=int(cos^(2)x-sin^(2)x)/((sinx+cosx)^(2))dx` `" "=int((cosx-sinx)(cosx+sinx))/((sinx+cosx)(sinx+cosx))dx` `" "=int(cosx-sinx)/(sinx+cosx)dx` माना `" "sinx+cosx=t " "rArr" "(cosx-sinx)dx=dt` `therefore" "I=int(1)/(t)dt=log|t|+C` `" "=log|sinx+cosx|+C` |
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| 217. |
`int_(2)^(3)(1)/(x)dx` |
| Answer» `int_(2)^(3)(1)/(x)dx=[logx]_(2)^(3)=log2-log2=log.(3)/(2)` | |
| 218. |
`int_(1)^(2)(4x^(3)-5^(2)+6x+9)dx` |
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Answer» `int_(1)^(2)(4x^(3)-5x^(2)+6x+9)dx` `=[(4x^(4))/(4)-(5x^(3))/(3)+(6x^(2))/(2)+9x]_(1)^(2)` `=[x^(4)-(5x^(3))/(3)+5x^(2)+9x]_(1)^(2)` `=[(2)^(4)-(5(2)^(3))/(3)+3(2)^(2)+9(2)]-[(1)^(4)-(5(1)^(3))/(3)+3(1)^(2)+9(1)]` ltbr `=(16-(40)/(3)+12+18)-(1-(5)/(3)+3+9)` `=(98)/(3)-(34)/(3)=(64)/(3)` |
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| 219. |
`intsqrt(1+x^(2))dx` बराबर है :A. `(x)/(2)sqrt(1+x^(2))+(1)/(2)log|(x+sqrt(1+x^(2)))|+C`B. `(2)/(3)(1+x^(2))^((3)/(2))+C`C. `(2)/(3)x(1+x^(2))^((3)/(2))+C`D. `(x^(2))/(2)sqrt(1+x^(2))+(1)/(2)x^(2)log|x+sqrt(1+x^(2))|+C` |
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Answer» Correct Answer - a माना `I=int sqrt(1+x^(2))dx` `rArr" "I=(x)/(2)sqrt(1+x^(2))+(1)/(2)log||x+sqrt(1+x^(2))|+C` |
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| 220. |
`sqrt(x^(2)-4x-x^(2))` |
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Answer» माना `I=intsqrt(x^(2)+4x+1)dx` `=intsqrt(x^(2)+4x+1-2^(2)+2^(2))dx` `=intsqrt((x+2)^(2)-(sqrt3)^(2))dx` `rArr" "I=(x+2)/(2)sqrt(x^(2)+4x+1)-(3)/(2)log|(x+2)+sqrt(x^(2)+4x+1)|+C` |
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| 221. |
`sqrt(1-4x^(2))` |
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Answer» माना `I=intsqrt(1-4x^(2))dx=intsqrt(4((1)/(4)-x^(2)))dx` `=2intsqrt([((1)/(2))^(2)-x^(2)])dx` `=2.[(x)/(2)sqrt((1)/(4)-x^(2))+(1)/(4).sin^(-1).(x)/(1//2)]+C` `rArr" "I=(x)/(2)sqrt(1-4x^(2))+(1)/(4)sin^(-1)(2x)+C` |
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| 222. |
`int(dx)/(sqrt(9x-4x^(2)))` बराबर है :A. `(1)/(9)sin^(-1)((9x-8)/(8))+C`B. `(1)/(2)sin^(-1)((8x-9)/(9))+C`C. `(1)/(3)sin^(-1)((9x-8)/(8))+C`D. `(1)/(2)sin^(-1)((9x-8)/(9))+C` |
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Answer» Correct Answer - B `int(1)/(sqrt(9x-4x^(2)))dx=(1)/(2)int(1)/(sqrt((9)/(4)x-x^(2)))dx` `" "=(1)/(2)int(1)/(sqrt(((9)/(8))^(2)-(x-(9)/(8))^(2)))dx` `" "=(1)/(2)sin^(-1)((x-(9)/(8))/((9)/(8)))+c` `" "=(1)/(2)sin^(-1)((8x-9)/(9))+c` |
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| 223. |
`sqrt(x^(2)+4x+6)` |
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Answer» माना `I=intsqrt(x^(2)+4x+6dx)` `=intsqrt(x^(2)+4x+2^(2)+6-4)dx` `=ntsqrt((x+2)^(2)+(sqrt2)^(2))dx` `=(x+2)/(2)sqrt(x^(2)+4x+6)+(2)/(2)log|(x+2)+sqrt(x^(2)+4x+6)|+C` `rArr" "I=(x+2)/(2)sqrt(x^(2)+4x+6)+log|(x+2)+sqrt(x^(2)+4x+6)|+C` |
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| 224. |
`sqrt(1-4x-x^(2))` |
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Answer» `I=intsqrt(1-4x-x^(2))dx` `=intsqrt(-(x^(2)+4x-1-2^(2)+2^(2))dx` `=intsqrt(-[(x+2)^(2)-(sqrt5)^(2)])dx` `=intsqrt((sqrt5)^(2)-(x+2)^(2))dx` `rArr" "I=(x+2)/(2)sqrt(1-4x-x^(2))+(5)/(2)sin^(-1).((x+2))/(sqrt5)+C` |
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| 225. |
`sqrt(x^(2)+4x-5)` |
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Answer» `intsqrt(x^(2)+4x-5)dx` `=intsqrt(x^(2)+4x-5-2^(2)+2^(2))dx` `=intsqrt((x+2)^(2)-5-4)dx` `=intsqrt((x+2)^(2)-(3)^(2))dx` `=(x+2)/(2)sqrt(x^(2)+4x-5)-(9)/(2)log|(x+2)+sqrt(x^(2)+4x-5)|+C` |
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| 226. |
`sqrt(1+(x^(2))/(9))` |
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Answer» माना `I=intsqrt(1+(x^(2))/(9))dx` `=intsqrt((9+x^(2))/(9))dx=(1)/(3)intsqrt((3)^(2)+x^(2))dx` `=(1)/(3)[(x)/(2)sqrt((3)^(2)+x^(2))+(9)/(2)log|x+sqrt(9+x^(2))|]+C` `=(x)/(6)sqrt(x^(2)+9)+(3)/(2)log|x+sqrt(9+x^(2))|+C` |
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| 227. |
`int_(2)^(3)(x dx)/(x^(2)+1)` |
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Answer» माना `I=int_(2)^(3)(x)/(x^(2)+1)dx` माना `" "x^(2)+1=t` `rArr" "2x dx = dt " "rArr" "xdx=(dt)/(2)` `x = 2` तो t = 5 तथा x = 3 तो t = 10 `I=(1)/(2)int_(5)^(10)(1)/(t)dt=(1)/(2)[log|t|]_(5)^(10)` `=(1)/(2)[log|10|-log|5|]` `=(1)/(2)log|(10)/(5)|=(1)/(2)log2` |
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| 228. |
निम्नलिखित समाकलनों के मान ज्ञात कीजिए- `int(sinx)/(sin3x)dx` |
| Answer» `(1)/(2sqrt3)log|(sqrt3+tanx)/(sqrt3-tanx)|+c` | |
| 229. |
योगफल की सीमा के रूप में निम्नलिखित समाकलनों के मान ज्ञात कीजिए - `int_(1)^(4)(2x^(2)+1)dx` |
| Answer» Correct Answer - 45 | |
| 230. |
योगफल की सीमा के रूप में निम्नलिखित समाकलनों के मान ज्ञात कीजिए - `int_(0)^(3)(x^(2)+1)dx` |
| Answer» Correct Answer - 12 | |
| 231. |
`int_(1)^(4)(1)/(sqrtx)dx` का मान ज्ञात कीजिए। |
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Answer» `int_(1)^(4)(1)/(sqrtx)dx` `=int_(1)^(2)x^(-1//2)dx=[(x^(1//2))/(1//2)]_(1)^(4)=2[sqrtx]_(1)^(4)` `=[sqrt4-sqrt1]=2(2-1)=2` |
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| 232. |
निम्नलिखित समाकलनों के मान ज्ञात कीजिए- `int(x^(2))/((a+bx)^(2))dx` |
| Answer» `(1)/(b^(3))[(a+bx)-2alog(a+bx)-(a^(2))/(a+bx)]+c` | |
| 233. |
निम्नलिखित समाकलनों के मान ज्ञात कीजिए- `int(1)/(1+x+x^(2)+x^(3))dx` |
| Answer» `(1)/(2)log|1+x|-(1)/(4)log|1+x^(2)|+(1)/(2)tan^(-1)x+c` | |
| 234. |
`int_(0^(1)(e^(sqrtx))/(sqrtx)dx` का मान ज्ञात कीजिए । |
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Answer» माना `sqrtx=t` `rArr" "(1)/(2sqrtx)dx=dt` `rArr" "(1)/(sqrtx)dx=2dt` `{:(x=0," पर ",t=sqrt0=0),(x=1," पर ",t=sqrt1=t):}` `therefore" "int_(0)^(1)(e^(sqrtx))/(sqrtx)dx=int_(0)^(1)e^(t).2dt=2[e^(t)]_(0)^(1)` `" "=2(e^(-1)-e^(0))=2(e-1)` |
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| 235. |
योगफल की सीमा के रूप में निम्नलिखित समाकलनों के मान ज्ञात कीजिए - `int_(2)^(4)xdx` |
| Answer» Correct Answer - 6 | |
| 236. |
सिद्ध कीजिए कि `int_(a//4)^(3a//4)(sqrtx)/(sqrt(a-x)+sqrtx)dx=(a)/(4)` |
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Answer» माना `I=int_(a//4)^(3a//4)(sqrtx)/(sqrt(a-x)+sqrtx)dx` `rArr" "I=int_(a//4)^(3a//4)(sqrt(a-x))/(sqrt(a-(a-x))+sqrt(a-x))dx` [प्रगुण (7 ) से ] `rArr" "I=int_(a//4)^(3a//4)(sqrt(a-x))/(sqrtx+sqrt(a-x))dx" …(2)"` समीकरण (1 ) और (2 ) को जोड़ने पर `2I=int_(a//4)^(3a//4)(sqrtx+sqrt(a-x))/(sqrt(a-x)+sqrtx)dx` `=int_(a//4)^(3a//4)1dx` `=[x]_(a//4)^(3a//4)=(3a)/(4)-(a)/(4)=(2a)/(4)` `rArr" "I=(a)/(4)" "` यही सिद्ध करना था । |
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| 237. |
निम्नलिखित को सिद्ध कीजिए । `int_(0)^((pi)/(2))tan^(3)x dx = 1-log2` |
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Answer» माना `I=int_(0)^(pi//4)2 tan^(3)x dx` `" "=2 int_(0)^(pi//4)tan^(2)x. tanx dx` `" "=2int_(0)^(pi//4)(sec^(2)x-1)tanxdx` `=2[int_(0)^(pi//4)sec^(2)x tanx dx - int_(0)^(pi//4)tanx dx]` `=2int_(0)^(pi//4)(tanx)sec^(2)x dx-2[-log|cosx|]_(0)^(pi//4)` `=2[(tan^(2)x)/(2)]_(0)^(pi//4)+2[log|cos.(pi)/(4)|-log|cos0|]` `=tan^(2)((pi)/(4))-0+2[log((1)/(sqrt2))-log1]` `=1+2log2^(-1//2)-0=1-2xx(1)/(2)log2 = 1-log2" "` यही सिद्ध करना था । |
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| 238. |
निम्नलिखित समाकलनों के मान ज्ञात कीजिए- `int(x+1)/(x(1+xe^(x))^(2))dx` |
| Answer» `log|(xe^(x))/(xe^(x)+1)|+(1)/(1+xe^(x))+c` | |
| 239. |
`(xe^(x))/((1+x)^(2))` |
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Answer» `int(x.e^(x))/((1+x)^(2))dx` `=int({(1+x)-1})/((1+x)^(2)).e^(x)dx` `=int(1+x)/((1+x)^(2)).e^(x)dx-int(1)/((1+x)^(2)).e^(x)dx` `=int(1)/(1+x).e^(x)dx- int(1)/((1+x)^(2)).e^(x)dx` `=(1)/(+x).inte^(x)dx-int{(d)/(dx).(1)/(1+x).inte^(x)dx}dx-int(1)/((1+x)^(2)).e^(x)dx` `=(e^(x))/(1+x)-int(-1)/((1+x)^(2)).e^(x)dx-int(1)/((1+x)^(2)).e^(x)dx+C` `=(e^(x))/(1+x)+int(1)/((1+x)^(2)).e^(x)dx-int(1)/((1+x)^(2)).e^(x)dx+C` `=(e^(x))/(1+x)+C` |
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| 240. |
निम्नलिखित को सिद्ध कीजिए । `int_(0)^((pi)/(2))sin^(3)x dx =(2)/(3)` |
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Answer» माना `I=int_(0)^(pi//2)sin^(3)x dx` `" "=int_(0)^(pi//2)sin^(2)x. sinx dx` `" "=int_(0)^(pi//2)(1-cos^(2)x)sinxdx` माना `cosx=t` `rArr" "-sinx dx=dt` `rArr" "sinx dx = 0dt` तथा`" "x=0" "rArr" "t=cos 0 =1` और `" "x=(pi)/(2)" "rArr" "t=cos.(pi)/(2)=0` `therefore" "I=int_(0)^(pi//2)(1-cos^(2)x)sinx dx` `" "=int_(1)^(0)(1-t^(2))(-dt)=-[t-(t^(3))/(3)]_(1)^(0)` `" "=-{(0-0)-(1-(1)/(3))}=(2)/(3)` |
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| 241. |
`int_(0)^(1)x(1-x)^(5)dx` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(1)/(42)` | |
| 242. |
`int_(0)^(4)x(4-x)^(3//2)dx` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(512)/(35)` | |
| 243. |
`int_(-pi//4)^(pi//4)|sinx|dx` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(2-sqrt2)` | |
| 244. |
`int_(0)^(1)x(1-x)^(3//2)dx` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(4)/(35)` | |
| 245. |
निम्नलिखित को सिद्ध कीजिए । `int_(-1)^(1)x^(17)cos^(4)xdx=0` |
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Answer» माना `f(x)=x^(17)cos^(4)x` `rArr" "f(-x)=(-x)^(17)cos^(4)(-x)` `" "=-x^(17)cos^(4)x=-f(x)` अतः f(x) एक विषम फलन है । हम जानते हैं कि यदि f(x) एक विषम फलन हो, तो `int_(-a)^(a)f(x)dx=0` `therefore" "int_(-1)^(1)x^(17)cos^(4)xdx=0" "` यही सिद्ध करना था । |
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| 246. |
सिद्ध कीजिए कि `int_(-a)^(a)x^(3)sqrt(a^(2)-x^(2))dx=0` |
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Answer» माना `f(x)=x^(3)sqrt(a^(2)-x^(2))` `rArr" "f(-x)=(-x)^(3)sqrt(a^(2)-(-x)^(2))` `" "=-x^(3)sqrt(a^(2)-x^(2))=-f(x)` अतः f(x) एक विषम फलन है। `rArr" "int_(-a)^(a)f(x)=0` `rArr" "int_(-a)^(a)f(x)=0` `rArr" "int_(-a)^(a)x^(3)sqrt(a^(3)-x^(2))dx=0` यही सिद्ध करना था । |
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| 247. |
सिद्ध कीजिए कि `int_(0)^(oo)(x)/((1+x)(1+x^(2))dx=(pi)/(4)` |
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Answer» माना `x=tan theta` `rArr" "dx=sec^(2)thetad theta` `{:(x=0,"पर ",tan theta=0, rArr.,theta=0),(x=oo,"पर ",tan theta=oo,rArr.,theta=(pi)/(2)):}` माना `I=int_(0)^(oo)(x)/((1+x)(1+x^(2)))dx` `rArr" "I=int_(0)^(pi//2)(tan theta)/((1+tan theta)(1+tan^(2)theta)).sec^(2) theta d theta` `rArr" "I=int_(0)^(pi//2)(tan theta)/(1+tan theta)d theta" ...(1)"` `rArr" "I=int_(0)^(pi//2)(tan((pi)/(2)-theta))/(1+tan((pi)/(2)-theta))d theta" "` [प्रगुण (4 ) से ] `=int_(0)^(pi//2)(cot theta)/(1+cot theta)d theta=int_(0)^(pi//2)(1//tan theta)/(1+(1)/(tan theta))d theta` `rArr" "I=int_(0)^(pi//2)(1)/(tantheta+1)d theta" ...(2)"` समीकरण (1 ) और (2 ) को जोड़ने पर `2I=int_(0)^(pi//2)(tan theta+1)/(tan theta+1)d theta=int_(0)^(pi//2)d theta=[theta]_(0)^(pi//2)=(pi)/(2)` `rArr" "I=(pi)/(4)" "`यही सिद्ध करना था । |
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| 248. |
निम्नलिखित को सिद्ध कीजिए । `int_(0)^(1)xe^(x)dx=1` |
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Answer» `int_(0)^(1)xe^(x)dx=[xinte^(x)dx-int1.e^(x)dx]_(0)^(1)` `" "=[xe^(x)-e^(x)]_(0)^(1)=[e^(x)(x-1)]_(0)^(1)` `" "=e(1-1)-e^(0)(0-1)=0+1=1` `" "` यही सिद्ध करना था । |
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| 249. |
`int_(0)^(1)xe^(x^(2))dx` |
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Answer» माना `int_(0)^(pi//2)cos^(2)x dx=int_(0)^(pi//2)((cos2x+1)/(2))dx` `=(1)/(2)int_(0)^(pi//2)cos 2x dx +(1)/(2)int_(0)^(pi//2)1dx` `=(1)/(2)[(sin 2x)/(2)]_(0)^(pi//2)+(1)/(2)[x]_(0)^(pi//2)` `=(1)/(2)((sinpi)/(2)-(sin0)/(2))+((pi)/(2)-0)]=(pi)/(4)` |
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| 250. |
समाकलन कीजिए - `int tan^(-1)(secx+tanx)dx` |
| Answer» `(pix)/(4)+(1)/(4)x^(2)+c` | |