Practice Start Test

Definite Integration

JEE Mathematics · 230 questions · Page 1 of 23 · Click an option or "Show Solution" to reveal answer

Q1
If ∫0x \int\limits_0^x \, f(t) dt = x2 + ∫x1 \int\limits_x^1 \, t2f(t) dt then f '(12)\left( {{1 \over 2}} \right) is -
A 1825{{18} \over {25}}
B 625{{6} \over {25}}
C 2425{{24} \over {25}}
D 45{{4} \over {5}}
Correct Answer
Option C
Solution
∫0x \int\limits_0^x \,

f(t) dt = x2 +

∫x1 \int\limits_x^1 \,

t2f(t) dt f '

(12)\left( {{1 \over 2}} \right)

= ? Differentiate w.r.t. 'x' f(x) = 2x + 0 −- x2 f(x) f(x) =

2x1+x2{{2x} \over {1 + {x^2}}}

⇒\Rightarrow f '(x) =

(1+x2)2−2x(2x)(1+x2)2{{\left( {1 + {x^2}} \right)2 - 2x\left( {2x} \right)} \over {{{\left( {1 + {x^2}} \right)}^2}}}

f '(x) =

2x2−4x2+2(1+x2)2{{2{x^2} - 4{x^2} + 2} \over {{{\left( {1 + {x^2}} \right)}^2}}}

f '

(12)=2−2(14)(1+14)2=(32)2516=4850=2425\left( {{1 \over 2}} \right) = {{2 - 2\left( {{1 \over 4}} \right)} \over {{{\left( {1 + {1 \over 4}} \right)}^2}}} = {{\left( {{3 \over 2}} \right)} \over {{{25} \over {16}}}} = {{48} \over {50}} = {{24} \over {25}}
Q2
∫020π(∣sin⁡x∣+∣cos⁡x∣)2dx is equal to  \int\limits_{0}^{20 \pi}(|\sin x|+|\cos x|)^{2} d x \text{ is equal to }
A 10(π+4)10(\pi+4)
B 10(π+2)10(\pi+2)
C 20(π−2)20(\pi-2)
D 20(π+2)20(\pi+2)
Correct Answer
Option D
Solution
I=∫020π(∣sin⁡x∣+∣cos⁡x∣)2 dxI = \int\limits_0^{20\pi } {{{\left( {|\sin x| + |\cos x|} \right)}^2}\,dx}
=20∫0π(1+∣sin⁡2x∣) dx= 20\int\limits_0^\pi {\left( {1 + |\sin 2x|} \right)\,dx}
=40∫0π2(1+sin⁡2x) dx= 40\int\limits_0^{{\pi \over 2}} {(1 + \sin 2x)\,dx}
=40(x−cos⁡2x2)∣0π2= \left. {40\left( {x - {{\cos 2x} \over 2}} \right)} \right|_0^{{\pi \over 2}}
=40(π2+12+12)=20(π+2)= 40\left( {{\pi \over 2} + {1 \over 2} + {1 \over 2}} \right) = 20(\pi + 2)
Q3
lim⁡n→∞1p+2p+3p+.....+npnp+1\mathop {\lim }\limits_{n \to \infty } {{{1^p} + {2^p} + {3^p} + ..... + {n^p}} \over {{n^{p + 1}}}} is
A 1p+1{1 \over {p + 1}}
B 11−p{1 \over {1 - p}}
C 1p−1p−1{1 \over p} - {1 \over {p - 1}}
D 1p+2{1 \over {p + 2}}
Correct Answer
Option A
Solution

We have

lim⁡x→∞1p+2p+....+npnp+1;\mathop {\lim }\limits_{x \to \infty } {{{1^p} + {2^p} + .... + {n^p}} \over {{n^{p + 1}}}};
lim⁡x→∞∑r=1nrpnp.n=∫01xpdx\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^n {{{{r^p}} \over {{n^p}.n}}} = \int\limits_0^1 {{x^p}} dx
=[xp+1p+1]0=1p+1= {\left[ {{{{x^{p + 1}}} \over {p + 1}}} \right]_0} = {1 \over {p + 1}}
Q4
The value of ∫02π[sin⁡2x(1+cos⁡3x)]dx\int\limits_0^{2\pi } {\left[ {\sin 2x\left( {1 + \cos 3x} \right)} \right]} dx, where [t] denotes the greatest integer function is :
A 2π\pi
B π\pi
C -2π\pi
D -π\pi
Correct Answer
Option D
Solution
I=∫02π[sin⁡2x(1+cos⁡3x)]dxI = \int\limits_0^{2\pi } {\left[ {\sin 2x(1 + \cos 3x)} \right]} dx

.... (i)

∴∫0af(x)=∫0af(a−x)dx\therefore\int\limits_0^a {f(x)} = \int\limits_0^a {f(a - x)} dx
∵I=∫02π[−sin⁡2x(1+cos⁡3x)]dx\because I = \int\limits_0^{2\pi } {\left[ { - \sin 2x(1 + \cos 3x)} \right]} dx

By (i) + (ii)

⇒2I=∫02π(−1)dx\Rightarrow 2I = \int\limits_0^{2\pi } {( - 1)dx}
⇒2I=−(x)02π\Rightarrow 2I = - (x)_0^{2\pi }

⇒\Rightarrow I = –π\pi

Q5
∫−ππ2x(1+sin⁡x)1+cos⁡2xdx\int_{ - \pi }^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} dx is
A π24{{{\pi ^2}} \over 4}
B π2{{\pi ^2}}
C zero
D π2{\pi \over 2}
Correct Answer
Option B
Solution
∫−ππ2x(1+sin⁡ x)1+cos⁡2xdx\int_{ - \pi }^\pi {{{2x\left( {1 + \sin \,x} \right)} \over {1 + {{\cos }^2}x}}} dx
=∫−ππ2x dx1+cos⁡2x+2∫−ππx sin⁡x1+cos⁡2xdx= \int_{ - \pi }^\pi {{{2x\,dx} \over {1 + {{\cos }^2}x}} + 2\int_{ - \pi }^\pi {{{x\,\sin x} \over {1 + {{\cos }^2}x}}} } dx
=0+4∫0πxsin⁡x dx1+cos⁡2x;= 0 + 4\int_0^\pi {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} ;
[ \left[ \, \right.

as

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

if

f(x)f(x)

is odd

=2∫0af(x)dx= 2\int\limits_0^a {f\left( x \right)} dx

if

f(x)f(x)

is even.

I=4∫0π(π−x)sin⁡(π−x)1+cos⁡2(π−x)dxI = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \left( {\pi - x} \right)} \over {1 + {{\cos }^2}\left( {\pi - x} \right)}}} dx
I=4∫0π(π−x)sin⁡ x1+cos⁡2xI = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \,x} \over {1 + {{\cos }^2}x}}}
⇒I=4π∫0πsin⁡x dx1+cos⁡2x−4∫xsin⁡x dx1+cos⁡2x\Rightarrow I = 4\pi \int_0^\pi {{{\sin x\,dx} \over {1 + {{\cos }^2}x}}} - 4\int {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}}
⇒2I=4π∫0πsin⁡x1+cos⁡2xdx\Rightarrow 2I = 4\pi \int_0^\pi {{{\sin x} \over {1 + {{\cos }^2}x}}} dx

put

cos⁡x=t⇒−sin⁡xdx=dt\cos x = t \Rightarrow - \sin xdx = dt

∴\therefore

I=−2π∫1−111+t2dtI = - 2\pi \int\limits_1^{ - 1} {{1 \over {1 + {t^2}}}} dt
=2π∫−1111+t2dt= 2\pi \int\limits_{ - 1}^1 {{1 \over {1 + {t^2}}}} dt
=2π[tan⁡−1t]−11= 2\pi \left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1
=2π[tan⁡−11−tan⁡−1(−1)]= 2\pi \left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}\left( { - 1} \right)} \right]
=2π[π4−(−π4)]=2ππ2=π2= 2\pi \left[ {{\pi \over 4} - \left( {{{ - \pi } \over 4}} \right)} \right] = 2\pi {\pi \over 2} = {\pi ^2}
Q6
In=∫0π/4tan⁡nx dx{I_n} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx} then  lim⁡n→∞ n[In+In+2]\,\mathop {\lim }\limits_{n \to \infty } \,n\left[ {{I_n} + {I_{n + 2}}} \right] equals
A 12{1 \over 2}
B 11
C ∞\infty
D zero
Correct Answer
Option B
Solution
In+In+2{I_n} + {I_{n + 2}}
=∫0π/4tan⁡n x(1+tan⁡2x)dx= \int\limits_0^{\pi /4} {{{\tan }^n}} \,x\left( {1 + {{\tan }^2}x} \right)dx
=∫0π/4tan⁡nx sec⁡2x dx= \int\limits_0^{\pi /4} {{{\tan }^n}} x\,{\sec ^2}x\,dx
=[tan⁡n+1xn+1]0π/4= \left[ {{{{{\tan }^{n + 1}}x} \over {n + 1}}} \right]_0^{\pi /4}
=1−0n+1=1n+1= {{1 - 0} \over {n + 1}} = {1 \over {n + 1}}

∴\therefore

In+In+2=1n+1{I_n} + {I_{n + 2}} = {1 \over {n + 1}}
⇒lim⁡n→∞n[In+In+2]\Rightarrow \mathop {\lim }\limits_{n \to \infty } n\left[ {{I_n} + {I_{n + 2}}} \right]
=lim⁡n→∞ n.1n+1=lim⁡n→∞nn+1= \mathop {\lim }\limits_{n \to \infty } \,n.{1 \over {n + 1}} = \mathop {\lim }\limits_{n \to \infty } {n \over {n + 1}}
=lim⁡n→∞nn(1+1n)=1= \mathop {\lim }\limits_{n \to \infty } {n \over {n\left( {1 + {1 \over n}} \right)}} = 1
Q7
∫02[x2]dx\int\limits_0^2 {\left[ {{x^2}} \right]dx} is
A 2−22 - \sqrt 2
B 2+22 + \sqrt 2
C  2−1\,\sqrt 2 - 1
D −2−3+5 - \sqrt 2 - \sqrt 3 + 5
Correct Answer
Option D
Solution
∫02[x2]dx=∫01[x2]dx+∫12[x2]dx+\int\limits_0^2 {\left[ {{x^2}} \right]} dx = \int\limits_0^1 {\left[ {{x^2}} \right]dx} + \int\limits_1^{\sqrt 2 } {\left[ {{x^2}} \right]} dx +
                 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
∫23[x2]+∫32[x2]dx\int\limits_{\sqrt 2 }^{\sqrt 3 } {\left[ {{x^2}} \right]} + \int\limits_{\sqrt 3 }^2 {\left[ {{x^2}} \right]} dx
=∫010dx+∫121dx+∫232dx+∫323dx= \int\limits_0^1 {0dx} + \int\limits_1^{\sqrt 2 } {1dx} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {2dx} + \int\limits_{\sqrt 3 }^2 {3dx}
=[x]1n+[2x]23+[3x]32= \left[ x \right]_1^{\sqrt n } + \left[ {2x} \right]_{\sqrt 2 }^{\sqrt 3 } + \left[ {3x} \right]_{\sqrt 3 }^2
=2−1+23−22+6−33= \sqrt 2 - 1 + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3
=5−3−2= 5 - \sqrt 3 - \sqrt 2
Q8
If y=f(x)y=f(x) makes +veve intercept of 22 and 00 unit on xx and yy axes and encloses an area of 3/43/4 square unit with the axes then ∫02xf′(x)dx\int\limits_0^2 {xf'\left( x \right)dx} is
A 3/23/2
B 11
C 5/45/4
D −3/4-3/4
Correct Answer
Option D
Solution

We have

∫02f(x)dx=34;Now,\int\limits_0^2 {f\left( x \right)} dx = {3 \over 4};Now,
∫02xf′(x)dx\int\limits_0^2 {xf'\left( x \right)} dx
=x∫02f′(x)dx−∫02f(x)dx= x\int\limits_0^2 {f'\left( x \right)dx} - \int\limits_0^2 {f\left( x \right)} dx
=[x f(x)]02−34= \left[ {x\,f\left( x \right)} \right]_0^2 - {3 \over 4}
=2f(2)−34= 2f\left( 2 \right) - {3 \over 4}
=0−34= 0 - {3 \over 4}
(\left( {} \right.

As

f(2)=0f\left( 2 \right) = 0
)\left. {} \right)
=−34.= - {3 \over 4}.
Q9
The value of the integral I=∫01x(1−x)ndxI = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} is
A 1n+1+1n+2{1 \over {n + 1}} + {1 \over {n + 2}}
B 1n+1{1 \over {n + 1}}
C 1n+2{1 \over {n + 2}}
D 1n+1−1n+2{1 \over {n + 1}} - {1 \over {n + 2}}
Correct Answer
Option D
Solution
I=∫01x(1−x)ndxI = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}} dx
=∫01(1−x)(1−1+x)ndx= \int\limits_0^1 {\left( {1 - x} \right){{\left( {1 - 1 + x} \right)}^n}dx}
=∫01(1−x)xndx= \int\limits_0^1 {\left( {1 - x} \right)} {x^n}dx
=[xn+1n+1−xn+2n+2]01= \left[ {{{{x^{n + 1}}} \over {n + 1}} - {{{x^{n + 2}}} \over {n + 2}}} \right]_0^1
=1n+1−1n+2= {1 \over {n + 1}} - {1 \over {n + 2}}
Q10
lim⁡n→∞1+24+34+....+n4n5\mathop {\lim }\limits_{n \to \infty } {{1 + {2^4} + {3^4} + .... + {n^4}} \over {{n^5}}} - lim⁡n→∞1+23+33+....+n3n5\mathop {\lim }\limits_{n \to \infty } {{1 + {2^3} + {3^3} + .... + {n^3}} \over {{n^5}}}
A 15{1 \over 5}
B 130{1 \over 30}
C zero
D 14{1 \over 4}
Correct Answer
Option A
Solution

The given expression can be written as

lim⁡n→∞1n∑r=1n(rn)4−lim⁡n→∞1n.lim⁡n→∞1n(rn)3\mathop {\lim }\limits_{n \to \infty } {1 \over n}{\sum\limits_{r = 1}^n {\left( {{r \over n}} \right)} ^4} - \mathop {\lim }\limits_{n \to \infty } {1 \over n}.\mathop {\lim }\limits_{n \to \infty } {1 \over n}{\left( {{r \over n}} \right)^3}
=∫01x4  dx−lim⁡n→∞1n×∫01x3  dx= \int\limits_0^1 {{x^4}} \,\,dx - \mathop {\lim }\limits_{n \to \infty } {1 \over n} \times \int\limits_0^1 {{x^3}} \,\,dx
=[x55]01−0=15= \left[ {{{{x^5}} \over 5}} \right]_0^1 - 0 = {1 \over 5}
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