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Definite Integration

JEE Mathematics · 230 questions · Page 8 of 23 · Click an option or "Show Solution" to reveal answer

Q71
Let f(x)=∫0xetf(t)dt+exf(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} be a differentiable function for all x∈\inR. Then f(x) equals :
A e(ex−1){e^{({e^{x - 1}})}}
B 2eex−12{e^{{e^x}}} - 1
C 2eex−1−12{e^{{e^x} - 1}} - 1
D eex−1{e^{{e^x}}} - 1
Correct Answer
Option C
Solution
f(x)=∫0xetf(t)dt+exf(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}}

.... (1) Differentiating both sides w.r.t. x

f′(x)=ex.f(x)+exf'(x) = {e^x}.f(x) + {e^x}

(Using Newton L:eibnitz Theorem)

⇒f′(x)f(x)+1=ex\Rightarrow {{f'(x)} \over {f(x) + 1}} = {e^x}

Integrating w.r.t. x

∫f′(x)f(x)+1dx=∫exdx\int {{{f'(x)} \over {f(x) + 1}}dx = \int {{e^x}dx} }
⇒ln⁡(f(x)+1)=ex+c\Rightarrow \ln (f(x) + 1) = {e^x} + c

Put x = 0 ln 2 = 1 + c (∵\because f(0) = 1, from equation (1)) ∴\therefore

ln⁡(f(x)+1)=ex+ln⁡2−1\ln (f(x) + 1) = {e^x} + \ln 2 - 1
⇒f(x)+1=2. eex−1\Rightarrow f(x) + 1 = 2.\,{e^{{e^x} - 1}}
⇒f(x)=2eex−1−1\Rightarrow f(x) = 2{e^{{e^x} - 1}} - 1
Q72
For x > 0, if f(x)=∫1xlog⁡et(1+t)dtf(x) = \int\limits_1^x {{{{{\log }_e}t} \over {(1 + t)}}dt} , then f(e)+f(1e)f(e) + f\left( {{1 \over e}} \right) is equal to :
A 12{1 \over 2}
B −-1
C 0
D 1
Correct Answer
Option A
Solution
f(x)=∫1xln⁡t1+tdtf(x) = \int_1^x {{{\ln t} \over {1 + t}}dt}

then

f(1x)=∫11/xln⁡t1+tdtf\left( {{1 \over x}} \right) = \int_1^{1/x} {{{\ln t} \over {1 + t}}dt}

Let

t=1u⇒dt=−1u2dut = {1 \over u} \Rightarrow dt = - {1 \over {{u^2}}}du
⇒f(1x)=∫1xln⁡1u1+1u(−1u2)dx\Rightarrow f\left( {{1 \over x}} \right) = \int_1^x {{{\ln {1 \over u}} \over {1 + {1 \over u}}}\left( { - {1 \over {{u^2}}}} \right)dx}
f(1x)=∫1xln⁡uu(1+u)du=∫1xln⁡tt(1+t)dtf\left( {{1 \over x}} \right) = \int_1^x {{{\ln u} \over {u(1 + u)}}} du = \int_1^x {{{\ln t} \over {t(1 + t)}}dt}

∴\therefore

f(x)+f(1x)=∫1xln⁡t(11+t+1t(1+t))dtf(x) + f\left( {{1 \over x}} \right) = \int_1^x {\ln t\left( {{1 \over {1 + t}} + {1 \over {t(1 + t)}}} \right)} dt
=∫1xln⁡t(11+t+1t−1t+1)dt= \int_1^x {\ln t\left( {{1 \over {1 + t}} + {1 \over t} - {1 \over {t + 1}}} \right)} dt
=∫1xln⁡ttdt=12(ln⁡x)2= \int_1^x {{{\ln t} \over t}dt = {1 \over 2}{{(\ln x)}^2}}

∴\therefore

f(e)+f(1e)=12(ln⁡e)2=12f(e) + f\left( {{1 \over e}} \right) = {1 \over 2}{(\ln e)^2} = {1 \over 2}
Q73
If f(x)={∫0x(5+∣1−t∣)dt,x>25x+1,x≤2f(x) = \left\{ \begin{array}{ll}{\int\limits_0^x {\left( {5 + \left| {1 - t} \right|} \right)dt,} } & {x > 2} \\ {5x + 1,} & {x \le 2} \end{array} \right., then
A f(x) is not continuous at x = 2
B f(x) is everywhere differentiable
C f(x) is continuous but not differentiable at x = 2
D f(x) is not differentiable at x = 1
Correct Answer
Option C
Solution
f(x)=∫01(5+(1−t))dt+∫1x(5+(t−1))dtf(x) = \int\limits_0^1 {(5 + (1 - t))dt + \int\limits_1^x {(5 + (t - 1))dt} }
=6−12+(4t+t22)∣1x= \left. {6 - {1 \over 2} + \left( {4t + {{{t^2}} \over 2}} \right)} \right|_1^x
=112+4x+x22−4−12= {{11} \over 2} + 4x + {{{x^2}} \over 2} - 4 - {1 \over 2}
=x22+4x−1= {{{x^2}} \over 2} + 4x - 1
f(2+)=2+8+1=11f({2^ + }) = 2 + 8 + 1 = 11
f(2)=f(2−)=5×2+1=11f(2) = f({2^ - }) = 5 \times 2 + 1 = 11

⇒\Rightarrow continuous at x = 2 Clearly differentiable at x = 1 Lf' (2) = 5 Rf' (2) = 6 ⇒\Rightarrow Not differentiable at x = 2

Q74
Which of the following statements is correct for the function g(α\alpha) for α\alpha ∈\in R such that g(α)=∫π6π3sin⁡αxcos⁡αx+sin⁡αxdxg(\alpha ) = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{{{\sin }^\alpha }x} \over {{{\cos }^\alpha }x + {{\sin }^\alpha }x}}dx}
A g(α)g(\alpha ) is a strictly increasing function
B g(α)g(\alpha ) is an even function
C g(α)g(\alpha ) has an inflection point at α\alpha = −-12{1 \over 2}
D g(α)g(\alpha ) is a strictly decreasing function
Correct Answer
Option B
Solution
g(α)=∫π/6π/3sin⁡α(π2−x)cos⁡α(π2−x)x+sin⁡α(π2−x)dxg(\alpha ) = \int\limits_{\pi /6}^{\pi /3} {{{{{\sin }^\alpha }\left( {{\pi \over 2} - x} \right)} \over {{{\cos }^\alpha }\left( {{\pi \over 2} - x} \right)x + {{\sin }^\alpha }\left( {{\pi \over 2} - x} \right)}}dx}
=∫π/6π/3cos⁡αxsin⁡αx+cos⁡αxdx= \int\limits_{\pi /6}^{\pi /3} {{{{{\cos }^\alpha }x} \over {{{\sin }^\alpha }x + {{\cos }^\alpha }x}}dx}

∴\therefore

2.g(α)=∫π/6π/3sinαx+cos⁡αxsin⁡αx+cos⁡αxdx=∫π/6π/3dx=π3−π6=π62.g(\alpha ) = \int\limits_{\pi /6}^{\pi /3} {{{si{n^\alpha }x + {{\cos }^\alpha }x} \over {{{\sin }^\alpha }x + {{\cos }^\alpha }x}}dx} = \int\limits_{\pi /6}^{\pi /3} {dx} = {\pi \over 3} - {\pi \over 6} = {\pi \over 6}

⇒\Rightarrow

g(α)=π12g(\alpha ) = {\pi \over {12}}

i.e. a constant function hence an even function.

Q75
Let f : R → \to R be defined as f(x) = e−-xsinx. If F : [0, 1] → \to R is a differentiable function with that F(x) = ∫0xf(t)dt\int_0^x {f(t)dt} , then the value of ∫01(F′(x)+f(x))exdx\int_0^1 {(F'(x) + f(x)){e^x}dx} lies in the interval
A [331360,334360]\left[ {{{331} \over {360}},{{334} \over {360}}} \right]
B [330360,331360]\left[ {{{330} \over {360}},{{331} \over {360}}} \right]
C [335360,336360]\left[ {{{335} \over {360}},{{336} \over {360}}} \right]
D [327360,329360]\left[ {{{327} \over {360}},{{329} \over {360}}} \right]
Correct Answer
Option B
Solution

F(x) =

∫0xf(t)dt\int_0^x {f(t)dt}

⇒\Rightarrow F'(x) = f(x) by Leibnitz theorem I =

∫01(F′(x)+f(x))exdx=∫012f(x)exdx\int\limits_0^1 {(F'(x) + f(x)){e^x}dx = \int\limits_0^1 {2f(x){e^x}dx} }
I=∫012sin⁡x dxI = \int\limits_0^1 {2\sin x\,dx}
I=2(1−cos⁡1)I = 2(1 - \cos 1)
=2{1−(1−122!+144!−16!+...)}= 2\left\{ {1 - \left( {1 - {{{1^2}} \over {2!}} + {{{1^4}} \over {4!}} - {1 \over {6!}} + ...} \right)} \right\}
=2{1−(1−12+124)}<2(1−cos⁡1)<2{1−(1−12+124−1720)}= 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}}} \right)} \right\} < 2(1 - \cos 1) < 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}} - {1 \over {720}}} \right)} \right\}
330360<2(1−cos⁡1)<331360{{330} \over {360}} < 2(1 - \cos 1) < {{331} \over {360}}
330360<1<331360{{330} \over {360}} < 1 < {{331} \over {360}}
Q76
If the integral ∫010[sin⁡2πx]ex−[x]dx=αe−1+βe−12+γ\int_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}} dx = \alpha {e^{ - 1}} + \beta {e^{ - {1 \over 2}}} + \gamma , where α\alpha, β\beta, γ\gamma are integers and [x] denotes the greatest integer less than or equal to x, then the value of α\alpha + β\beta + γ\gamma is equal to :
A 0
B 10
C 20
D 25
Correct Answer
Option A
Solution

Given integral

∫010[sin⁡2πx]ex−[x]dx=10∫01[sin⁡2πx]e{x}dx\int\limits_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}dx = 10\int\limits_0^1 {{{[\sin 2\pi x]} \over {{e^{\{ x\} }}}}dx} }

(using property of definite in.)

=10[∫01/20.dx+∫1/21−1exdx]= 10\left[ {\int\limits_0^{1/2} {0.dx} + \int\limits_{1/2}^1 {{{ - 1} \over {{e^x}}}dx} } \right]

=

−10[e−x−1]1/21=10[e−1−e−1/2]- 10\left[ {{{{e^{ - x}}} \over { - 1}}} \right]_{1/2}^1 = 10\left[ {{e^{ - 1}} - {e^{ - 1/2}}} \right]
=10e−1−10e−1/2= 10{e^{ - 1}} - 10{e^{ - 1/2}}

comparing with the given relation, α\alpha = 10, β\beta = −-10, γ\gamma = 0 α\alpha + β\beta + γ\gamma = 0 Therefore, the correct answer is (A).

Q77
Let g(x) = ∫0xf(t)dt\int_0^x {f(t)dt} , where f is continuous function in [ 0, 3 ] such that 13{1 \over 3} ≤ \le f(t) ≤ \le 1 for all t∈\in [0, 1] and 0 ≤ \le f(t) ≤ \le 12{1 \over 2} for all t∈\in (1, 3]. The largest possible interval in which g(3) lies is :
A [−1,−12]\left[ { - 1, - {1 \over 2}} \right]
B [−32,−1]\left[ { - {3 \over 2}, - 1} \right]
C [1, 3]
D [13,2]\left[ {{1 \over 3},2} \right]
Correct Answer
Option D
Solution

Given, g(x)=∫0xf(t)dtg(x)=\int_0^x f(t) d t ∴g(3)=∫03f(t)dt=∫01f(t)dt+∫13f(t)dt\therefore g(3)=\int_0^3 f(t) d t=\int_0^1 f(t) d t+\int_1^3 f(t) d t ⇒∫0113dt+∫130⋅dt≤g(3)≤∫011dt+∫1312dt\Rightarrow \int_0^1 \dfrac{1}{3} d t+\int_1^3 0 \cdot d t \leq g(3) \leq \int_0^1 1 d t+\int_1^3 \dfrac{1}{2} d t ⇒13≤g(3)≤1+1\Rightarrow \dfrac{1}{3} \leq g(3) \leq 1+1 ⇒13≤g(3)≤2\Rightarrow \dfrac{1}{3} \leq g(3) \leq 2

Q78
Let a be a positive real number such that ∫0aex−[x]dx=10e−9\int_0^a {{e^{x - [x]}}} dx = 10e - 9 where [ x ] is the greatest integer less than or equal to x. Then a is equal to:
A 10−log⁡e(1+e)10 - {\log _e}(1 + e)
B 10+log⁡e210 + {\log _e}2
C 10+log⁡e310 + {\log _e}3
D 10+log⁡e(1+e)10 + {\log _e}(1 + e)
Correct Answer
Option B
Solution

a > 0 Let

n≤a<n+1,n∈Wn \le a < n + 1,n \in W
a=[a]+{a}⇓⇓G.I.F.Fractional parta=\begin{array}{lll}{[a]} & + & {\{ a\} } \\ \Downarrow & {} & \Downarrow \\ {G.I.F.} & {} & {Fractional\,part} \end{array}

Here [ a ] = n Now,

∫0aex−[x]dx=10e−9\int_0^a {{e^{x - [x]}}} dx = 10e - 9
⇒∫0ne{x}dx+∫naex−[x]dx=10e−9\Rightarrow \int\limits_0^n {{e^{\{ x\} }}dx} + \int\limits_n^a {{e^{x - [x]}}dx} = 10e - 9

∴\therefore

n∫01exdx+∫naex−ndx=10e−9n\int\limits_0^1 {{e^x}dx} + \int\limits_n^a {{e^{x - n}}dx} = 10e - 9
⇒n(e−1)+(ea−n−1)=10e−9\Rightarrow n(e - 1) + ({e^{a - n}} - 1) = 10e - 9

∴\therefore n = 0 and {a} = loge 2 So,

a=[a]+{a}=(10+log⁡e2)a = [a] + \{ a\} = (10 + {\log _e}2)

⇒\Rightarrow Option (2) is correct.

Q79
The value of the integral ∫−11log⁡e(1−x+1+x)dx\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt {1 - x} + \sqrt {1 + x} )dx} is equal to:
A 12log⁡e2+π4−32{1 \over 2}{\log _e}2 + {\pi \over 4} - {3 \over 2}
B 2log⁡e2+π4−12{\log _e}2 + {\pi \over 4} - 1
C log⁡e2+π2−1{\log _e}2 + {\pi \over 2} - 1
D 2log⁡e2+π2−122{\log _e}2 + {\pi \over 2} - {1 \over 2}
Correct Answer
Option C
Solution
∫−11log⁡e(1−x+1+x)dx\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} )dx}

We know,

∫−aaf(x)dx=2∫0af(x)dx\int\limits_{ - a}^a {f(x)dx = 2\int\limits_0^a {f(x)dx} }

So,

2∫01log⁡e(1−x+1+x)dx2\int\limits_0^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} )dx}
l=2∫01log⁡e(1−x+1+x).1dxl = 2\int\limits_0^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} ).1dx}
⇒l2=log⁡e(1−x+1+x). x]01−∫01121+x−121−x1−x+1+x. x dx\Rightarrow {l \over 2} = {\log _e}(\sqrt {1 - x} + \sqrt {1 + x} ).\,x]_0^1 - \int\limits_0^1 {{{{1 \over {2\sqrt {1 + x} }} - {1 \over {2\sqrt {1 - x} }}} \over {\sqrt {1 - x} + \sqrt {1 + x} }}.\,x\,dx}
⇒l2=log⁡e2−12∫01(1−x−1+x1−x+1+x)x1−x2dx\Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {\left( {{{\sqrt {1 - x} - \sqrt {1 + x} } \over {\sqrt {1 - x} + \sqrt {1 + x} }}} \right){x \over {\sqrt {1 - {x^2}} }}dx}
⇒l2=log⁡e2−12∫01((1−x)+(1+x)−21−x2(1−x)−(1+x))x1−x2dx\Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {\left( {{{(1 - x) + (1 + x) - 2\sqrt {1 - {x^2}} } \over {(1 - x) - (1 + x)}}} \right){x \over {\sqrt {1 - {x^2}} }}dx}
⇒l2=log⁡e2−12∫012(1−1−x2)−2x.x1−x2dx\Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {{{2(1 - \sqrt {1 - {x^2}} )} \over { - 2x}}.{x \over {\sqrt {1 - {x^2}} }}dx}
⇒l2=log⁡e2+12∫01(11−x2−1)dx\Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}\int\limits_0^1 {\left( {{1 \over {\sqrt {1 - {x^2}} }} - 1} \right)dx}
⇒l2=log⁡e2+12[sin⁡−1x−x]01\Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}[{\sin ^{ - 1}}x - x]_0^1
⇒l2=log⁡e2+12(π2−1)\Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}\left( {{\pi \over 2} - 1} \right)

∴\therefore

l=log⁡e2+π2−1l = {\log _e}2 + {\pi \over 2} - 1
Q80
If [x] denotes the greatest integer less than or equal to x, then the value of the integral ∫−π/2π/2[[x]−sin⁡x]dx\int_{ - \pi /2}^{\pi /2} {[[x] - \sin x]dx} is equal to :
A −- π\pi
B π\pi
C 0
D 1
Correct Answer
Option A
Solution
I=∫−π/2π/2[[x]−sin⁡x]dxI = \int_{ - \pi /2}^{\pi /2} {[[x] - \sin x]dx}
=∫−π/2π/2([x]+[−sin⁡x])dx= \int_{ - \pi /2}^{\pi /2} {\left( {[x] + [ - \sin x]} \right)dx}
=∫0π/2([x]+[−sin⁡x]+[−x]+[sin⁡x])dx= \int_0^{\pi /2} {\left( {[x] + [ - \sin x] + [ - x] + [\sin x]} \right)} dx
=∫0π/2(−2)dx= \int_0^{\pi /2} {( - 2)dx}
=−π= - \pi
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