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Definite Integration

JEE Mathematics · 230 questions · Page 2 of 23 · Click an option or "Show Solution" to reveal answer

Q11
The value of lim⁡x→0∫0x2sec⁡2tdtxsinx\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {{{\sec }^2}tdt} } \over xsinx} is
A 0
B 3
C 2
D 1
Correct Answer
Option D
Solution
lim⁡x→0ddx∫0x2sec⁡2tdtdx(xsin⁡x)\mathop {\lim }\limits_{x \to 0} {{{d \over {dx}}\int\limits_0^{{x^2}} {{{\sec }^2}tdt} } \over {{d \over x}\left( {x\sin x} \right)}}
=lim⁡x→0sec⁡2x2.2xsin⁡ x+x cos⁡ x= \mathop {\lim }\limits_{x \to 0} {{{{\sec }^2}{x^2}.2x} \over {\sin \,x + x\,\cos \,x}}

(by

L′L'

Hospital rule)

lim⁡x→02sec⁡2x2(sin⁡xx+cos⁡ x)=2×11+1=1\mathop {\lim }\limits_{x \to 0} {{2{{\sec }^2}{x^2}} \over {\left( {{{\sin x} \over x} + \cos \,x} \right)}} = {{2 \times 1} \over {1 + 1}} = 1
Q12
If f(a+b−x)=f(x)f\left( {a + b - x} \right) = f\left( x \right) then ∫abxf(x)dx\int\limits_a^b {xf\left( x \right)dx} is equal to
A a+b2∫abf(a+b+x)dx{{a + b} \over 2}\int\limits_a^b {f\left( {a + b + x} \right)dx}
B a+b2∫abf(b−x)dx{{a + b} \over 2}\int\limits_a^b {f\left( {b - x} \right)dx}
C a+b2∫abf(x)dx{{a + b} \over 2}\int\limits_a^b {f\left( x \right)dx}
D  b−a2∫abf(x)dx\,{{b - a} \over 2}\int\limits_a^b {f\left( x \right)dx}
Correct Answer
Option C
Solution
I=∫abxf(x)dxI = \int\limits_a^b {xf\left( x \right)} dx
=∫ab(a+b−x)f(a+b−x)dx= \int\limits_a^b {\left( {a + b - x} \right)} f\left( {a + b - x} \right)dx
=(a+b)∫abf(a+b−x)dx−∫abxf(a+b−x)dx= \left( {a + b} \right)\int\limits_a^b {f\left( {a + b - x} \right)} dx - \int\limits_a^b {xf} \left( {a + b - x} \right)dx
=(a+b)∫abf(x)dx−∫abxf(x)dx= \left( {a + b} \right)\int\limits_a^b {f\left( x \right)dx} - \int\limits_a^b {xf\left( x \right)dx}
[\left[ {} \right.

As given that

f(a+b−x)=f(x)f\left( {a + b - x} \right) = f\left( x \right)
]\left. {} \right]
2I=(a+b)∫abf(x)dx2I = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)} dx
⇒I=(a+b)2∫abf(x)dx\Rightarrow I = {{\left( {a + b} \right)} \over 2}\int\limits_a^b {f\left( x \right)} dx
Q13
If f(y)=ey,f\left( y \right) = {e^y}, g(y)=y;y>0g\left( y \right) = y;y > 0 and F(t)=∫0tf(t−y)g(y)dy,F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,} then :
A F(t)=te−tF\left( t \right) = t{e^{ - t}}
B F(t)=1t−te−1(1+t)F\left( t \right) = 1t - t{e^{ - 1}}\left( {1 + t} \right)
C F(t)=et−(1+t)F\left( t \right) = {e^t} - \left( {1 + t} \right)
D F(t)=tetF\left( t \right) = t{e^t}.
Correct Answer
Option C
Solution
F(t)=∫0tf(t−y)g(y)dyF\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)} dy
=∫0tet−yydy=et∫0te−y ydy= \int\limits_0^t {{e^{t - y}}} ydy = {e^t}\int\limits_0^t {{e^{ - y}}} \,ydy
=et[−ye−y−e−y]0t= {e^t}\left[ { - y{e^{ - y}} - {e^{ - y}}} \right]_0^t
=−et[ye−y+e−y]0t= - {e^t}\left[ {y{e^{ - y}} + {e^{ - y}}} \right]_0^t
=−et[t e−t+e−t−0−1]= - {e^t}\left[ {t\,{e^{ - t}} + {e^{ - t}} - 0 - 1} \right]
=−et[t+1−etet]= - {e^t}\left[ {{{t + 1 - {e^t}} \over {{e^t}}}} \right]
=et−(1+t)= {e^t} - \left( {1 + t} \right)
Q14
Limn→∞∑r=1n1nern\mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{e^{{r \over n}}}} is
A e+1e+1
B e−1e-1
C 1−e1-e
D ee
Correct Answer
Option B
Solution
Limn→∞∑r=1n1nern  \mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}} {e^{{r \over n}}}\,\,
[\left[ {} \right.

Using definite integrals as limit of sum

]\left. {} \right]
=∫θ1exdx=e−1= \int\limits_\theta ^1 {{e^x}} dx = e - 1
Q15
The value of I=∫0π/2(sin⁡x+cos⁡x)21+sin⁡2xdxI = \int\limits_0^{\pi /2} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}dx} is
A 33
B 11
C 22
D 00
Correct Answer
Option C
Solution
I=∫0π2(sin⁡ x+cos⁡x)21+sin⁡2xdxI = \int\limits_0^{{\pi \over 2}} {{{{{\left( {\sin \,x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}} dx

We know

[(sin⁡x+cos⁡x)2=1+sin⁡2x], \left[ {{{\left( {\sin x + \cos x} \right)}^2} = 1 + \sin 2x} \right],\,

So

I=∫0π2(sin⁡x+cos⁡x)2(sin⁡x+cos⁡x)dxI = \int\limits_0^{{\pi \over 2}} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\left( {\sin x + \cos x} \right)}}} dx
=∫0π2(sin⁡x+cos⁡x)dx= \int\limits_0^{{\pi \over 2}} {\left( {\sin x + \cos x} \right)dx}
[\left[ {} \right.
sin⁡x+cos⁡x>0\sin x + \cos x > 0
  if  0<x<π2\,\,if\,\,0 < x < {\pi \over 2}
]\left. {} \right]

or

I=[−cos⁡x+sin⁡x]0π2I = \left[ { - \cos x + \sin x} \right]_0^{{\pi \over 2}}

= - cos

π2{\pi \over 2}

+ sin

π2{\pi \over 2}

+ cos 0 - sin 0 = - 0 + 1 + 1 - 0 = 2

Q16
The value of ∫−23∣1−x2∣dx\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|dx} is
A 13{1 \over 3}
B 143{14 \over 3}
C 73{7 \over 3}
D 283{28 \over 3}
Correct Answer
Option D
Solution
∫−23∣1−x2∣dx=∫−23∣x2−1∣dx\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|} dx = \int\limits_{ - 2}^3 {\left| {{x^2} - 1} \right|} dx

Now

∣x2−1∣={x2−1ifx≤−11−x2if−1≤x≤1x2−1ifx≥1\left| {{x^2} - 1} \right| = \left\{ \begin{array}{lll}{{x^2} - 1} & {if} & {x \le - 1} \\ {1 - {x^2}} & {if} & { - 1 \le x \le 1} \\ {{x^2} - 1} & {if} & {x \ge 1} \end{array} \right.

∴\therefore Integral is

∫−2−1(x2−1)dx+∫−11(1−x2)dx+∫13(x2−1)dx\int\limits_{ - 2}^{ - 1} {\left( {{x^2} - 1} \right)dx + \int\limits_{ - 1}^1 {\left( {1 - {x^2}} \right)} } dx + \int\limits_1^3 {\left( {{x^2} - 1} \right)dx}
[x33−x]−2−1+[x−x33]−11+[x33−x]13\left[ {{{{x^3}} \over 3} - x} \right]_{ - 2}^{ - 1} + \left[ {x - {{{x^3}} \over 3}} \right]_{ - 1}^1 + \left[ {{{{x^3}} \over 3} - x} \right]_1^3
=(−13+1)−(−83+2)+(2−23)+(273−3)−(13−1)= \left( { - {1 \over 3} + 1} \right) - \left( { - {8 \over 3} + 2} \right) + \left( {2 - {2 \over 3}} \right) + \left( {{{27} \over 3} - 3} \right) - \left( {{1 \over 3} - 1} \right)
=23+23+43+6+23=283= {2 \over 3} + {2 \over 3} + {4 \over 3} + 6 + {2 \over 3} = {{28} \over 3}
Q17
If f(x)=ex1+ex,I1=∫f(−a)f(a)xg{x(1−x)}dxf\left( x \right) = {{{e^x}} \over {1 + {e^x}}},{I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}dx} and I2=∫f(−a)f(a)g{x(1−x)}dx,{I_2} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {g\left\{ {x\left( {1 - x} \right)} \right\}dx} , then the value of I2I1{{{I_2}} \over {{I_1}}} is
A 11
B −3-3
C −1-1
D 22
Correct Answer
Option D
Solution
f(x)=ex1+exf\left( x \right) = {{{e^x}} \over {1 + {e^x}}}
⇒f(−x)=e−x1+e−x=1ex+1\Rightarrow f\left( { - x} \right) = {{{e^{ - x}}} \over {1 + {e^{ - x}}}} = {1 \over {{e^x} + 1}}

∴\therefore

f(x)+f(−x)=1∀xf\left( x \right) + f\left( { - x} \right) = 1\forall x

Now

I1=∫f(−a)f(a)xg{x(1−x)}dx{I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}} dx
=∫f(−a)f(a)(1−x)g{x(1−x)}dx= \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {\left( {1 - x} \right)} g\left\{ {x\left( {1 - x} \right)} \right\}dx
[\left[ {} \right.

using

∫abf(x)dx a\int\limits_a^b {f\left( x \right)} dx\,a
=∫abf(a+b−x)dx= \int\limits_a^b {f\left( {a + b - x} \right)dx}
 ]\left. \, \right]
=I2−I1⇒2I1=I2= {I_2} - {I_1} \Rightarrow 2{I_1} = {I_2}

∴\therefore

I2I1=2{{{I_2}} \over {{I_1}}} = 2
Q18
lim⁡n→∞[1n2sec⁡21n2+2n2sec⁡24n2....+1nsec⁡21]\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}}.... + {1 \over n}{{\sec }^2}1} \right] equals
A 12sec⁡1{1 \over 2}\sec 1
B 12{1 \over 2}cosec 1
C tan 1
D 12{1 \over 2}tan 1
Correct Answer
Option D
Solution
lim⁡n→∞[1n2sec⁡21n2+2n2sec⁡24n2+3n2sec29n2+...+1nsec⁡21]\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}} + {3 \over {{n^2}}}se{c^2}{9 \over {{n^2}}} + ... + {1 \over n}{{\sec }^2}1} \right]
=lim⁡n→∞rn2sec⁡2r2n2=lim⁡n→∞1n.πnsec⁡2r2n2= \mathop {\lim }\limits_{n \to \infty } {r \over {{n^2}}}{\sec ^2}{{{r^2}} \over {{n^2}}} = \mathop {\lim }\limits_{n \to \infty } {1 \over n}.{\pi \over n}{\sec ^2}{{{r^2}} \over {{n^2}}}

⇒\Rightarrow Given limit is equal to value of integral

              ∫01xsec⁡2 x2 dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_0^1 {x{{\sec }^2}} \,{x^2}\,dx

or

12∫012x sec⁡  x2dx=12∫01sec⁡2tdl{1 \over 2}\int\limits_0^1 {2x\,\sec } \,\,{x^2}dx = {1 \over 2}\int\limits_0^1 {{{\sec }^2}} tdl
[\left[ {} \right.

put

x2=t]\left. {{x^2} = t} \right]
=12(tan⁡t)01=12tan⁡ 1.= {1 \over 2}\left( {\tan t} \right)_0^1 = {1 \over 2}\tan \,1.
Q19
If I1=∫012x2dx,I2=∫012x3dx, I3=∫122x2dx{I_1} = \int\limits_0^1 {{2^{{x^2}}}dx,{I_2} = \int\limits_0^1 {{2^{{x^3}}}dx,\,{I_3} = \int\limits_1^2 {{2^{{x^2}}}dx} } } and I4=∫122x3dx{I_4} = \int\limits_1^2 {{2^{{x^3}}}dx} then
A I2>I1{I_2} > {I_1}
B I1>I2{I_1} > {I_2}
C I3=I4{I_3} = {I_4}
D I3>I4{I_3} > {I_4}
Correct Answer
Option B
Solution
I1=∫012x2dx, I2=∫012x3dx,{I_1} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,{I_2} = \int\limits_0^1 {{2^{{x^3}}}} dx,
=I3=∫012x2dx, = {I_3} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,
I4=∫012x3dx  {I_4} = \int\limits_0^1 {{2^{{x^3}}}} dx\,\,
∀0<x<1, x2>x3\forall 0 < x < 1,\,{x^2} > {x^3}
⇒∫012x2 dx>∫012x3dx\Rightarrow \int\limits_0^1 {{2^{{x^2}}}} \,dx > \int\limits_0^1 {{2^{{x^3}}}} dx

and

∫122x3dx>∫122x2dx\int\limits_1^2 {{2^{{x^3}}}dx} > \int\limits_1^2 {{2^{{x^2}}}dx}
⇒I1>I2\Rightarrow {I_1} > {I_2}

and

I4>I3{I_4} > {I_3}
Q20
The value of ∫−ππcos⁡21+axdx,  a>0,\int\limits_{ - \pi }^\pi {{{{{\cos }^2}} \over {1 + {a^x}}}dx,\,\,a > 0,} is
A a πa\,\pi
B π2{\pi \over 2}
C πa{\pi \over a}
D 2π{2\pi }
Correct Answer
Option B
Solution

Let

I=∫−ππcos⁡2x1+axdx             ...(1)I = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {a^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)
=∫−ππcos⁡2(−x)1+a−xdx= \int\limits_{ - \pi }^\pi {{{{{\cos }^2}\left( { - x} \right)} \over {1 + {a^{ - x}}}}} dx
[ \left[ \, \right.

Using

∫abf(x)dx=∫abf(a+b−x)dx\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)dx}
 ]\left. \, \right]
=∫−ππcos⁡2x1+αxdx             ...(2)= \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {\alpha ^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)

Adding equations

(1)(1)

and

(2)(2)

we get

2I=∫−ππcos⁡2x(1+ax1+ax)dx2I = \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\left( {{{1 + {a^x}} \over {1 + {a^x}}}} \right)dx
=∫−ππcos⁡2x dx= \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\,dx
=2∫0πcos⁡2x dx= 2\int\limits_0^\pi {{{\cos }^2}} x\,dx
=2×2∫0π2cos⁡2x dx= 2 \times 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} x\,dx
=4∫0π2sin⁡2x dx= 4\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} x\,dx
⇒I=2∫0π2sin⁡2 x dx\Rightarrow I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} \,x\,dx
=2∫0π2(1−cos⁡2x dx)= 2\int\limits_0^{{\pi \over 2}} {\left( {1 - {{\cos }^2}x\,dx} \right)}
⇒I  =2∫0π2dx−2∫0π2cos⁡2 x dx\Rightarrow I\,\, = 2\int\limits_0^{{\pi \over 2}} {dx - 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} } \,x\,dx
⇒I+I=2(π2)=π\Rightarrow I + I = 2\left( {{\pi \over 2}} \right) = \pi
⇒I=π2\Rightarrow I = {\pi \over 2}
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