InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 151. |
`int_(0)^(pi//2) x^(2) cos xdx` का मान ज्ञात कीजिएः| |
|
Answer» `underset(0)overset(pi//2)int underset(I)(x^(2)) underset(II)(cos) xdx` `=x^2underset(0)overset(pi//2)int cosdx-int [d/dxx^(2)int cos x]dx` `=x^(2) sin x-iint [2x.sin x]dx` `=x^(2) sin x-2[x.int sin xdx-int {d/dxx int sin xdx }dx` `=x^(2) sin x-2[x.(-cos x)-int 1(-cos x)dx]` `=x^(2)sin x+2x cos x-2int cos xdx` `=x^(2)sinx+2x cos x-2 sin x` `=x^(2) sin x+2x cos x -2 sin x` `=(x^(2)-2)sin x+2x. cos x` `therefore underset(0)overset(pi//2)int x^(2)cosxdx` `=[(x^(2)-2)sin x+2x. cosx]_(0)^(pi//2)` `=[((pi^(2))/(4)-2).sin""pi/2+2xxpi/2xxcos""pi/2]-[(0-2)sin 0+2(0) cos 0]` `=[((pi^(2))/(4)-2).1+pi/(0)]=[0+0]` `=(pi^2)/(4)-2` |
|
| 152. |
`int_(0)^(pi) x sin^(2) xdx` का मान ज्ञात कीजिएः |
|
Answer» `int x. sin^(2) xdx =1/ int x(1-cos 2x)dx` `=1/2 int (x-xcos 2x)dx` `=1/4x^(2)-1/2int x cos 2 xdx` `=1/4x^(2)-1/2 [(x sin 2x)/(2)-int (1. sin 2x)/(2)dx]` `=1/4x^(2)-1/4x sin 2x-1/8 cos 2x` `therefore underset(0)overset(pi)int x.sin^(2)xdx=1/4 [x^(2)-x sin 2x-1/2 cos 2x]_(0)^(pi)` `=1/4 [pi^(2)-pisin 2pi-1/2 cos 2pi-0+0.sin 0+1/2 cos 0]` `=1/4[pi-0+1/2+1/2]=(pi^(2))/(4)` |
|
| 153. |
इन समाकलनो का मान ज्ञात कीजिएः `int_(0)^(pi//2) x sin xdx` |
|
Answer» यहाँ `underset(0)overset(pi//2)int underset(I)(x) underset(II)(sin) xdx` `=[x(-cosx)]_(0)^(pi//2)-underset(0)overset(pi//2)int 1.(-cos x)dx` `=[x cos x]_(0)^(pi//2) +[sin x]_(0)^(pi//2)` `=[-pi/2cos""pi/2+0 cos 0]+[sin""pi/2-sin 0]` `=0+1-0=1` |
|
| 154. |
सिद्ध कीजिएः की `int_(0)^(1) tan^(-1)xdx=pi/4-1/2log2.` |
|
Answer» माना `I=underset(0)overset(1)int tan^(-1)xdx " "x=tan theta ` रखने पर `dx=sec^(2) theta d theta` जब x=0 तब `theta=0` तथा जब x=1 तब `theta=pi/4` `therefore I=underset(0)overset(pi/4)int tan^(-1)(tan theta) sec^(2) theta d theta ` `=underset(0) overset(pi/4)int theta sec^(2) d theta` `=[theta int theta sec^(2) d theta]_(0)^(pi)-underset(0)overset(pi/4)int {(d)/(d theta) intsec^(2) theta d theta}d theta` `=[theta tan theta]_(0)^(pi//4)-underset(0)overset(pi//4)int tan theta d theta` `=[pi/4tan""pi/4-0]+[log cos theta]_(0)^(pi//4)` `=pi/4. 1 log cos""pi/4-log cos theta` `=pi/4-log ""1/sqrt2-log 0` `[therefore log 1=0" तथा "pi/4log ""1/A=-logA]` `therefore I=pi/4-1/2log2 ` |
|
| 155. |
इन संकलनो का मान ज्ञात कीजिएः `int_(0)^(2) xsqrt(x+2)dx` |
|
Answer» यहाँ `underset(0)overset(2)int xsqrt(x+2)dx` माना `x+2=t rArr x=t -2 rArr dx=dt` जब `x=0 rArr t=0+2=2` और `x=2 rArr t=2+2=4` `therefore I=underset(2)overset(4)int (t-2)sqrtt dt` `rArr I=underset(2)overset(4)int (t^(3//2)-2t^(1//2))dt` `rArr I=[2/5t^(5//2)-2xx2/3t^(3//2)]_2^(4)` `rArr I=[2/5t^(2)xx(4)^(5//2)-(4)/(3)xx(4)^(3//2)]-[2/5xx(2)^(5//2)-4/3(2)^(3//2)]` `rArr I= [2/5xx32-4/3xx8]-[2/5xx4sqrt2-4/3xx2sqrt2]` `rArr I=2/5[32-4sqrt2]-4/3[8-2sqrt2]` `rArr I=(64)/(5)-(8sqrt2)/(5)-(32)/(3)+(8sqrt2)/(3)` `rArr I=((192-160)/(15))+8sqrt2 ((5-3)/(15))` `rArr I=32/15 +(16sqrt2)/(15)=(16sqrt2)/(15)(sqrt2+1)` |
|
| 156. |
`int_(0)^(1)(e^(tan^(-1))x)/(1+x^(2))dx` किसके तुल्य है ?A. `e^(pi//4)-1`B. `e^(pi//4)+1`C. e-1D. e |
| Answer» Correct Answer - A | |
| 157. |
`int_(0)^(1)(tan^(-1)x)/(1+x^(2))dx` का मान किसके बराबर है ?A. `(pi^(2))/(8)`B. `(pi^(2))/(32)`C. `(pi)/(4)`D. `(pi)/(8)` |
| Answer» Correct Answer - B | |
| 158. |
`int_(-1)^(1) e^(x)dx` का मान हैA. `e+1/e`B. `e-1/e`C. 0D. 2 |
| Answer» Correct Answer - B | |
| 159. |
`int_(-pi//4)^(pi//4) x^(3).sin^(4)xdx` का मान है |
| Answer» Correct Answer - A | |
| 160. |
`int_(0)^(2)e^(logx)dx` किसके बराबर है ?A. `1//4`B. 2C. 4D. इनमें से कोई नहीं |
| Answer» Correct Answer - B | |
| 161. |
संकलनो का मान ज्ञात कीजिएः `int_(0)^(1) cos^(-1)x dx` |
|
Answer» यहाँ `underset(0)overset(1)int cos^(-1)x dx` माना `cos^(-1)x=t rArr x=cos t rArr dx=-sin t dt` जब `x=0 rArr t=cos^(-1) 0=pi/2` और `x=1 rArr t=cos^(-1) 1=0` `therefore I=underset(pi//2)overset(0)int cos^(-1)xdx` `=-underset(pi//2)overset(0)int t sin tdt` `=[{-t cos t}_(pi//2)^(0)+underset(pi//2)overset(0)int cos t dt]` `=-[0+{sin t}_(pi//2)^(-)]=-[0-1]=1` |
|
| 162. |
`int_(a)^(b) f(x)dx=`A. `underset(a)overset(b)int f(a+b+x)dx`B. `underset(a)overset(b)int f(b-x)dx`C. `underset(a)overset(b)int f(a-x)dx`D. `underset(a)overset(b)int f(a+b-x)dx` |
| Answer» Correct Answer - D | |
| 163. |
`int_(0)^(2)(dx)/(x^(2)+4)` किसके बराबर है ?A. `(pi)/(2)`B. `(pi)/(4)`C. `(pi)/(8)`D. इनमें से कोई नहीं |
| Answer» Correct Answer - C | |
| 164. |
संकलनो का मान ज्ञात कीजिएः `int_(0)^(1) sin^(-1)x dx` |
|
Answer» यहाँ `I=underset(0)overset(1)int sin^(-1)x dx` माना `sin^(-1)x=t rArr x=sin t rArr dx =cos tdt` जब `x=0 rArr t=sin^(-1) 0=0` और `x=1 rArr t=sin^(-1)1 =(pi)/(2)` `therefore I=underset(0)overset(pi//2)int underset(I)(t). underset(II)(cos t) dt` `=[t sin t]_(0)^(pi//2)-underset(0)overset(pi//2)int sin t dt` `=[t sin t]_(0)^(pi//2)-[-cos t]_(0)^(pi//2)` `=[pi/2sin""pi/2-0]-[-cos""pi/2+cos0]` `=pi/2xx1-(0+1)=pi/2-1` |
|
| 165. |
इन संकलनो का मान ज्ञात कीजिएः `int_(0)^(4) (1)/(x+sqrtx)dx` |
|
Answer» यहाँ `underset(0)overset(4)int (1)/(x+sqrtx)dx` माना `x=t^(2) rArr dx=2tdt` जब `x=0 rArr t^(2)=0 rArr t=0` और `x=4 rArr t^(2)=4 rArr t=2` `therefore I=underset(0)overset(2)int (2dt)/(t^(2)+t)` `rArr I=2underset(0)overset(2)int (dt)/(t+1)` `rArr I=2[log(t+1)]_(0)^(2)` `rArr I=2 [log 3-log1]` `rArr I=2log3` |
|
| 166. |
संकलनो का मान ज्ञात कीजिएः `int_(0)^(pi//2) sqrt(sin phi) cos^(5) phi dphi` |
|
Answer» यहाँ `underset(0)overset(pi//2)int sqrt(sin phi) cos^(5) phi dphi` `rArr I=underset(0)overset(pi//2)int sqrt(sinphi).cos^(4) phi. cos phi d phi` `rArr I=underset(0)overset(pi//2)int sqrt(sinphi)(1-sin^(2) phi)^(2)cos phi d phi` माना `sin phi=t rArr cos phi=dt ` जब `phi=0rArr t=sin 0=0` और `phi=pi/2 rArr t=sin "pi/2=1` `therefore I=underset(0)overset(1)int sqrtt (1-t)^(2)dt` `=underset(0)overset(1)int sqrtt(1-2t^(2)+t^(4))dt` `=underset(0)overset(1)int (sqrtt-2t^(5//2)+t^(9//2))dt` `=underset(0)overset(1)int (t^(1//2)-2t^(5//2)+9^(//2))dt` `=[2/3t^(3//2)-2. 2/7t^(7//2)+2/11t^(11//2)]_(0)^(1)` `=(2/3-4/7+2/11)-0=64/231` |
|
| 167. |
इन समक़लानो का मान ज्ञात कीजिएः `int_(1)^(2)(4x^(3)-5x^(2)+6x+9)dx` |
|
Answer» `underset(1)overset(1)int (4x^(3)-5x^(2)+6x+9)dx` `=[x^(4)-(5x^3)/(3)+3x^(2)+9x]_(1)^(2)` `=[(2)^(4)-(5)/(3)(2)^(3)+3(2)^(2)+9xx2]-[(1)^(4)-5/3(1)^(3)+3(1)^(2)+9xx1]` `=[16-(40)/(3)+12+18]-[1-(5)/(3)+3+9]` `=(46-(40)/(3))-(13-5/3)` `=(98)/(3)-(34)/(3)=(64)/(3)` |
|
| 168. |
इन समक़लानो का मान ज्ञात कीजिएः `int_(-1)^(1) (x+1)dx` |
|
Answer» `underset(-1)overset(1)int dx=[(x^(2))/(2)+x]` `=[(1)^(2)/(2)+1]-[(-1)^(2)/(2)-1]` `=3/2-(-1/2)` `=3/2+1/2=2` |
|
| 169. |
इन समक़लानो का मान ज्ञात कीजिएः `int_(0)^(pi) (sin^(2)""(x)/(2)-cos^(2)""x/2)dx` |
|
Answer» `underset(0)overset(pi)int(sin^(2)""x/2-cos^(2)""x/2)dx` `=-underset(0)overset(pi)int (cos^(2)""pi/2-sin^(2)""x/2)dx` `=-underset(0)overset(pi)int cosx=-[sinx]_(0)^(x)` `=-[sin pi-sin 0]=0` |
|
| 170. |
इन समक़लानो का मान ज्ञात कीजिएः `int_(0)^(pi//4) (2sec^(2)x+x^(3)+2)dx` |
|
Answer» `underset(0)overset(pi//4)int (2sec^(2)x+x^(3)+2)dx` `=[2 tanx+(x^(4))/(4)+2x]_(0)^(pi//4)` `=[2 tan ""pi/4+((pi)/(4))^(4)/(4)+2(pi/4)]-[2tan0+0/4+2xx0]` `=2xx1+(pi^(4))/(1024)+pi/2` |
|
| 171. |
इन समक़लानो का मान ज्ञात कीजिएः `int_(2)^(3) x^(-7)dx` |
|
Answer» `underset(2)overset(3)int x^(-7)dx=[(x^(-6))/(-6)]_(2)^(3)=-1/6[(3)^(-6)-(2)^(-6)]` `=-1/6[(1)/(729)-(1)/(64)]=(665)/(67)` |
|
| 172. |
इन समक़लानो का मान ज्ञात कीजिएः `int_(2)^(3) 1/x dx` |
|
Answer» `underset(2)overset(3)int 1/xdx=[logx]_(2)^(3)` `=log3-log2` `=log""3/2` |
|
| 173. |
इन समक़लानो का मान ज्ञात कीजिएः `int_(0)^(1) 2^(x)dx` |
|
Answer» `underset(0)overset(1)int 2^(x)dx=[(2^(x))/(log_(e)2]]_(0)^(1)=1/(log_(e)2)[2^(1)-2^(0)]` `=(2-1)/(log_(e)2)=(1)/(log_(e)2)` |
|
| 174. |
इन समक़लानो का मान ज्ञात कीजिएः `int_(4)^(5) e^(x)dx` |
| Answer» `underset(4)overset(5)int e^(x)dx=[e^(x)]_(4)^(5)=e^(5)-e^(4)=e^(4)(e-1)` | |
| 175. |
योगफल की सिमा के रूप में `int_(-1)^(1) e^(x)dx` का मान ज्ञात कीजिएः |
|
Answer» यहाँ `a=-1, b=1, f(x)=e^(x), h=(b-a)/(n)=2/n` या योगफल की सिमा की परीभाषा से, `underset(a)overset(b)intf(x)dx=underset(h to 0)limh[f(a)+f(a+h)+f(a+2h)+.....+f{a+(n-1)h}]` `underset(-1)overset(1)intf(x)dx` `=underset(h to 0)limh [e^(-1)+e^(-1+h)+e^(-1+2h)+......+e^(-1(n-1)h)]` `=underset(h to 0)limh [e^(-1){1+e^(h)+2^(h)+....+e^((n-1)h)}]` `=underset(h to 0)limhe^(-1) [1+e^(h)+2^(h)+....+e^((n-1)h)]` `=underset(h to 0)limhe^(-1) [1+((e^(h))^(n)-1)/(e^(h)-1)] [therefore a+ar+.....+ar^(n-1)=(a(r^(n)-1))/(r-1)]` `=underset(h to 0)lim [(e^(nh)-1)/(((e^(h)-1)/(h)))]=underset(h to 0)lime^(-1)((e^(2)-1))/(((e^(h)-1)/(h)))` `=(e^(-1)(e^(2)-1))/(1)" "[therefore underset(h to 0)lim (e^(h)-1)/(h)=1]` `=e-e^(-1)` |
|
| 176. |
योगफल की सिमा के रूप में `underset(1)overset(3)(3x^(2)+1)dx` का मान ज्ञात कीजिएः |
|
Answer» यहाँ a=1, b=3, `f(x)=3x^(2)+1,h=(b-a)/(n)=2/n"या " nh=2` योगफल की सिमा की परिभाषा से, `underset(0)overset(b)int f(x)dx=underset(h to 0)lim h[(f(a)+f(a+h)+f(a+2h)+...+f{a+(n-1)h}]` `underset(1)overset(3)int (3x^(2)+1)dx=underset(h to 0)lim h[(f(1)+f(1+h)+f(1+2h)+...+f{1+(n-1)h}]........(1)` अब `f(1)=3(1)^(2)+1=4` `f(1+h)=3(1+h)^(2)+1=3(1+2h+h^(2))+1` `=3+6h+3h^(2)+1=4+6h+3h^(2)` `f(1+2h)=3(1+2h)^(2)+1=3(1+4h+4h^(2))+1` `=3+12h+12h^(2)+1=4+12h+12h^(2)` `f(1+(n-1)h]=3[{1+(n-1)h}]^(2)+1` `=3(1+2(n-1)h+(n-1)^(2)h+(n-1)^(2)h^(2)]+1` `=4+6h(n-1)+3h^(2)(n-1)^(2)` उपरोक्त मनो को समी, में रखने पर `underset(1)overset(3)int (3x^(2)+1)dx` `=underset(h to 0)limh[4+(4+6h+3h^2)+(4+12h+12h^(2))+{4+6h(n-1)+3h^(2)(n-1)^(2)}]` `=underset(h to 0)limh[4+4+....+4("n बार ")+{6h+12h+....+6h(n-1)}+{3h^(2)+12h^(2)+....+3h^(2)(n-1)^(2)}]` `=underset(h to 0)limh[4+4+....+4("n बार ")+6h(1+2+...+(n-1)}+3h^(2){1^(2)+2^(2)+.....+(n-1)^(2)}]` `=underset(h to 0)lim h[4n+6h xx (n(n-1))/(2)+3h^(2)xx(n(n-1)(2n-1))/(6)]` `=underset(h to 0)lim h[4n+6(nh(nh-h))/(2)+3(nh(nh-h)(2nh-h))/(6)]` `=underset(h to 0)lim h[4xx2+(6xx2xx(2-h))/(2)+(3xx2xx(2-h)(2xx2-h))/(6)]` `=8+(12(2-0))/(2)+(6(2-0)(4-0))/(6)` =8+12+8=28 |
|
| 177. |
योगफल की सिमा के रूप में `int_(0)^(2) e^(x)dx` का मान ज्ञात कीजिएः |
|
Answer» यहाँ `=0,b=2, f(x)=e^(x), h=(b-a)/(n)=2/n" या "nh=2` योगफल की सिमा की परिभाषा से, `underset(a)overset(b)int f(x)dx=underset(h to 0)lim h[f(a+h)+f(a+2h)+....+f{a+(n-1)h}]` `Rightarrow underset(0)overset(2)int e^(x)edx=underset(h to0)lim [f(0)+f(0+h)+f(0+2h)+....+f{0+(n-1)h}]` `Rightarrow underset(0)overset(2)int e^(x)edx=underset(h to0)lim [f(0)+f(h)+f(2h)+....+f{(n-1)h}]......(1)` ltBRgt अब `f(0)=e^(0)=1, f(h)=e^(h),f(2h)=e^(2h),f{n-1)h}=e^((n-1)h)` उपरोक्त मनो को समी, (1) में रखने पर `underset(0)overset(2)inte^(x)dx=underset(h to 0)lim h[e^(0)+e^(h)+e^(2h)+....+e^((n-1)h)]` `underset(0)overset(2)inte^(x)dx=underset(h to 0)lim h[e^(0)+e^(h)+(e^(h))^(2)+....+(e^(h))^(n-1)]` `=underset(h to 0)lim h[e^(0){(e^(h))^(n)-1)/(e^(h)-1)]` `[{:(,therefore a+ar+...+ar^(n-1)=(a(r^(n-1)))/(r-1)),(,"यहाँ "r=e^(h)):}]` `=underset(h to 0)lim h {(e^(nh)-1)/(e^(h)-1)}=underset(h to 0)limh{(e^(2)-1)/((e^(h)-1))}` `=underset(h to 0)lim h ((e^(2)-1))/(((e^(2)-1)/(h)))=(e^(2)-1)/(1) " "[therefore underset(h to 0)lim (e^(h)-1)/(h)=1]` `=e^(2)-1)` |
|
| 178. |
समाकलनो का मान ज्ञात कीजिएः `int_(pi//2)^(pi) (1-sin x)/(1-cos x)dx` |
|
Answer» यहाँ `I=underset(pi//2)overset(pi)int (1-sin x)/(1-cos x)dx` `=underset(pi//2)overset(pi)int (1-sin""x/2cos""x/2)/(2sin^(2)""x/2)` `=underset(pi//2)overset(pi)int ((1)/(2)"cosec"^(2)pi/2-cot""x/2)dx` `=underset(pi//2)overset(pi)int 1/2"cosec"^(2) x/2 dx-underset(pi//2)overset(pi)int cot ""x/2dx` `=[1/2-(-cot""x/2)/(1/2)]_(pi//2)^(pi)-[(log|sin""x/2|)/(1/2)]_(pi//2)^(pi)` `=[cot""x/2]_(pi//2)^(pi)-[2log|sin""x/2|]_(pi//2)^(pi)` `=[-cot""pi/2+cot""pi/4]-[2 log sin ""pi/2-2 log sin ""pi/4]` `=(0+1)-[2log1-2log ""1/sqrt2]=1+2log""1/sqrt2` `=1-2log sqrt2=1-log 2` |
|
| 179. |
योगफल की सिमा के रूप में `int_(0)^(1) e^(2-3x)dx` का मान ज्ञात कीजिएः |
|
Answer» यहाँ `0,b=1, f(x)=e^(2-3x),h(b-a)/(n)=(1)/(n)"या "nh=1` योगफल की सिमा की परिभाषा से, `underset(a)overset(b)int f(x)dx` `=underset(h to 0)limh[f(a)+f(a+h)+f(a+2h)+.....+f{a+(n-1)h}]` `=underset(h to 0)lim h[f(0)+f(0+h)+f(0+2h)+.....+f{a+(n-1)h}]` `=underset(h to 0)lim h[f(0)+f(h)+f(2h)+.....+f{a+(n-1)h}]` `=underset(h to 0)lim f[e^(2)+2^(2-3h)+e^(2-3(2h))+....+e^(2-3(n-1)h)]` `=underset(h to 0)lim he^2[1+e^(-3h)+e^(-3(2h))+....+e^(-3(n-1)h)]` `=underset(h to 0)lim he^2[1+e^(-3h)+(e^(-3h))^(2)+....+(e^(-3h))^(n-1)]` `=underset(h to 0)lim he^2[((e^(-3))^(n)-1)/(e^(-3h)-1)]=underset(h to 0)lim he^(2)[(e^(-3nh)-1)/(e^(-3h)-1)]` `=underset(h to 0)lim he^2[(e^(-3)-1)/(((e^(-3h)-1)/(h)))]` `=e^(2)xx(e^(-3)-1)xx(1)/(underset(h to 0)lim((e^(-3h)-1)/(-3)h)xx-3)` `=e^(2)(e^(-3)-1)xx(1)/(1xx(-3)) " "[therefore underset(x to 0)lim (e^(x)-1)/(x)=1]` `=-1/3(e^(-1)-e^(2))=1/3(e^(2)-e^(-1))` |
|
| 180. |
योगफल की सिमा के रूप में `int_(0)^(2) 2^(x)dx` का मान ज्ञात कीजिएः। |
|
Answer» यहाँ `a=0,b=2,f(x)=2^(x), nh=b-a=2 ` योगफल की सिमा की परिभाषा से, `underset(a)overset(b)int f(x)dx=underset(h to 0)lim h[f(a)+f(a+h)+f(a+2h)+.....+f{a+(n-1)h}]` `underset(0)overset(2)int 2^(x)dx` `=underset(h to 0)limh[f(0)+f(h)+f(2h)+...f{(n-1)h}]` `=underset(h to 0)lim h[2^(0)+2^(h)+2^(2h)+...2^((n-1))]` `=underset( to 0)lim h[1+2^(h)+2^(2h)+...2^((n-1)h)]` `=underset( to 0)lim {{(2h)^(n)-1)/((2^(h)-1))} [therefore a+ar+...+ar^(n-1)=(a(r^(n-)1))/(r-1)` `=underset(h to 0)lim h((2^(nh)-1))/((2^(h)-1))=underset(h to 0)lim ((2^(2)-1))/(((2^(h)-1)/(n)))` `=((2^2-1))/(log_(e)2)=(3)/(log_(e)2), [therefore underset(h to 0)lim (2^(h)-1)/(h)=log2]` |
|
| 181. |
योगफल की सिमा के रूप में `int_(1)^(3) x^(3)dx` का मान ज्ञात कीजिएः |
|
Answer» यहाँ `a=1, b=3, f(x)=x^(3), nh=b-a=2` योगफल की सिमा की परिभाषा से, `underset(a)overset(b)int f(x)dx=underset(h to 0)limh[f(1)+f(1+h)+f(1+2h)+....+f(1+(n-1)h}]....(1)` अब `f(1)=(1)^(3)=1` `f(1+h)=(1+h)^(3)=1+h^(3)+3h+3h^(2)` `=1+3h+3h^(2)+h^(3)` `f(2+h)=(1+2h)^(3)=1+3(2h)+3(2h)^(2)+(2h)^(3)` `=1+6h+12h^2+8h^3` `f[1+(n-1)h]=1+3(n-1)h+3(n-1)^2h^2+(n-1)^(3)h^(3)` उपरोक्त मनो को समी में रखने पर `underset(1)overset(3)intx^(3)dx` `=underset(h to 0)lim h{1+(1+3h+3h^(2)+h^(3)) +(1+6h+12h^(2)+8h^(3)) +{1+3(n-1)h+3(n-1)^(2)h^(2)+(n-1)^(3)h^(3)}]` `=underset(h to 0)lim h[1+1+......+1("n बार ") +3h{1+2+......+(n-1)} +3h^(2){1^(2)+2^(2)+...+(n-1)^(2)} +h^(3){1^3+2^3+....+(n-1)^(3)}]` `=underset(h to 0)lim h[n+3hxx(n(n-1))/(2)+3h^(2)xx(n(n-1)(2n-1))/(6)+h^(3)xx {(n(n-1))/(2)}^(2)}` `=underset(h to 0)lim [nh+(3xxnh(nh-h))/(2)+(3xxnh(nh-h)(2nh-h))/(6)+{(nh(nh-h))/(2)}^2]` `=[2+(3xx2xx(2-0))/(2)+(3xx2xx(2-0)(2xx2-0))/(6)+{(2(2-0))/(2)}^(2)]` =2+6+8+4=20 |
|
| 182. |
योगफल की सिमा के रूप में `=int_(a)^(b) sin xdx` का मान ज्ञात कीजिएः। |
|
Answer» यहाँ f(x)=sin x, nh=b-a हम जानते है की `underset(a)overset(b)int f(x)dx=underset(x to 0)limh[f(a)+f(a+h)+f(a+2h)+.....+f{a+(n-1)h}]` `underset(a)overset(b)int sinx dx` `=underset(h to 0)lim h[sin a+sin(a+h)+sin(a+2h)+....+sin{a+(n-1)h}]` `=underset(h to 0)lim h[(sin {a+((n-1)/(2))h}sin((nh)/(2))}/(sin((h)/(2)))]` `=underset(h to 0)lim h[(sin(a+(nh)/(2)-(h)/(2))sin((nh)/(2)))/(sin((h)/(2)))]` `=underset(h to 0)lim h[(sin(a+(b-a)/(2)-(h)/(2))sin((b-a)/(2)))/(sin ((h)/(2)))]` `=underset(h to 0)lim [((h)/(2))/(sin((h)/(2)))xx2xxsin(a+(b-a)/(2)(h)/2)]sin((b-a)/(2))]` `=1xx2xxsin(a+(b-a)/(2))sin((b-a)/(2))` `=2sin ((a+b)/(2))sin ((b-a)/(2))` `=cosa-cosb [therefore 2 sinA sin B=cos(A-B)-cos(A+B)` |
|
| 183. |
योगफल की सिमा के रूप में `int_(1)^(4)(x^2-1)dx` का मान ज्ञात कीजिएः। |
|
Answer» यहाँ a=1, b=4 `f(x)=x^(2)-x, h=(b-a)/n=3/n "या" nh=3` योगफल की सिमा की परिभाषा से, `underset(a)overset(b)intf(x)dx=underset(h to 0)lim h[f(a)+f(a+h)+f(a+2h)+....+f{a+(n-1)h}]` `underset(1)overset(4)int f(x^(2)-x)dx=underset(h to 0)lim h[f(1)+f(1+h)+f(1+2h)+....+f{1+(n-1)h}]........(1)` अब `f(1)=(1)^(2)-1=0` `f(1+h)=(1+h^(2))-(1+h)` `=(1+2h+h^(2))-1-h=h^(2)+h` `f(1+2h)=(1+2h)^(2)-(1+2h)` `=1+4h+4h^(2)-1-2h=4h^(2)+2h` `f[1+(n-1)h]` `=[1+(n-1)h]^(2)-[1+(n-1)h]` `=1+2(n-1)h+(n-1)^(2)h^(2)-1-(n-1)h` `=(n-1)^(2)h^(2)+(n-1)h` उपरोक्त मनो को समी (1) में रखने पर `underset(1)overset(4)int(x^(2)-x)dx` `=underset(h to 0)lim h[0+(h^(2)+h)+(4h^(2)+2h)+.....+{(n-1)^(2)h^(2)+(n-1)h}]` `=underset(h to 0)lim h[h^(2){1^2+2^2+.....+(n-1)^2}+h{1+2+...+(n-1)}]` `=underset(h to 0)lim [h^2 xx(n(n-1))/6+hxx(n(n-1))/(2)]` `=underset(h to 0)lim [(nh(nh-h)(2nh-1))/(6)+(nh(nh-h))/(2)]` `=underset(h to 0)lim [(3(3-h)(6-h))/6+(3(3-h))/(2)]` `=(3(3-0)(6-0))/6+(3(3-0))/(2)=9+9/2=27` |
|
| 184. |
`int_(0)^(sqrt2) sqrt(2-x^(2)dx)` का मान ज्ञात कीजिएः। |
|
Answer» `underset(0)overset(sqrt2)int sqrt(2-x^(2)dx)` `underset(0)overset(sqrt2)int sqrt((sqrt2)^(2)-x^(2))dx` `[1/2xsqrt(2-x^(2))+((sqrt2)^(2))/(2)sin^(-1) ""(x)/(sqrt2)]_(0)^(sqrt2)` `=[sqrt2/2xx0+sin^(-1)1]-[0+0]` `=0+pi/2=pi/2` |
|
| 185. |
इन समक़लानो का मान ज्ञात कीजिएः `int_(2)^(3) x^(2)dx` |
|
Answer» `underset(2)overset(3)intx^(2)dx=[(x^(3))/(3)]_(2)^(3)` `" "=[(3^(3))/(3)-(2^3)/(3)]` `=(27)/(3)-(8)/(3)=(19)/(3)` |
|
| 186. |
योगफल की सिमा के रूप में `int_(0)^(pi//2) sin x dx` का मान ज्ञात कीजिएः। |
|
Answer» यहाँ `a=0,b=(pi)/(2),f(x)=sinx,nh=(pi)/(2)-0=(pi)/(2).` `therefore underset(0)overset(pi//2)intsinxdx=cos0-cos(pi)/(2)-1-0=1` |
|
| 187. |
`int_(0)^(1) (3x^(2)+2x+k)dx=0` तब k का मान ज्ञात कीजिएः। |
|
Answer» `underset(0)overset(1)int (3x^(2)+2x+k)dx=0` `rArr [x^(3)+x^(2)+kx]_(0)^(1)=0` `rArr 1+1+k-0=0` `rArr k=-2` |
|
| 188. |
संकलनो का मान ज्ञात कीजिएः। `int_(0)^(1) (tan^(-1)x)/(1+x^(2))dx` |
|
Answer» यहाँ `underset(0)overset(1)int (tan^(-1))/(1+x^(2))dx` माना `tan^(-1) x=t rArr (1)/(1+x^(2)) dx=dt` जब `x=0 rArr t=tan^(-1) 0=0` और `x=1 rArr t=tan^(-1) =pi/4` `therefore I=underset(0)overset(pi//4)int tdt=[(t^(2))/(2)]_(0)^(pi//4)` `rArr I=1/2[(pi/4)^2-0]` `=1/2xx(pi)^(2)/16=(pi^(2))/(32)` |
|
| 189. |
यदि `int_(0)^(a) (1)/(4+x^(2))dx=pi/8, ` तब a का मान ज्ञात कीजिएः। |
|
Answer» `underset(0)overset(a)int (1)/(4+x^(2))dx=pi/8, ` `rArr underset(0)overset(a)int (1)/(x^(2)+2^(2))dx=pi/8` `rArr [1/2tan^(-1)""pi/2]_(0)^(a)=pi/8` `rArr 1/2tan^(-1)""a/2-1/2tan^(-1)0=pi/8` `rArr 1/2tan^(-1)""a/2-0=pi/8` `rArr tan^(-1)""a/2=pi/4` `rArr a/2=tan(pi/4)=1` `rArr a=2` |
|
| 190. |
संकलनो का मान ज्ञात कीजिएः। `int_(0)^(1) xe^(x^(2))dx` |
|
Answer» यहाँ `underset(4)overset(9)int xe^(x^(2))dx` माना `x^2 = t rArr 2xdx=dt "या " xdx=(dt)/(2)` जब `x =0 rArr t=0` और `x=1 rArrr t=(1)^(2)=1`. `I=1/2 underset(0)overset(1) int e^(t)dt` `=1/2[e^(t)]_(0)^(1)=1/2[e-e^(0)]` |
|
| 191. |
संकलनो का मान ज्ञात कीजिएः। `int_(e)^(e^(2)) (dx)/(x log x)` |
|
Answer» यहाँ `underset(e)overset(e^(2))int (dx)/(x log x)` माना `log x=t rArr 1/x dx=dt` जब `x=e rArr t=log e=1` और `x=e^(2) rArr t=log e^(2)=2log e=2` `therefore I=underset(1)overset(2)int (dt)/t=[log t]_(1)^(2)-log 1=log 2` |
|
| 192. |
संमाकलनो का मान ज्ञात कीजिएः। `int_(4)^(9) (sqrtx)/((30-x^(3//2))^(2))d x` |
|
Answer» यहाँ `underset(4)overset(9)int (sqrtx)/((3-x^(3//2))^(2))dx` माना `30-x^(3//2)=t rArr -3/2sqrtx dx=dt` और `x=9 rArr 30=(9)^(3//2)=39-27=3` `therefore I=2/3 underset(22)overset(3)int =-2/3underset(22)overset(3)int t^(-2)dt` `=-2/3.[(-1)/(t)]_(22)^(3)=-2/3[-1/3+1/22]` `=-2/3 xx (-19)/(66)=(19)/99` |
|
| 193. |
`int_(0)^(pi//2) sin^(5)x dx`=A. `(8)/(15)`B. `(4sqrt(x))/(15)`C. `(8sqrt(x))/(15)`D. `(8pi)/(15)` |
| Answer» Correct Answer - A | |
| 194. |
`int_(1)^(sqrt(3)) (1)/(1+x^(2))dx=`A. `(pi)/(4)`B. `(pi)/(3)`C. `(pi)/(12)`D. इनमे से कोई नहीं |
| Answer» Correct Answer - C | |
| 195. |
`int_(0)^(pi//2) cos^(2)x dx `=A. `pi`B. `pi/2`C. `pi/4`D. इनमे से कोई नहीं |
| Answer» Correct Answer - C | |
| 196. |
`int_(0)^(a) sqrt(a^(2)-x^(2))dx=`A. `(a^(2))/(4)`B. `(a^(2))/(2)`C. `(pia^(2))/(2)`D. `(pia^(2))/(4)` |
| Answer» Correct Answer - D | |
| 197. |
`int_(0)^(pi//4) tan^(6)x sec^(2)x dx=`A. `1//7`B. `2//7`C. `1`D. इनमे से कोई नहीं |
| Answer» Correct Answer - A | |
| 198. |
`int_(0)^(pi//2) log tan x dx =`A. `pi/2 log_(e) 2`B. `-(pi)/(2)log_(e)2`C. `pilog_(e)2`D. 0 |
| Answer» Correct Answer - D | |
| 199. |
`int_(-pi//2)^(pi//2) sin^(9)xdx=` |
| Answer» Correct Answer - A | |
| 200. |
यदि `int_(0)^(a) 3x^(2)dx=8` तो a=A. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - B | |