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Definite Integration

JEE Mathematics · 230 questions · Page 18 of 23 · Click an option or "Show Solution" to reveal answer

Q171
The value of the integral ∫−12log⁡e(x+x2+1)dx\int\limits_{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x is
A 5−2+log⁡e(7+451+2)\sqrt{5}-\sqrt{2}+\log _e\left(\dfrac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)
B 2−5+log⁡e(7+451+2)\sqrt{2}-\sqrt{5}+\log _e\left(\dfrac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)
C 5−2+log⁡e(9+451+2)\sqrt{5}-\sqrt{2}+\log _e\left(\dfrac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)
D 2−5+log⁡e(9+451+2)\sqrt{2}-\sqrt{5}+\log _e\left(\dfrac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)
Correct Answer
Option D
Solution
∫−12log⁡e(x+x2+1)dx=[xlog⁡e(x+x2+1)]−12−∫−12x(x+x2+1)(1+xx2+1)dx=2log⁡2(2+5)+log⁡e(2−1)−∫−12xx2+1dx=log⁡e[(2+5)2(2−1)]−[x2+1]−12=log⁡e[9+451+2]−5+2=2−5+log⁡e[9+451+2]\begin{aligned} & \int\limits_{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x \\ & =\left[x \log _e\left(x+\sqrt{x^2+1}\right)\right]_{-1}^2-\int\limits_{-1}^2 \frac{x}{\left(x+\sqrt{x^2+1}\right)}\left(1+\frac{x}{\sqrt{x^2+1}}\right) d x \\ & =2 \log _2(2+\sqrt{5})+\log _e(\sqrt{2}-1)-\int\limits_{-1}^2 \frac{x}{\sqrt{x^2+1}} d x \\ & =\log _e\left[(2+\sqrt{5})^2(\sqrt{2}-1)\right]-\left[\sqrt{x^2+1}\right]_{-1}^2 \\ & =\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right]-\sqrt{5}+\sqrt{2} \\ & =\sqrt{2}-\sqrt{5}+\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right] \end{aligned}
Q172
$$\text { Let } f(x)=\left\{\begin{array}{lr} -2, & -2 \leq x \leq 0 \\ x-2, & 0
A 2
B 6
C 4
D 1
Correct Answer
Option A
Solution
f(x)={−2−2≤x≤0x−20<x≤2h(x)=f∣x∣+∣f(x)∣f(x)=\left\{\begin{array}{cc} -2 & -2 \leq x \leq 0 \\ x-2 & 0< x \leq 2 \end{array} h(x)=f|x|+|f(x)|\right.
h(x)={−x−2+2=−x−2≤x≤000<x≤2∴∫−22h(x)dx=∫−20−xdx+∫020dxx22∣−20=42=2\begin{aligned} & h(x)=\left\{\begin{array}{cc} -x-2+2=-x & -2 \leq x \leq 0 \\ 0 & 0< x \leq 2 \end{array}\right. \\ & \therefore \int\limits_{-2}^2 h(x) d x=\int\limits_{-2}^0-x d x+\int\limits_0^2 0 d x \\ & \left.\frac{x^2}{2}\right|_{-2} ^0=\frac{4}{2}=2 \end{aligned}
Q173
Let f(x)=∫0x(t+sin⁡(1−et))dt,x∈Rf(x)=\int_0^x\left(t+\sin \left(1-e^t\right)\right) d t, x \in \mathbb{R}. Then, lim⁡x→0f(x)x3\lim \limits_{x \rightarrow 0} \dfrac{f(x)}{x^3} is equal to
A 16\dfrac{1}{6}
B −16-\dfrac{1}{6}
C 23\dfrac{2}{3}
D −23-\dfrac{2}{3}
Correct Answer
Option B
Solution

Given

f(x)=∫0x(t+sin⁡(1−et))dtf(x)=\int\limits_0^x\left(t+\sin \left(1-e^t\right)\right) d t

Now,

lim⁡x→0f(x)x3(00 form )\lim \limits_{x \rightarrow 0} \frac{f(x)}{x^3}\left(\frac{0}{0} \text{ form }\right)
=lim⁡x→0∫0x(t+sin⁡(1−et))dtx3=lim⁡x→0x+sin⁡(1−ex)3x2(00)=lim⁡x→01+cos⁡(1−ex)(−ex)6x(00)=lim⁡x→0−sin⁡(1−ex)(ex)2+cos⁡(1−ex)(−ex)6=−16\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\int\limits_0^x\left(t+\sin \left(1-e^t\right)\right) d t}{x^3} \\ & =\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2}\left(\frac{0}{0}\right) \\ & =\lim _{x \rightarrow 0} \frac{1+\cos \left(1-e^x\right)\left(-e^x\right)}{6 x}\left(\frac{0}{0}\right) \\ & =\lim _{x \rightarrow 0} \frac{-\sin \left(1-e^x\right)\left(e^x\right)^2+\cos \left(1-e^x\right)\left(-e^x\right)}{6} \\ & =-\frac{1}{6} \end{aligned}
Q174
If ∫−π2π296x2cos⁡2x(1+ex)dx=π(απ2+β),α,β∈Z\int_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \dfrac{96 x^2 \cos ^2 x}{\left(1+e^x\right)} \mathrm{d} x=\pi\left(\alpha \pi^2+\beta\right), \alpha, \beta \in \mathbb{Z}, then (α+β)2(\alpha+\beta)^2 equals
A 196
B 100
C 64
D 144
Correct Answer
Option B
Solution

∫−π2π296x2cos⁡2x(1+ex)dx\int\limits_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \dfrac{96 x^2 \cos ^2 x}{\left(1+e^x\right)} d x \quad (Apply King Property)

∫0π296x2cos⁡2x=48∫0π2x2(1+cos⁡2x)dx48[(x33)0π/2+∫0π2x12cos⁡II2xdx]⇒ On solving π(2π2−12)⇒α=2,β=−12⇒(α+β)2=100\begin{aligned} & \int_0^{\frac{\pi}{2}} 96 x^2 \cos ^2 x=48 \int_0^{\frac{\pi}{2}} x^2(1+\cos 2 x) d x \\ & 48\left[\left(\frac{x^3}{3}\right)_0^{\pi / 2}+\int_0^{\frac{\pi}{2}} x_1^2 \cos _{I I} 2 x d x\right] \\ & \Rightarrow \text{ On solving } \pi\left(2 \pi^2-12\right) \\ & \Rightarrow \alpha=2, \beta=-12 \\ & \Rightarrow(\alpha+\beta)^2=100 \end{aligned}
Q175
Let ∫αlog⁡e4dxex−1=π6\int\limits_\alpha^{\log _e 4} \dfrac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}}=\dfrac{\pi}{6}. Then eα\mathrm{e}^\alpha and e−α\mathrm{e}^{-\alpha} are the roots of the equation :
A 2x2−5x+2=02 x^2-5 x+2=0
B x2−2x−8=0x^2-2 x-8=0
C 2x2−5x−2=02 x^2-5 x-2=0
D x2+2x−8=0x^2+2 x-8=0
Correct Answer
Option A
Solution
 Let ex−1=tex−1=t2ex=1+t2ex=0+2t−dtdxdtdx=ex2t=t2+12tI=∫2tt(1+t2)dt=2tan⁡−1t⇒I=∫αlog⁡4dxex−1I=2tan⁡−1ex−1∣αlog⁡4=2(tan⁡−13−tan⁡−1eα−1)=π6⇒π3−tan⁡−1(eα−1)=π12tan⁡−1(eα−1)=π4⇒eα−1=1eα=2⇒e−α=12\begin{aligned} & \text{ Let } \sqrt{e^x-1}=t \\ & e^x-1=t^2 \\ & e^x=1+t^2 \\ & e^x=0+2 t-\frac{d t}{d x} \\ & \frac{d t}{d x}=\frac{e^x}{2 t}=\frac{t^2+1}{2 t} \\ & I=\int \frac{2 t}{t\left(1+t^2\right)} d t=2 \tan ^{-1} t \\ & \Rightarrow \quad I=\int\limits_\alpha^{\log ^4} \frac{d x}{\sqrt{e^x-1}} \\ & I=\left.2 \tan ^{-1} \sqrt{e^x-1}\right|_\alpha ^{\log 4} \\ & =2\left(\tan ^{-1} \sqrt{3}-\tan ^{-1} \sqrt{e^\alpha-1}\right)=\frac{\pi}{6} \\ & \Rightarrow \frac{\pi}{3}-\tan ^{-1}\left(\sqrt{e^\alpha-1}\right)=\frac{\pi}{12} \\ & \tan ^{-1}\left(\sqrt{e^\alpha-1}\right)=\frac{\pi}{4} \\ & \Rightarrow e^\alpha-1=1 \\ & e^\alpha=2 \Rightarrow e^{-\alpha}=\frac{1}{2} \end{aligned}
∴\therefore \quad

Quadratic equation whose roots are

eae^a

&

e−αe^{-\alpha}

is

x2−(eα+e−α)x+eα×e−α=0x2−(2+12)x+1=02x2−5x+2=0\begin{aligned} & x^2-\left(e^\alpha+e^{-\alpha}\right) x+e^\alpha \times e^{-\alpha}=0 \\ & x^2-\left(2+\frac{1}{2}\right) x+1=0 \\ & 2 x^2-5 x+2=0 \end{aligned}
Q176
The value of k∈Nk \in \mathbb{N} for which the integral In=∫01(1−xk)ndx,n∈NI_n=\int_0^1\left(1-x^k\right)^n d x, n \in \mathbb{N}, satisfies 147I20=148I21147 I_{20}=148 I_{21} is
A 8
B 14
C 7
D 10
Correct Answer
Option C
Solution
I(21)=∫01(1−xk)21dx=∫01(1−xk)(1−xk)20dx=∫01(1−xk)20dx−∫0xk(1−xk)20dxI(21)=I(20)−∫01xk(1−xk)20dx\begin{aligned} & I(21)=\int\limits_0^1\left(1-x^k\right)^{21} d x \\ & =\int\limits_0^1\left(1-x^k\right)\left(1-x^k\right)^{20} d x \\ & =\int\limits_0^1\left(1-x^k\right)^{20} d x-\int_0 x^k\left(1-x^k\right)^{20} d x \\ & I(21)=I(20)-\int\limits_0^1 x^k\left(1-x^k\right)^{20} d x \end{aligned}
I(21)=I(20)−⌊(1−xk)21−21kx−∫01(1−xk)21−21kdx⌋I(21)=I(20)-\left\lfloor\frac{\left(1-x^k\right)^{21}}{-21 k} x-\int\limits_0^1 \frac{(1-x^k)^{21}}{-21 k} d x\right\rfloor
I(21)=I(20)−121kI(20)⇒[I(21)](21k+1)=21KI(20)⇒21K=147⇒K=7\begin{aligned} & I(21)=I(20)-\frac{1}{21 k} I(20) \\ & \Rightarrow[I(21)](21 k+1)=21 K I(20) \\ & \Rightarrow 21 K=147 \Rightarrow K=7 \end{aligned}
Q177
The integral ∫0π/4136sin⁡x3sin⁡x+5cos⁡x dx\int\limits_0^{\pi / 4} \dfrac{136 \sin x}{3 \sin x+5 \cos x} \mathrm{~d} x is equal to :
A 3π−50log⁡e2+20log⁡e53 \pi-50 \log _e 2+20 \log _e 5
B 3π−25log⁡e2+10log⁡e53 \pi-25 \log _e 2+10 \log _e 5
C 3π−10log⁡e(22)+10log⁡e53 \pi-10 \log _{\mathrm{e}}(2 \sqrt{2})+10 \log _{\mathrm{e}} 5
D 3π−30log⁡e2+20log⁡e53 \pi-30 \log _e 2+20 \log _e 5
Correct Answer
Option A
Solution
∫0π/4136sin⁡x3sin⁡x+5cos⁡xdx\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x
sin⁡x=A(3sin⁡x+5cos⁡x)+B(3cos⁡x−5sin⁡x)3A−5B=15A+3B=0>A=334B=−534\begin{aligned} & \sin x=A(3 \sin x+5 \cos x)+B(3 \cos x-5 \sin x) \\ & \begin{array}{l} 3 A-5 B=1 \\ 5 A+3 B=0 \end{array}>A=\frac{3}{34} \quad B=\frac{-5}{34} \end{aligned}
∫0π/4136[334(3sin⁡x+5cos⁡x)−534(3cos⁡x−5sin⁡x)]3sin⁡x+5cos⁡xdx∫0π/412dx−20∫0π/43cos⁡x−5sin⁡x3sin⁡x+5cos⁡xdx12×π4−20[ln⁡∣32+52∣−ln⁡5]3π−20ln⁡25/2+20ln⁡5⇒3π−50ln⁡2+20ln⁡5\begin{aligned} & \int_0^{\pi / 4} \frac{136\left[\frac{3}{34}(3 \sin x+5 \cos x)-\frac{5}{34}(3 \cos x-5 \sin x)\right]}{3 \sin x+5 \cos x} d x \\ & \int_0^{\pi / 4} 12 d x-20 \int_0^{\pi / 4} \frac{3 \cos x-5 \sin x}{3 \sin x+5 \cos x} d x \\ & 12 \times \frac{\pi}{4}-20\left[\ln \left|\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right|-\ln 5\right] \\ & 3 \pi-20 \ln 2^{5 / 2}+20 \ln 5 \\ & \Rightarrow 3 \pi-50 \ln 2+20 \ln 5 \end{aligned}
Q178
The value of ∫−ππ2y(1+sin⁡y)1+cos⁡2ydy\int\limits_{-\pi}^\pi \dfrac{2 y(1+\sin y)}{1+\cos ^2 y} d y is :
A π2\dfrac{\pi}{2}
B π2\pi^2
C π22\dfrac{\pi^2}{2}
D 2π22 \pi^2
Correct Answer
Option B
Solution
I=∫−ππ2y(1+sin⁡y)1+cos⁡2ydy=∫0π(2y(1+sin⁡y)1+cos⁡2y+−2y(1−sin⁡y)1+cos⁡2y)dy=∫0π(2y+2ysin⁡y−2y+2ysin⁡y1+cos⁡2y)dy\begin{aligned} & I=\int\limits_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y \\ & =\int\limits_0^\pi\left(\frac{2 y(1+\sin y)}{1+\cos ^2 y}+\frac{-2 y(1-\sin y)}{1+\cos ^2 y}\right) d y \\ & =\int\limits_0^\pi\left(\frac{2 y+2 y \sin y-2 y+2 y \sin y}{1+\cos ^2 y}\right) d y \end{aligned}
I=4∫0π(ysin⁡1+cos⁡2y)dy... (1)I=4 \int_0^\pi\left(\frac{y \sin }{1+\cos ^2 y}\right) d y \quad \text{... (1)}
I=4∫0π((π−y)sin⁡(π−)1+cos⁡2(π−y))dyI=4⌊∫0(πsin⁡y1+cos⁡2y)dy−∫0πysin⁡y1+cos⁡2ydy⌋... (2)\begin{aligned} & I=4 \int\limits_0^\pi\left(\frac{(\pi-y) \sin (\pi-)}{1+\cos ^2(\pi-y)}\right) d y \\ & I=4\left\lfloor\int\limits_0\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y-\int\limits_0^\pi \frac{y \sin y}{1+\cos ^2 y} d y\right\rfloor \quad \text{... (2)} \end{aligned}

Adding equation (1) and (2)

2I=4∫0π(πsin⁡y1+cos⁡2y)dyI=2π∫0πsin⁡y1+cos⁡2ydy=2π×π2=π2\begin{aligned} & 2 I=4 \int\limits_0^\pi\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y \\ & I=2 \pi \int\limits_0^\pi \frac{\sin y}{1+\cos ^2 y} d y \\ & =2 \pi \times \frac{\pi}{2} \\ & =\pi^2 \end{aligned}
Q179
Let β(m,n)=∫01xm−1(1−x)n−1 dx, m,n>0\beta(\mathrm{m}, \mathrm{n})=\int\limits_0^1 x^{\mathrm{m}-1}(1-x)^{\mathrm{n}-1} \mathrm{~d} x, \mathrm{~m}, \mathrm{n}>0. If ∫01(1−x10)20 dx=a×β(b,c)\int\limits_0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x=\mathrm{a} \times \beta(\mathrm{b}, \mathrm{c}), then 100(a+b+c)100(\mathrm{a}+\mathrm{b}+\mathrm{c}) equals _________.
A 2012
B 1021
C 1120
D 2120
Correct Answer
Option D
Solution

First, let's rewrite the given integral using the given form of the Beta function. The given integral is:

∫01(1−x10)20 dx\int\limits_0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x

To use the Beta function, let us make a substitution.

Let x10=t x^{10} = t .

Then, dx=110t−910dt dx = \dfrac{1}{10}t^{-\dfrac{9}{10}} dt or dx=110t−910dt dx = \dfrac{1}{10} t^{-\dfrac{9}{10}} dt .

The limits of integration change as follows: when x=0 x = 0 , t=0 t = 0 , and when x=1 x = 1 , t=1 t = 1 .

Substituting these into the integral, we have:

∫01(1−t)20⋅110t−910dt\int\limits_0^1 (1 - t)^{20} \cdot \frac{1}{10} t^{-\frac{9}{10}} dt

which simplifies to:

110∫01(1−t)20t−910dt\frac{1}{10} \int\limits_0^1 (1 - t)^{20} t^{-\frac{9}{10}} dt

We recognize this integral as a Beta function β(m,n) \beta(m, n) where m=1−910=110 m = 1 - \dfrac{9}{10} = \dfrac{1}{10} and n=20+1=21 n = 20 + 1 = 21 .

Therefore, we can write this as:

110β(110,21)\frac{1}{10} \beta \left( \frac{1}{10}, 21 \right)

Comparing this to a×β(b,c) a \times \beta(b, c) , we have a=110 a = \dfrac{1}{10} , b=110 b = \dfrac{1}{10} , and c=21 c = 21 .

Now we calculate 100(a+b+c) 100(a + b + c) :

100(110+110+21)=100(110+110+21)=100(15+21)=100(15+1055)=100(1065)=100×21.2=2120100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{5} + 21 \right) = 100 \left( \frac{1}{5} + \frac{105}{5} \right) = 100 \left( \frac{106}{5} \right) = 100 \times 21.2 = 2120

So, the answer is Option D, 2120.

Q180
∫0π/4cos⁡2xsin⁡2x(cos⁡3x+sin⁡3x)2dx is equal to \int\limits_0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x \text{ is equal to }
A 1/9
B 1/6
C 1/3
D 1/12
Correct Answer
Option B
Solution
∫0π/4cos⁡2x⋅sin⁡2x(cos⁡3x+sin⁡3x)2dx=∫0π/4tan⁡2x⋅sec⁡2x(1+tan⁡3x)2dx\begin{aligned} & \int\limits_0^{\pi / 4} \frac{\cos ^2 x \cdot \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x \\ & =\int\limits_0^{\pi / 4} \frac{\tan ^2 x \cdot \sec ^2 x}{\left(1+\tan ^3 x\right)^2} d x \end{aligned}

Let

tan⁡x=t\tan x=t
∫01t2dt(1+t3)2\int\limits_0^1 \frac{t^2 d t}{\left(1+t^3\right)^2}

Let

1+t3=z1+t^3=\mathrm{z}
3t2dt=dz13∫12dzz2=13(−1z)∣12=−13(12−1)=16\begin{gathered} 3 t^2 d t=d z \\ \frac{1}{3} \int\limits_1^2 \frac{d z}{z^2}=\left.\frac{1}{3}\left(-\frac{1}{z}\right)\right|_1 ^2 \\ =-\frac{1}{3}\left(\frac{1}{2}-1\right)=\frac{1}{6}\end{gathered}
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