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1.

ज्ञात करे [Evaluate ] ` int_(0)^(pi//2) (x)/(sin x +cos x) dx `

Answer» माना कि ` f(x) = x/(sin x + cos x) " " ` … (1)
तो , ` f(pi/2-x) = (pi/2-x)/(sin(pi/2-x)+cos(pi/2-x))`
या , ` f (pi/2 - x) = (pi/2-x)/(cos x +sinx) " "` (2)
अब ,(1) +(2) `rArr f(x) +f(pi/2-x) = pi/2 1/(2cos x+ sinx)`
` = pi/(2sqrt(2)cos (x-pi/4))=pi/(2sqrt(2)) sec (x- pi/4)`
अब , ` I = 1/2 int_(0)^(pi//2) [ f(x)+f(pi/2-x)]dx `
` =1/2 . pi/(2sqrt(2)) int_(0)^(pi//2) sec (x- pi/4) dx `
` = pi/(4sqrt(2))[log|sec(x-pi/4)+tan (x-pi/4)|]_(0)^(pi//2)`
` = pi/(4sqrt(2)) [ log | sec. pi/4 +tan pi/4 | - log |sec.pi/4 - tan .pi/4 |]`
` = pi/(4sqrt(2)) [ log (sqrt(2)+1)-log (sqrt(2)-1)] `
` = pi/(4sqrt(2)) log ((sqrt(2)+1)/(sqrt(2)-1)) = pi/(4sqrt(2))log (sqrt(2)+1)^(2) = pi/(2sqrt(2)) log (sqrt(2)+1)`
Discussion
2.

निम्नलिखित को ज्ञात करे : (ii) `int_(0)^(2)xsqrt(2-x)dx`

Answer» माना कि `I = int _(0)^(2) x sqrt(2-x) dx" "`…(1) ,
x कि जगह `(2-x)` रखने पर हमे मिलता है ,
` I = int _(0)^(2) (2-x) sqrt(2-(2-x))dx`
या ` I = int _(0)^(2)(2-x) sqrt(x) dx = int _(0)^(2)[ 2 sqrt(x)- x^(3//2)] dx `
` = [ 2. 2/3x^(3//2)-2/5 x^(5//2) ]_(0)^(2)= ( 4/3 2^(3//2) - 2/5 2^(5//2)) - 0 `
` = 4/3 sqrt(2^(3)) -2/5 sqrt(2^(5)) = 4/3 . 2 sqrt(2) - 2/5 4 sqrt(2)`
` = (8sqrt(2))/3 - (8sqrt(2))/5 = (8 sqrt(2))/15 (5-3) = (16sqrt(2))/15`
Discussion
3.

ज्ञात करे [ Evaluate] `int_(1)^(4)f(x)dx, "` जहाँ (where) ` f(x) = |x -1| + |+| x-2| + | x- 3|`

Answer» यहाँ `|x-1|,|x-2|` तथा`|x-3|` आता है
अब , ` x-1 = 0 rArr x = 1 , x - 2 = 0 rArr x = 2 " तथा " x - 3 = 0 rArr x = 3 `
अब, ` = int _(1)^(4) f(x) dx = int_(1)^(2) f(x) dx + int _(2)^(3)f(x) dx + int _(3)^(4) f(x) dx `
= ` int _(1)^(2) [{ (x-1)- (x-2) - (x-3) } dx + int_(2)^(3) {(x-1)+(x-2) - (x-3) }] dx `
` + int _(3)^(4) {(x-1)+(x-2)+(x-3)dx}`
`= int _(1)^(2) (-x+4) dx + int _(2)^(3) x dx + int _(3)^(4) (3x-6)dx `
` = [ (-x^(2))/2 + 4x]_(1)^(2) +[ (x^(2))/2] _(2)^(3) + [ (3x^2)/2-6x]_(3)^(4) = ( 5/2 +5/2 + 9/2 )=19/2`
Discussion
4.

ज्ञात करे ` int _(-1)^(1)x^(3) e^(x^(4))dx`

Answer» माना कि ` f (x) = x^(3) e^(x^(4))`
तो , ` f (-x) = (-x)^(3) e^((-x)^(4)) = - x^(3) e^(x^(4)) = - f (x)`
अतः f(x) एक विषम फलन है
` :. int _(-1)^(1) f(x) dx = 0 " या " int _(-1)^(1) x^(3) e^(x^(4)) dx = 0 `
Discussion
5.

` int _(1)^(3)x^(3)dx` का मान ज्ञात कीजिए ।

Answer» परिभाषा से , ` int _(a)^(b) f(x) dx = lim _(h to 0) sum_(r=1)^(n) h f (a+rh)" "` …(1)
जहाँ, `nh = b-a` तथा `n to oo`
यहाँ a = 1, b=3 ` :. nh = b- a = 3 -1 = 2`
` f(x) = x^(3) :. f(a+rh) = f(1+rh) = (1+rh)^(3)`
` = 1^(3) +3rh + 3r^(2) h^(2) + r^(3) h^(3)`
अब (1 ) से , ` int _(1)^(3) x^(3) dx = lim _(h to 0 ) sum_(r=1)^(n) h f(a+rh)`
` = lim _(h to 0 ) sum_(r=1)^(n) h (1^(3) + 3rh + 3h^(2) r^(2) +h^(3)r^(3))`
` = lim _(h to 0 ) ( h sum _(r=1)^(n) 1+ 3h^(2) sum_(r=1)^(n)r + h^(4) sum_(r=1)^(n) r^(3))`
` = lim _(h to 0) [ nh + 3h^(2) (n(n+1))/2 3 h^(2) + (n(n+h)(2n+1))/1 + h^(2) (n^(2)(n+1)^(2))/4 ] `
` = lim _( h to 0 ) [ nh + 3/2 nh (nh+h) + (nh(nh+h)(2nh+h))/2 + ((nh)^(2)(nh+h)^(2))/4]`
` lim _( h to 0 ) [ 2 + 3/2 xx 2 (2+h) + (2(2+h)(4+h))/h + ((2)^(2).(2+h)^(2))/4] = 20 `
Discussion
6.

निम्नलिखित को ज्ञात करे । (v) ` int _(-1)^(1) (|x|+|x -1|) dx`

Answer» Correct Answer - 3
`I = int _(-1)^(1) | x| dx + int _(-1)^(1) | x-1| dx `
Discussion
7.

माना कि ` I = int _(-pi)^(pi) (2x (1+ sin x))/(1+cos^(2) x)dx `

Answer» तो , ` I = int _(pi)^(pi) (2x)/(1+cos^(2) x ) dx + int _(-pi)^(pi) (2x sin x)/(1+cos^(2) x ) dx = I _(1) + I_(2)`
जहाँ ` I _(1) = int _(-pi)^(pi) (2x)/(1+cos^(2) x ) dx " तथा " I_(2) = int _(-pi)^(pi) (2x sin x )/(1+cos^(2) x) dx " "` …(2)
चूँकि `(2x)/(1+cos^(2)x) ` एक विषम फलन है तथा ` (2x sin x)/(1+cos^(2)x ) dx ` एक सम फलन है ।
` :. I _(1) = 0 " " I_(2) = 2 int _(0)^(pi) (2x sin x)/(1+cos^(2) x ) dx `
अब , ` I _(2) = 4 int _(0)^(pi) (x sin x)/(1+cos^(2) x ) dx " "` ..(3) ,
` :. I _(2) = 4 int _(0)^(pi) ((pi-x)sin (pi-x))/(1+cos^(2) (pi-x))dx`
(3 ) और (4 ) को जोड़ने पर हमे मिलता है ,
` 2I_(2) = 4pi int _(0)^(pi) 1/(1+cos^(2) x ) sin xdx = - 4x int_(1)^(-1) 1/(1+t^(2)) dt , " जहाँ " t = cos x `
` = - 4pi [ tan^(-1) t ] _(1)^(-1) = - 4pi [ -pi//4 - pi//4 ] = 2pi^(2)`
` :. 2I_(2) = 2pi^(2) rArr I_(2) = pi^(2) `
अतः ( 2 ) से , ` I = 0 + pi^(2) = pi^(2) `
Discussion
8.

ज्ञात करे `int _(-pi//4)^(pi//4) x^(3) sin^(4) x dx `

Answer» माना कि ` f (x)= x^(3) sin^(4) x " "` …(1)
तो , ` f(-x) = (-x) ^(3) sin^(4) (-x) = -x^(3) (-sin x)^(4) = - x^(3) sin^(4) x = - f (x)`
अतः f(x) एक विषम फलन है ।
` :. int _(-pi//4)^(pi//4) f (x) dx = 0 `
Discussion
9.

ज्ञात करे[ Evalute]` int _(0)^(3//2) | x cos pi x | dx `

Answer» ` x cos pi x = 0 rArr {{:(x =0 ),(cos pix = 0 " या " pi x = (2n+1)pi/2 " , " n in Z ):}`
` rArr {{:( x=0 ),(x=1/2","0 " तथा " 3/2 " के बीच ") :}`
अब , ` int _(0)^(3//2) | x cos pi x | dx = int _(0)^(1//2) | x cos pix| dx + int _(1//2)^(3//2) | x cos pix | dx `
` = int _(0)^(1//2) x cos pi x dx - int _(1//2)^(3//2) x cos pix //dx`
` = [ (x sin pi x)/pi + (cos pix)/pi ]_(0)^(1//2) - [ (xsin pix)/pi+ (cos pix)/(pi^(2))]_(1//2)^(3//2)`
` = (1/(2pi)-1/(pi^(2)))-(-3/(2pi)-1/(2pi))= 1/(2pi) - 1/(pi^(2)) + 2/pi = 5/(2pi) - 1/(pi^(2))`
Discussion
10.

निम्नलिखित को ज्ञात करे [ Evalute the following ] : (iii)` int _(0)^(1)x(1-x)^(n) dx `

Answer» माना कि ` I = int _(0)^(1) x (1-x)^(n) dx " "` (1)
x कि जगह `(1-x )` रखने पर हमे मिलता है ,
या ` I = int _(0)^(1) (1-x) x^(n) dx = int _(0)^(1) (x^(n) - x^(n+1) )dx `
` = [ (x^(n+1))/(n+1) - (x^(n+2))/(n+2)] _(0)^(1) = ( 1/(n+1) - 1/(n+2)) - 0`
` = (n+2-n-1)/((n+1)(n+2))=1/((n+1)(n+2))`
Discussion
11.

निम्नलिखित को ज्ञात करे [ Evalute the following ] : (i) ` int _(0)^(a) (sqrt(x))/(sqrt(x)+sqrt(a-x))dx `

Answer» माना कि ` I = int _(0)^(a) (sqrt(a))/(sqrt(x)+sqrt(a-x))dx" "` (i)
समाकल्य में x कि जगह `(a-x )` रखने पर हमे मिलता है ,
` I = int_(0)^(a)(sqrt(a-x))/(sqrt(a-x)+sqrt(a-(a-x)))dx`
` I = int _(0)^(a) (sqrt(a-x))/(sqrt(a-x)+sqrt(x))dx" " ` …(2)
(1) और (2) को जोड़ने पर हमे मिलता है ,
` 2I = int _(0)^(a) (sqrt(x)+sqrt(a-x))/(sqrt(x)+sqrt(a-x)) dx = int _(0)^(a) 1 dx = [ x ] _(0)^(a) = a - 0 = a`
` :. I = 1/2 a `
Discussion
12.

ज्ञात करे[ Evalute] ` int_(0)^(pi//4) log (1+tan x)dx`

Answer» माना कि `f(x) = log (1+tanx)`
तो , ` f(pi/4-x) = log [ 1+ tan (pi/4 -x) = log (1+(1-tanx)/(1+tanx))`
या ` f (pi/4-x) = log. 2/(1+tanx) " "` …(2) ,
अब `(1) + (2) rArr f(x) (pi/4-x) = log (1+tan x) log (2/(1+tan x)) = log 2 `
` :. I = 1/2 int _(0)^(pi//4)[ f(x) + f (pi/4 -x)] dx = 1/2 int_(0)^(pi//4)log2 dx`
` = 1/2 * pi/4 log 2 = pi/8 log 2 `
Discussion
13.

ज्ञात करे । ` int _(pi//4)^(3pi//4) (phi)/(1+sin phi) d phi = `A. ` pi(sqrt(2)+1)`B. ` pi (sqrt(2)-1)`C. ` pi`D. `- pi`

Answer» Correct Answer - B
Discussion
14.

(iii)सिद्ध करे कि , ` int _(0)^(1) tan^(-1). ((2x-1)/(1+x-x^(2)))dx = 0`

Answer» ` tan^(-1) ((2x-1)/(1+x-x^(2))) = tan^(-1).(x-(1-x))/(1+x(1-x)) = tan^(-1) x - tan^(-1) (1-x)`
माना कि ` I = int _(0)^(1) [ tan^(-1) x - tan^(-1) (1-x) ] dx `
तो, ` I = int _(0)^(1) [ tan^(-1) (1-x) - tan^(-1) x ] dx " " :. 2I = 0 rArr I = 0 `
Discussion
15.

` int _(-1//2)^(1//2) cos x log ((1+x)/(1-x)) dx = `A. log 3B. `cos 1. log 2 `C. 0D. 1

Answer» Correct Answer - C
Discussion
16.

ज्ञात करे `int _(-1)^(1) log ((2-x)/(2+x)) dx = 0`

Answer» माना कि ` f(x) = log ((2-x)/(2+x))`
तो ` f (-x) = log ((2+x)/(2-x)) = log ((2-x)/(2+x))^(-1) = - log ((2-x)/(2+x)) = - f(x)`
अतः f(x) एक विषम फलन है ।
` :. int _(-1)^(1) log ((2-x)/(2+x)) dx = 0 `
Discussion
17.

ज्ञात करे[ Evalute] `int _(0)^(1) (log (1+x))/(1+x^(2))dx`

Answer» x = tan `theta ` , रखे , तो ` I = int _(0)^(pi//4) log (1+tan theta ) d theta = pi/8 log 2 ` [ 16 (i) कि तरह ]
Discussion
18.

The value of ` int _(-1)^(2) (|x|)/x dx ` isA. 1B. `-1`C. 0D. 2

Answer» Correct Answer - A
Discussion
19.

`int _(-pi)^(pi) (cos^(2)x)/(1+a^(x)) dx, a gt 0`A. `pi`B. `api`C. `pi/2`D. `2pi`

Answer» Correct Answer - C
Discussion
20.

`int _(0)^(pi/2) log tan xdx =….`A. `pi/4`B. `pi/2`C. 0D. `pi`

Answer» Correct Answer - C
Discussion
21.

Let f, `R to R and g : R to R` be continous functions . Then the value of the integral ` int _(-pi//2)^(pi//2) [ f(x) + f( -x) ] [ g(x) - g (-x) ] dx ` isA. `pi`B. 1C. `-1`D. 0

Answer» Correct Answer - D
Discussion
22.

यदि`int _(-1)^(4) g(x) dx = 4 तथा int _(2)^(4) (3- g(x)) dx = 7` , तो इसका मान क्या होगा ` int _(-1)^(2) g(x) dx`A. `-2`B. 3C. 5D. none of these

Answer» Correct Answer - C
Discussion
23.

` int _(-pi//2)^(pi//2) sin^(2) x cos^(2) x ( sin x + cos x ) dx` =A. `2/5`B. `2/15`C. `4/15`D. `8/15`

Answer» Correct Answer - C
Discussion
24.

यदिf(x) एक विषम फलन तथा` int _(0)^(5) f(x) dx = 10`., फिर` int _(-5)^(5) f(x) dx = ….`A. 10B. 0C. 20D. `-10`

Answer» Correct Answer - B
Discussion
25.

If f(x) is a function such that `f(20 -x) = f (x)`, then ` int _(0)^(20) f(x) dx =….`A. `2 int _(0)^(10) f(x) dx`B. 0C. `int _(0)^(10) f(x) dx `D. 20

Answer» Correct Answer - A
Discussion
26.

` int _(0)^(pi//2) 1/(1+tan theta ) d theta` =A. log 3B. 1C. `pi/2`D. `pi/4`

Answer» Correct Answer - D
Discussion
27.

निम्न समाकलन का मान `int _(-2)^(2) (ax^(3) + bx + c)` dx निर्भर करेगाA. the value of aB. the value of bC. the value of c,D. none of these

Answer» Correct Answer - C
Discussion
28.

The value of the integral ` int _(0)^(pi//2) (sqrt(cos x))/(sqrt(cosx)+ sqrt(sin x)) dx `

Answer» Correct Answer - C
Discussion
29.

The value of ` int _(-pi//2)^(pi//2) sqrt((1-cos 2theta )/2 ) d theta ` isA. `1/2`B. 1C. 2D. 0

Answer» Correct Answer - C
Discussion
30.

` int _(-1)^(2) x| x | dx `के बराबर हैA. ` int _(-1)^(2) x^(2) dx`B. `7/3`C. 3D. none of these

Answer» Correct Answer - B
Discussion
31.

यदि`g(x) = int _(0)^(x) cos4tdt ` फिर`g(x + pi)` बराबर होगाA. `g(x) +g(pi)`B. `g(x)-g(pi)`C. `g(x)+g(pi)`D. `(g(x))/(g(pi))`

Answer» Correct Answer - A
Discussion
32.

` int _(0)^sqrt(2)[x^(2)] dx` बराबर है

Answer» Correct Answer - sqrt(2)-1
Discussion