Practice Start Test

Definite Integration

JEE Mathematics · 230 questions · Page 4 of 23 · Click an option or "Show Solution" to reveal answer

Q31
If f(x)=∫0xt(sin⁡x−sin⁡t)dt   f(x) = \int\limits_0^x {t\left( {\sin x - \sin t} \right)dt\,\,\,} then :
A f'''(x) + f''(x) = sinx
B f'''(x) + f''(x) −- f'(x) = cosx
C f'''(x) + f'(x) = cosx −- 2x sinx
D f'''(x) −- f''(x) = cosx −- 2x sinx
Correct Answer
Option C
Solution

f(x) =

∫0xt(sin⁡x−sin⁡t).dt\int_0^x {t(\sin x - \sin t).dt}

= sin x

∫0xt.dt−∫0xtsin⁡t.dt\int_0^x {t.dt - \int_0^x {t\sin t.dt} }

=

x22{{{x^2}} \over 2}

sin x +

[tcos⁡t]0x\left[ {t\cos t} \right]_0^x

+ sin x ⇒\Rightarrowf(x) =

x22{{{x^2}} \over 2}

sinx + xcosx + sinx f'(x) =

x22{{{x^2}} \over 2}

cosx + 2cos x f''(x) = x cos x −-

x22{{{x^2}} \over 2}

sin x −- 2sin x f'''(x) = cos x −- 2x sin x −-

x22{{{x^2}} \over 2}

cos x −- 2cos x

∴   \therefore\,\,\,

f'''(x) + f'(x) = cos x −- 2x sin x

Q32
lim⁡n→∞((n+1)1/3n4/3+(n+2)1/3n4/3+.......+(2n)1/3n4/3)\mathop {\lim }\limits_{n \to \infty } \left( {{{{{(n + 1)}^{1/3}}} \over {{n^{4/3}}}} + {{{{(n + 2)}^{1/3}}} \over {{n^{4/3}}}} + ....... + {{{{(2n)}^{1/3}}} \over {{n^{4/3}}}}} \right) is equal to :
A 43(2)3/4{4 \over 3}{\left( 2 \right)^{3/4}}
B 34(2)4/3−34{3 \over 4}{\left( 2 \right)^{4/3}} - {3 \over 4}
C 43(2)4/3{4 \over 3}{\left( 2 \right)^{4/3}}
D 34(2)4/3−43{3 \over 4}{\left( 2 \right)^{4/3}} - {4 \over 3}
Correct Answer
Option B
Solution
lim⁡n→∞(n+1)13+(n+2)13+.....+(n+n)13n(n)13\mathop {\lim }\limits_{n \to \infty } {{{{(n + 1)}^{{1 \over 3}}} + {{(n + 2)}^{{1 \over 3}}} + ..... + {{\left( {n + n} \right)}^{{1 \over 3}}}} \over {n{{(n)}^{{1 \over 3}}}}}
⇒lim⁡n→∞∑r=1n(n+r)13n.n13   rn→x and 1n→dx \Rightarrow \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{{(n + r)}^{{1 \over 3}}}} \over {n.{n^{{1 \over 3}}}}}\,\,\,{r \over n} \to x\,and\,{1 \over n} \to dx\,}
⇒∫01(1+x)13dx\Rightarrow \int\limits_0^1 {{{(1 + x)}^{{1 \over 3}}}dx}
⇒[34(1+x)43]01=34(2)43−34\Rightarrow \left[ {{3 \over 4}{{(1 + x)}^{{4 \over 3}}}} \right]_0^1 = {3 \over 4}{(2)^{{4 \over 3}}} - {3 \over 4}
Q33
If ∫0π2cot⁡xcot⁡x+cos⁡ecxdx\int\limits_0^{{\pi \over 2}} {{{\cot x} \over {\cot x + \cos ecx}}} dx = m(π\pi + n), then m.n is equal to
A - 1
B 1
C −12 - {1 \over 2}
D 12{1 \over 2}
Correct Answer
Option A
Solution
∫0π/2cot⁡xcot⁡x+cosec xdx=∫0π/2cos⁡xcos⁡x+1dx\int\limits_0^{\pi /2} {{{\cot x} \over {\cot x + {\mathop{\rm cosec}\nolimits} \,x}}dx = } \int\limits_0^{\pi /2} {{{\cos x} \over {\cos x + 1}}dx}
⇒∫0π/22cos⁡2x2−12cos⁡2x2dx=∫0π/2(1−12sec⁡2x2)dx\Rightarrow \int\limits_0^{\pi /2} {{{2{{\cos }^2}{x \over 2} - 1} \over {2{{\cos }^2}{x \over 2}}}dx = } \int\limits_0^{\pi /2} {\left( {1 - {1 \over 2}{{\sec }^2}{x \over 2}} \right)dx}
⇒(x−tan⁡x2)0π/2=π2−1=12(π−2)\Rightarrow \left( {x - \tan {x \over 2}} \right)_0^{\pi /2} = {\pi \over 2} - 1 = {1 \over 2}\left( {\pi - 2} \right)

mn =

12x−2=−1{1 \over 2}x - 2 = - 1
Q34
The value of the integral ∫01xcot⁡−1(1−x2+x4)dx\int\limits_0^1 {x{{\cot }^{ - 1}}(1 - {x^2} + {x^4})dx} is :-
A π2−12log⁡e2{\pi \over 2} - {1 \over 2}{\log _e}2
B π4−log⁡e2{\pi \over 4} - {\log _e}2
C π4−12log⁡e2{\pi \over 4} - {1 \over 2}{\log _e}2
D π2−log⁡e2{\pi \over 2} - {\log _e}2
Correct Answer
Option C
Solution

I =

∫01xcot⁡−1(1−x2+x4)dx\int\limits_0^1 {x{{\cot }^{ - 1}}(1 - {x^2} + {x^4})dx}

Let x2 = t ⇒\Rightarrow 2xdx = dt At x = 0, t = 0 At x = 1, t = 1 I =

12∫01cot⁡−1(1−t+t2)dt{1 \over 2}\int\limits_0^1 {{{\cot }^{ - 1}}\left( {1 - t + {t^2}} \right)dt}

=

12∫01tan⁡−1(11−t+t2)dt{1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{1 \over {1 - t + {t^2}}}} \right)dt}

=

12∫01tan⁡−1(t+(1−t)1−t+t2)dt{1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{t + \left( {1 - t} \right)} \over {1 - t + {t^2}}}} \right)dt}

=

12∫01tan⁡−1(t+(1−t)1−t(1−t))dt{1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{t + \left( {1 - t} \right)} \over {1 - t\left( {1 - t} \right)}}} \right)dt}

=

12∫01[tan⁡−1(1−t)+tan⁡−1(t)]dt{1 \over 2}\int\limits_0^1 {\left[ {{{\tan }^{ - 1}}\left( {1 - t} \right) + {{\tan }^{ - 1}}\left( t \right)} \right]dt}

=

12∫01tan⁡−1(t)dt+12∫01tan⁡−1(1−t)dt{1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} + {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - t} \right)dt}

=

12∫01tan⁡−1(t)dt+12∫01tan⁡−1(t)dt{1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} + {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt}

[ As

12∫01tan⁡−1(1−t)dt{1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - t} \right)dt}

=

12∫01tan⁡−1[0+1−(1−t)]dt{1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left[ {0 + 1 - \left( {1 - t} \right)} \right]dt}

=

12∫01tan⁡−1(t)dt{1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt}

] =

2×12∫01tan⁡−1(t)dt2 \times {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt}

=

∫01tan⁡−1(t)dt\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt}

Using integration by parts rule =

[t.tan⁡−1(t)]01\left[ {t.{{\tan }^{ - 1}}\left( t \right)} \right]_0^1
−∫01t.11+t2dt- \int\limits_0^1 {t.{1 \over {1 + {t^2}}}} dt

=

π4−12[log⁡(1+t2)]01{\pi \over 4} - {1 \over 2}\left[ {\log \left( {1 + {t^2}} \right)} \right]_0^1

=

π4−12log⁡e2{\pi \over 4} - {1 \over 2}{\log _e}2
Q35
The value of ∫0π/2sin⁡3xsin⁡x+cos⁡xdx\int\limits_0^{\pi /2} {{{{{\sin }^3}x} \over {\sin x + \cos x}}dx} is
A π−28{{\pi - 2} \over 8}
B π−24{{\pi - 2} \over 4}
C π−12{{\pi - 1} \over 2}
D π−14{{\pi - 1} \over 4}
Correct Answer
Option D
Solution

I =

∫0π/2sin⁡3xsin⁡x+cos⁡xdx\int\limits_0^{\pi /2} {{{{{\sin }^3}x} \over {\sin x + \cos x}}dx}

.....(1) ⇒\Rightarrow I =

∫0π2cos⁡3xsin⁡x+cos⁡xdx\int\limits_0^{{\pi \over 2}} {{{{{\cos }^3}x} \over {\sin x + \cos x}}} dx

......(2) Adding those two 2I =

∫0π2sin3x+cos⁡3xsin⁡x+cos⁡xdx\int\limits_0^{{\pi \over 2}} {{{si{n^3}x + {{\cos }^3}x} \over {\sin x + \cos x}}} dx

=

∫0π2(sin⁡x+cos⁡x)(sin2x+cos⁡2x−sin⁡xcos⁡x)sin⁡x+cos⁡xdx\int\limits_0^{{\pi \over 2}} {{{\left( {\sin x + \cos x} \right)\left( {si{n^2}x + {{\cos }^2}x - \sin x\cos x} \right)} \over {\sin x + \cos x}}} dx

=

∫0π2(sin2x+cos⁡2x−sin⁡xcos⁡x)dx\int\limits_0^{{\pi \over 2}} {\left( {si{n^2}x + {{\cos }^2}x - \sin x\cos x} \right)} dx

=

∫0π2(1−sin⁡2x2)dx\int\limits_0^{{\pi \over 2}} {\left( {1 - {{\sin 2x} \over 2}} \right)} dx

=

[x+cos⁡2x4]0π2\left[ {x + {{\cos 2x} \over 4}} \right]_0^{{\pi \over 2}}

=

(π2−14)−(14)\left( {{\pi \over 2} - {1 \over 4}} \right) - \left( {{1 \over 4}} \right)

∴\therefore 2I =

(π2−12)\left( {{\pi \over 2} - {1 \over 2}} \right)

∴\therefore I =

π−14{{\pi - 1} \over 4}
Q36
Let f(x)=∫0xg(t)dtf(x) = \int\limits_0^x {g(t)dt} where g is a non-zero even function. If ƒ(x + 5) = g(x), then ∫0xf(t)dt \int\limits_0^x {f(t)dt} equals-
A 5∫x+55g(t)dt\int\limits_{x + 5}^5 {g(t)dt}
B ∫x+55g(t)dt\int\limits_{x + 5}^5 {g(t)dt}
C ∫5x+5g(t)dt\int\limits_{5}^{x+5} {g(t)dt}
D 2∫5x+5g(t)dt\int\limits_{5}^{x+5} {g(t)dt}
Correct Answer
Option B
Solution
f(x)=∫0xg(t)dtf(x) = \int\limits_0^x {g(t)dt}
f(−x)=∫0−xg(t)dtf\left( { - x} \right) = \int\limits_0^{ - x} {g\left( t \right)} dt

Put t = -v =

−∫0xg(−v)dv- \int\limits_0^x {g\left( { - v} \right)} dv

=

−∫0xg(v)d(v)- \int\limits_0^x {g\left( v \right)} d(v)

[ as g(v) is an even function.] = - f(x) ⇒\Rightarrow f(-x) = -f(x) ∴\therefore f(x) is an odd function.

Given ƒ(x + 5) = g(x) ∴\therefore g(- x) = ƒ(- x + 5) ⇒\Rightarrow g(x) = - f(x - 5) [as g(x) is even and f(x) is an odd function] Replacing x by x + 5, we get f(x) = - g(x + 5) ......(

1) Now

∫0xf(t)dt\int\limits_0^x {f(t)dt}

=

−∫0xg(t+5)dt- \int\limits_0^x {g\left( {t + 5} \right)} dt

=

−∫5x+5g(t)dt- \int\limits_5^{x + 5} {g\left( t \right)} dt

=

∫x+55g(t)dt\int\limits_{x + 5}^5 {g(t)dt}
Q37
lim⁡x→∞(nn2+12+nn2+22+nn2+32+.....+15n)\mathop {\lim }\limits_{x \to \infty } \left( {{n \over {{n^2} + {1^2}}} + {n \over {{n^2} + {2^2}}} + {n \over {{n^2} + {3^2}}} + ..... + {1 \over {5n}}} \right) is equal to :
A tan–1 (2)
B tan–1 (3)
C π4{\pi \over 4}
D π2{\pi \over 2}
Correct Answer
Option A
Solution
lim⁡x→∞∑r=12nnn2+r2\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {{n \over {{n^2} + {r^2}}}}
lim⁡x→∞∑r=12n1n(1+r2n2)=∫02dx1+x2=tan⁡−12\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {{1 \over {n\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}} = \int\limits_0^2 {{{dx} \over {1 + {x^2}}}} } = {\tan ^{ - 1}}2
Q38
The integral ∫1e{(xe)2x−(ex)x} \int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \, loge x dx is equal to :
A −12+1e−12e2 - {1 \over 2} + {1 \over e} - {1 \over {2{e^2}}}
B 32−e−12e2{3 \over 2} - e - {1 \over {2{e^2}}}
C 12−e−1e2{1 \over 2} - e - {1 \over {{e^2}}}
D 32−1e−12x2{3 \over 2} - {1 \over e} - {1 \over {2{x^2}}}
Correct Answer
Option B
Solution
∫1e{(xe)2x−(ex)x} \int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,

Let

(xe)2x=t,(ex)x=v{\left( {{x \over e}} \right)^{2x}} = t,{\left( {{e \over x}} \right)^x} = v
=12∫(1e)21dt+∫e1dv= {1 \over 2}\int\limits_{{{\left( {{1 \over e}} \right)}^2}}^1 {dt + \int\limits_e^1 {dv} }
=12(1−1e2)+(1−e)=32−12e2−e= {1 \over 2}\left( {1 - {1 \over {{e^2}}}} \right) + \left( {1 - e} \right) = {3 \over 2} - {1 \over {2{e^2}}} - e
Q39
Let f and g be continuous functions on [0, a] such that f(x) = f(a – x) and g(x) + g(a – x) = 4, then ∫0a \int\limits_0^a \, f(x) g(x) dx is equal to :
A 4∫0a \int\limits_0^a \, f(x)dx
B −- 3∫0a \int\limits_0^a \, f(x)dx
C ∫0a \int\limits_0^a \, f(x)dx
D 2∫0a \int\limits_0^a \, f(x)dx
Correct Answer
Option D
Solution
I=∫0af(x)g(x)dx{\rm I} = \int_0^a {f\left( x \right)g\left( x \right)dx}
I=∫0af(a−x)g(a−x)dx{\rm I} = \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx}
I=∫0af(x)(4−g(x)dx{\rm I} = \int_0^a {f\left( x \right)\left( {4 - g\left( x \right.} \right)dx}
I=4∫0af(x)dx−I{\rm I} = 4\int_0^a {f\left( x \right)dx - {\rm I}}
⇒I=2∫0af(x)dx\Rightarrow {\rm I} = 2\int_0^a {f\left( x \right)dx}
Q40
The value of the integral ∫−11log⁡(x+x2+1)dx\int\limits_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx} is :
A 2
B 0
C −-1
D 1
Correct Answer
Option B
Solution

Let

I=∫−11log⁡(x+x2+1)dxI = \int\limits_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx}

∵\because

log⁡(x+x2+1)\log \left( {x + \sqrt {{x^2} + 1} } \right)

is an odd function ∴\therefore I = 0

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