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Definite Integration

JEE Mathematics · 230 questions · Page 21 of 23 · Click an option or "Show Solution" to reveal answer

Q201
The integral ∫π12π4  8cos⁡2x(tan⁡x+cot⁡x)3 dx\int_{{\pi \over {12}}}^{{\pi \over 4}} {\,\,{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,dx equals :
A 15128{{15} \over {128}}
B 1564{{15} \over {64}}
C 1332{{13} \over {32}}
D 13256{{13} \over {256}}
Correct Answer
Option A
Solution

tan x + cot x =

sin⁡xcos⁡x{{\sin x} \over {\cos x}}

+

cos⁡xsin⁡x{{\cos x} \over {\sin x}}

=

sin⁡2x+cos⁡2xsin⁡x  cos⁡x{{{{\sin }^2}x + {{\cos }^2}x} \over {\sin x\,\,\cos x}}

=

1sin⁡x  cos⁡x{1 \over {\sin x\,\,\cos x}}

=

2sin⁡2x{2 \over {\sin 2x}}
∴   \therefore\,\,\,
∫π12π48cos⁡2x(tan⁡x+cot⁡x)3  dx\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,\,dx

=

∫π12π48cos⁡2x(2sin⁡2x)3  dx\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {{2 \over {\sin 2x}}} \right)}^3}}}} \,\,dx

=

∫π12π48sin⁡3 2x.cos⁡  2x8  dx\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8{{\sin }^3}\,2x.\cos \,\,2x} \over 8}} \,\,dx

=

∫π12π4sin⁡32x.cos⁡ 2x  dx\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{\sin }^3}2x.\cos \,2x} \,\,dx

Let sin 2x = t ⇒\Rightarrow

   \,\,\,

2 cos2x dx = dt At x =

π12{{\pi \over {12}}}

, t = sin

π6{{\pi \over {6}}}

=

12{1 \over 2}

At x =

π4{{\pi \over 4}}

, t = sin

π2{{\pi \over 2}}

= 1. =

12{1 \over 2}
∫121t3.dt\int\limits_{{1 \over 2}}^1 {{t^3}.dt}

=

12{{1 \over 2}}
[t44]121\left[ {{{{t^4}} \over 4}} \right]_{{1 \over 2}}^1

=

12{{1 \over 2}}

×\times

14{{1 \over 4}}

×\times [ 14 −- (

12{{1 \over 2}}

)4] =

18{{1 \over 8}}

×\times

1516{{15 \over 16}}

=

15128{{15 \over 128}}
Q202
∫010π∣sin⁡x∣dx\int\limits_0^{10\pi } {\left| {\sin x} \right|dx} is
A 2020
B 88
C 1010
D 1818
Correct Answer
Option A
Solution
I=∫010π∣sin⁡x∣dxI = \int\limits_0^{10\pi } {\left| {\sin x} \right|} dx
=10∫0π∣sin⁡x ∣ dx= 10\int\limits_0^\pi {\left| {\sin x\,} \right|} \,dx
=10∫0πsin⁡ x dx= 10\int\limits_0^\pi {\sin \,x\,dx}
[ \left[ \, \right.

as

∣sin⁡x∣\left| {\sin x} \right|

is periodic with period π\pi and

sinsin
x>0x > 0

if

0<x0 < x
<π< \pi
 ]\left. \, \right]
I=20∫0π/2sin⁡x dxI = 20\int\limits_0^{\pi /2} {\sin x} \,dx
=20[−cos⁡ x0π/2]=20= 20\left[ { - \cos \,x_0^{\pi /2}} \right] = 20
Q203
The value of ∫−π/2π/2sin⁡2x1+2xdx\int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}x} \over {1 + {2^x}}}} dx is
A π4{\pi \over 4}
B π8{\pi \over 8}
C π2{\pi \over 2}
D 4π{4\pi }
Correct Answer
Option A
Solution

As we know,

∫ab+(x)dx=∫abf(a+b−x)dx\int\limits_a^b { + \left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)} dx

Let

I=∫−π2π2  sin⁡2x1+2x dxI = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}x} \over {1 + {2^x}}}} \,dx
⇒    I=∫−π2π2  sin⁡2(−π2+π2−x)1+2−π2+π2−x\Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}\left( { - {\pi \over 2} + {\pi \over 2} - x} \right)} \over {1 + {2^{ - {\pi \over 2} + {\pi \over 2} - x}}}}}
⇒    I=∫−π2π2sin⁡2x1+2−xdx\Rightarrow\,\,\,\, I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\sin }^2}x} \over {1 + {2^{ - x}}}}} dx
⇒    I=∫−π2π22xsin⁡2x1+2x  dx\Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \,\,dx
∴   \therefore\,\,\,
2I=∫−π2π2(sin⁡2x1+2x+2xsin⁡2x1+2x)  dx2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{{{\sin }^2}x} \over {1 + {2^x}}} + {{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \right)} \,\,dx
⇒    2I=∫−π2π2sin⁡2x(1+2x1+2x)  dx\Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x\left( {{{1 + {2^x}} \over {1 + {2^x}}}} \right)} \,\,dx
⇒    2I=∫−π2π2sin⁡2x  dx\Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx
⇒    2I=2∫0π2sin⁡2x  dx   \Rightarrow \,\,\,\,2I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,

[ as sin2 x is an even function ]

⇒    I=∫0π2sin⁡2x  dx    \Rightarrow \,\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,\,

[ as

    \,\,\,\,
∫0af(a)  dx=∫0af(a−x) dx]\int\limits_0^a {f\left( a \right)\,\,dx = \int\limits_0^a {f\left( {a - x} \right)\,dx]} }

So,

   \,\,\,
I=∫0π2sin⁡2(b2−x)  dx    I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}\left( {{b \over 2} - x} \right)} \,\,dx\,\,\,\,
   I=∫0π2cos⁡2x  dx    \,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,\,dx\,\,\,\,
∴    \therefore\,\,\,\,
    2I=∫0π2(sin⁡2x+cos⁡2x)  dx    \,\,\,\,2I = \int\limits_0^{{\pi \over 2}} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \,\,dx\,\,\,\,
⇒    2I=∫0π2   dx    \Rightarrow \,\,\,\,2I = \int\limits_0^{{\pi \over 2}} \, \,\,dx\,\,\,\,
⇒    2I=[x]0π2\Rightarrow \,\,\,\,2I = {\left[ x \right]_0}^{{\pi \over 2}}
⇒    2I=π2\Rightarrow \,\,\,\,2I = {\pi \over 2}
⇒    I=π4\Rightarrow \,\,\,\,I = {\pi \over 4}
Q204
Let f(x)f(x) be a function satisfying f′(x)=f(x)f'(x)=f(x) with f(0)=1f(0)=1 and g(x)g(x) be a function that satisfies f(x)+g(x)=x2f\left( x \right) + g\left( x \right) = {x^2}. Then the value of the integral ∫01f(x)g(x)dx,\int\limits_0^1 {f\left( x \right)g\left( x \right)dx,} is
A e+e22+52e + {{{e^2}} \over 2} + {5 \over 2}
B e−e22−52e - {{{e^2}} \over 2} - {5 \over 2}
C e+e22−32e + {{{e^2}} \over 2} - {3 \over 2}
D e−e22−32e - {{{e^2}} \over 2} - {3 \over 2}
Correct Answer
Option D
Solution

Given

f′(x)=f(x)⇒f′(x)f(x)=1f'\left( x \right) = f\left( x \right) \Rightarrow {{f'\left( x \right)} \over {f\left( x \right)}} = 1

Integrating log

f(x)=x+c⇒f(x)=ex+cf\left( x \right) = x + c \Rightarrow f\left( x \right) = {e^{x + c}}
f(0)=1⇒f(x)=exf\left( 0 \right) = 1 \Rightarrow f\left( x \right) = {e^x}

∴\therefore

∫01f(x)g(x)dx\int\limits_0^1 {f\left( x \right)g\left( x \right)} dx
=∫01ex(x2−ex)dx= \int\limits_0^1 {{e^x}} \left( {{x^2} - {e^x}} \right)dx
=∫01x2exdx−∫01e2xdx= \int\limits_0^1 {{x^2}} {e^x}dx - \int\limits_0^1 {{e^{2x}}} dx
=[x2ex]01−2[xex−ex]01−120[e2x]01= \left[ {{x^2}{e^x}} \right]_0^1 - 2\left[ {x{e^x} - {e^x}} \right]_0^1 - {1 \over {20}}\left[ {{e^{2x}}} \right]_0^1
=e−[e22−12]−2[e−e+1]= e - \left[ {{{{e^2}} \over 2} - {1 \over 2}} \right] - 2\left[ {e - e + 1} \right]
=e−e22−32= e - {{{e^2}} \over 2} - {3 \over 2}
Q205
A value of α\alpha such that ∫αα+1dx(x+α)(x+α+1)=log⁡e(98)\int\limits_\alpha ^{\alpha + 1} {{{dx} \over {\left( {x + \alpha } \right)\left( {x + \alpha + 1} \right)}}} = {\log _e}\left( {{9 \over 8}} \right) is :
A 2
B - 2
C 12{1 \over 2}
D −12-{1 \over 2}
Correct Answer
Option B
Solution
I=∫αα+1dx(x+α)(x+α+1)I = \int\limits_\alpha ^{\alpha + 1} {{{dx} \over {(x + \alpha )(x + \alpha + 1)}}}

Let x + α\alpha = t and dx = dt

I=∫2α2α+1dtt(t+1)=∫2α2α+1(1t−1t+1)dxI = \int\limits_{2\alpha }^{2\alpha + 1} {{{dt} \over {t(t + 1)}}} = \int\limits_{2\alpha }^{2\alpha + 1} {\left( {{1 \over t} - {1 \over {t + 1}}} \right)dx}
⇒(ln⁡t−ln⁡(t+1))∣2α2α+1=ln⁡(tt+1)∣2α2α+1\Rightarrow \left( {\ln t - \ln (t + 1)} \right)|_{2\alpha }^{2\alpha + 1} = \ln \left( {{t \over {t + 1}}} \right)|_{2\alpha }^{2\alpha + 1}
⇒ln⁡(2α+12α+2)−ln⁡(2α2α+1)\Rightarrow \ln \left( {{{2\alpha + 1} \over {2\alpha + 2}}} \right) - \ln \left( {{{2\alpha } \over {2\alpha + 1}}} \right)
⇒ln⁡(2α+12α+2×2α+12α)=ln⁡(98)\Rightarrow \ln \left( {{{2\alpha + 1} \over {2\alpha + 2}} \times {{2\alpha + 1} \over {2\alpha }}} \right) = \ln \left( {{9 \over 8}} \right)
⇒(2α+1)2(2α+2)=9α4\Rightarrow {{{{\left( {2\alpha + 1} \right)}^2}} \over {\left( {2\alpha + 2} \right)}} = {{9\alpha } \over 4}
4(4α2+1+4α)=18α2+18α4(4{\alpha ^2} + 1 + 4\alpha ) = 18{\alpha ^2} + 18\alpha
⇒2α2−2α−4=0\Rightarrow 2{\alpha ^2} - 2\alpha - 4 = 0
⇒α2+α−2=0\Rightarrow {\alpha ^2} + \alpha - 2 = 0
⇒(α+2)(α−1)=0\Rightarrow (\alpha + 2)(\alpha - 1) = 0

α\alpha = 1 or α\alpha = -2

Q206
Let f:R→Rf:R \to R be a differentiable function having f(2)=6f\left( 2 \right) = 6, f′(2)=(148)f'\left( 2 \right) = \left( {{1 \over {48}}} \right). Then lim⁡x→2∫6f(x)4t3x−2dt\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}dt} equals :
A 2424
B 3636
C 1212
D 1818
Correct Answer
Option D
Solution
lim⁡x→2∫6f(x)4t3x−2dt\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}} dt
=lim⁡x→0∫6f(x)4t3dtx−2= \mathop {\lim }\limits_{x \to 0} {{\int\limits_6^{f\left( x \right)} {4{t^3}dt} } \over {x - 2}}

This limit resembles a derivative because the fraction has the form 0/00/0 as x→2x \to 2 since both the numerator (integral from 66 to f(x)f(x)) and the denominator (x−2x - 2) are zero when x=2x = 2.

Applying

L′L'

Hospital rule

lim⁡x→2[4f(x)3f′(x)]1\mathop {\lim }\limits_{x \to 2} {{\left[ {4f{{\left( x \right)}^3}f'\left( x \right)} \right]} \over 1}
=4(f(2))3f′(2)= 4{\left( {f\left( 2 \right)} \right)^3}f'\left( 2 \right)
=4×63×148=18= 4 \times {6^3} \times {1 \over {48}} = 18
Q207
Let p(x)p(x) be a function defined on RR such that p′(x)=p′(1−x),p'(x)=p'(1-x), for all x∈[0,1],p(0)=1x \in \left[ {0,1} \right],p\left( 0 \right) = 1 and p(1)=41.p(1)=41. Then ∫01p(x)dx\int\limits_0^1 {p\left( x \right)dx} equals :
A 2121
B 4141
C 4242
D 41\sqrt {41}
Correct Answer
Option A
Solution
p′(x)=p′(1−x)p'\left( x \right) = p'\left( {1 - x} \right)
⇒p(x)=−p(1−x)+c\Rightarrow p\left( x \right) = - p\left( {1 - x} \right) + c

at

x=0x=0
p(0)=−p(1)+c⇒42=cp\left( 0 \right) = - p\left( 1 \right) + c \Rightarrow 42 = c

Now,

p(x)=−p(1−x)+42p\left( x \right) = - p\left( {1 - x} \right) + 42
⇒p(x)+p(1−x)=42\Rightarrow p\left( x \right) + p\left( {1 - x} \right) = 42
⇒I=∫01p(x)dx          ...(i)\Rightarrow I = \int\limits_0^1 {p\left( x \right)dx\,\,\,\,\,\,\,\,\,\,...\left( i \right)}
⇒I=∫01p(1−x)dx      ...(ii)\Rightarrow I = \int\limits_0^1 {p\left( {1 - x} \right)} dx\,\,\,\,\,\,...\left( {ii} \right)

on adding

(i)(i)

and

(ii),(ii),
           \,\,\,\,\,\,\,\,\,\,\,
2I=∫01(42)dx⇒I=212I = \int\limits_0^1 {\left( {42} \right)dx \Rightarrow I = 21}
Q208
If f : R → \to R is a differentiable function and f(2) = 6, then lim⁡x→2∫6f(x)2tdt(x−2)\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}} is :-
A 2f'(2)
B 24f'(2)
C 0
D 12f'(2)
Correct Answer
Option D
Solution
lim⁡x→2∫6f(x)2tdt(x−2)\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}

This is

00{0 \over 0}

form so we use L – Hospital Rule. =

lim⁡x→22f(x).f′(x)−01\mathop {\lim }\limits_{x \to 2} {{2f\left( x \right).f'\left( x \right) - 0} \over 1}

=

2f(2).f′(2){2f\left( 2 \right).f'\left( 2 \right)}

= 12f'(2) Note : Newton - Leibnitz rule of differentiation under integration

ddx∫α(x)β(x)f(u)du{d \over {dx}}\int\limits_{\alpha \left( x \right)}^{\beta \left( x \right)} {f\left( u \right)} du

=

f(β(x))β′(x)−f(α(x))α′(x)f\left( {\beta \left( x \right)} \right)\beta '\left( x \right) - f\left( {\alpha \left( x \right)} \right)\alpha '\left( x \right)
Q209
The integral ∫π/6π/3sec⁡2/3xcos⁡ec4/3xdx\int\limits_{\pi /6}^{\pi /3} {{{\sec }^{2/3}}} x\cos e{c^{4/3}}xdx is equal to :
A 353−313{3^{{5 \over 3}}} - {3^{{1 \over 3}}}
B 356−323{3^{{5 \over 6}}} - {3^{{2 \over 3}}}
C 343−313{3^{{4 \over 3}}} - {3^{{1 \over 3}}}
D 376−356{3^{{7 \over 6}}} - {3^{{5 \over 6}}}
Correct Answer
Option D
Solution
∫π/6π/3sec⁡2/3x cosec4/3x dx\int\limits_{\pi /6}^{\pi /3} {{{\sec }^{2/3}}x\,{{{\mathop{\rm cosec}\nolimits} }^{4/3}}x\,dx}
=∫sec⁡2xtan⁡4/3xdx= \int {{{{{\sec }^2}x} \over {{{\tan }^{4/3}}x}}dx}

Let tan x = t, sec2 x dx = dt

=∫dtt4/3= \int {{{dt} \over {{t^{4/3}}}}}
I=−3(t−1/3)I = - 3\left( {{t^{ - 1/3}}} \right)
⇒(−3(tan⁡x)−13)π/6π/3\Rightarrow \left( { - 3{{\left( {\tan x} \right)}^{{{ - 1} \over 3}}}} \right)_{\pi /6}^{\pi /3}
⇒3(313−1316)\Rightarrow 3\left( {{3^{{1 \over 3}}} - {1 \over {{3^{{1 \over 6}}}}}} \right)

= 37/6 - 35/6

Q210
Let P(x) = x2 + bx + c be a quadratic polynomial with real coefficients such that ∫01P(x)dx\int_0^1 {P(x)dx} = 1 and P(x) leaves remainder 5 when it is divided by (x −- 2). Then the value of 9(b + c) is equal to :
A 9
B 11
C 7
D 15
Correct Answer
Option C
Solution
(x−2)Q(x)+5=x2+bx+c(x - 2)Q(x) + 5 = {x^2} + bx + c

Put x = 2 5 = 2b + c + 4 .... (1)

∫01(x2+bx+c)dx=1\int_0^1 {({x^2} + bx + c)dx} = 1
⇒13+b2+c=1\Rightarrow {1 \over 3} + {b \over 2} + c = 1
b2+c=23{b \over 2} + c = {2 \over 3}

.... (2) Solve (1) & (2)

b=29b = {2 \over 9}
c=59c = {5 \over 9}

9(b + c) = 7

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