Definite Integration — Page 20 | JEE Mathematics

Q191

The integral

\int_0^\pi \dfrac{(x+3) \sin x}{1+3 \cos ^2 x} d x

is equal to

A

\dfrac{\pi}{\sqrt{3}}(\pi+1)

B

\dfrac{\pi}{3 \sqrt{3}}(\pi+6)

C

\dfrac{\pi}{\sqrt{3}}(\pi+2)

D

\dfrac{\pi}{2 \sqrt{3}}(\pi+4)

Correct Answer

Option B

Solution

\begin{aligned} & I=\int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x \quad\text{..... (i)}\\ & =\int_0^\pi \frac{(\pi-x+3) \sin (\pi-x)}{1+3 \cos ^2(\pi-x)} d x \\ & =\int_0^\pi \frac{(\pi+3) \sin x-x \sin x}{1+3 \cos ^2 x} d x \quad\text{..... (ii)} \end{aligned}

Add (i) & (ii)

2 I=\int_0^\pi \frac{(\pi+6) \sin x}{1+3 \cos ^2 x} d x

Let $\cos x=t \Rightarrow-\sin x d x=d t$ $(\pi+6) \int_1^{-1} \dfrac{-d t}{\left(1+3 t^2\right)}=\left(\dfrac{\pi+6}{3}\right) \int_{-1}^1 \dfrac{d t}{t^2+\left(\dfrac{1}{\sqrt{3}}\right)^2}$ $\Rightarrow \quad 2 I=\left.\left(\dfrac{\pi+6}{3}\right) \cdot \dfrac{1}{\left(\dfrac{1}{\sqrt{3}}\right)} \cdot \tan ^{-1} \dfrac{t}{\dfrac{1}{\sqrt{3}}}\right|_{-1} ^1$

\begin{aligned} & \Rightarrow \quad 2 I=\left(\frac{\pi+6}{\sqrt{3}}\right)\left[\tan ^{-1}(\sqrt{3})-\tan ^{-1}(-\sqrt{3})\right] \\ &=\left(\frac{\pi+6}{\sqrt{3}}\right) \cdot\left(\frac{\pi}{3}-\left(-\frac{\pi}{3}\right)\right) \\ &=\left(\frac{\pi+6}{\sqrt{3}}\right) \cdot \frac{2 \pi}{3} \\ & \Rightarrow \quad I=\frac{\pi(\pi+6)}{3 \sqrt{3}} \end{aligned}

Q192

Let

f:[1, \infty) \rightarrow[2, \infty)

be a differentiable function. If

10 \int_1^1 f(\mathrm{t}) \mathrm{dt}=5 x f(x)-x^5-9

for all

x \geqslant 1

, then the value of

f(3)

is :

A 22

B 26

C 32

D 18

Correct Answer

Option C

Solution

To solve the problem, we start with the given equation: $10 \int_1^x f(t) \, dt = 5x f(x) - x^5 - 9$ By differentiating both sides with respect to $x$ , we have: $\dfrac{d}{dx}\left(10 \int_1^x f(t) \, dt\right) = \dfrac{d}{dx}(5x f(x) - x^5 - 9)$ The left side simplifies to $10 f(x)$ .

For the right side, using the product rule and the power rule, we get: $10 f(x) = 5f(x) + 5x \dfrac{d}{dx} f(x) - 5x^4$ Rearranging terms, we obtain: $5 f(x) = 5x \dfrac{d}{dx} f(x) - 5x^4$ Let $y = f(x)$ .

Thus: $5 y = 5x \dfrac{dy}{dx} - 5x^4$ Dividing by 5, we have: $y = x \dfrac{dy}{dx} - x^4$ Rewriting, we get: $\dfrac{dy}{dx} - \dfrac{y}{x} = x^3$ This is a linear differential equation.

The integrating factor (I.F.) is calculated as: $\text{I.F.} = e^{\int \dfrac{-1}{x} \, dx} = e^{-\ln x} = \dfrac{1}{x}$ Multiplying through by the integrating factor, we have: $y \cdot \dfrac{1}{x} = \int x^3 \cdot \dfrac{1}{x} \, dx = \int x^2 \, dx = \dfrac{x^3}{3} + C$ Thus, we solve for $y$ : $y = \dfrac{x^4}{3} + Cx$ Substituting back, we need to find $C$ using the condition at $x = 1$ : Since $\int_1^1 f(t) \, dt = 0$ , we substitute: $10 \cdot 0 = 5 \cdot 1 \cdot f(1) - 1^5 - 9 \quad \Rightarrow \quad 0 = 5f(1) - 1 - 9$ $5f(1) = 10 \quad \Rightarrow \quad f(1) = 2$ Now use $f(1) = 2$ in the function: $2 = \dfrac{1}{3} + C \cdot 1 \quad \Rightarrow \quad C = \dfrac{5}{3}$ The function $f(x)$ is given by: $f(x) = \dfrac{x^4}{3} + \dfrac{5}{3}x$ To find $f(3)$ : $f(3) = \dfrac{3^4}{3} + \dfrac{5}{3} \cdot 3 = 27 + 5 = 32$ Therefore, the value of $f(3)$ is 32.

Q193

Let

(a, b)

be the point of intersection of the curve

x^2=2 y

and the straight line

y-2 x-6=0

in the second quadrant. Then the integral

\mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}} \dfrac{9 x^2}{1+5^x} \mathrm{~d} x

is equal to :

A 27

B 18

C 24

D 21

Correct Answer

Option C

Solution

\begin{aligned} & x^2=2 y \text{ and } y-2 x-6=0 \\ & \frac{x^2}{2}-2 x-6=0 \\ & x^2-4 x-12=0 \\ & x^2-6 x+2 x-12=0 \\ & x(x-6)+2(x-6)=0 \\ & (x-6)(x+2)=0 \end{aligned}

Point of intersection are $(6,18)$ and $(-2,2)$ $(-2,2)$ is in second quadrant

\begin{aligned} & a=-2, b=2 \\ & I=\int_{-2}^2 \frac{9 x^2}{1+5^x} d x\quad\text{..... (i)} \end{aligned}

$I=\int_{-2}^2 \dfrac{9 x^2}{1+5^{-x}} d x\quad\text{..... (ii)}$

\begin{aligned} &\text{ Adding (i) and (ii) }\\ &\begin{aligned} & 2 I=\int_{-2}^2 9 x^2 d x \\ & I=9 \int_0^2 x^2 d x \\ & I=9\left(\frac{x^3}{3}\right)_0^2 \Rightarrow I=24 \end{aligned} \end{aligned}

Q194

4 \int_0^1\left(\dfrac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\right) d x-3 \log _e(\sqrt{3})

is equal to :

A

2-\sqrt{2}-\log _{\mathrm{e}}(1+\sqrt{2})

B

2+\sqrt{2}+\log _{\mathrm{e}}(1+\sqrt{2})

C

2+\sqrt{2}-\log _{\mathrm{e}}(1+\sqrt{2})

D

2-\sqrt{2}+\log _e(1+\sqrt{2})

Correct Answer

Option A

Solution

\begin{aligned} & I=4 \int_0^1 \frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}} d x \\ & =2 \int_0^1 \sqrt{3+x^2}-\sqrt{1+x^2} d x \end{aligned}

\begin{aligned} & =2\left[\int_0^1 \sqrt{3+x^2} d x-\int_0^1 \sqrt{1+x^2} d x\right] \\ & =2\left[\left(\frac{1}{2} x \sqrt{x^2+3}+\frac{3}{2} \ln \left|\sqrt{3+x^2}+x\right|\right)-\right. \\ & \left.\quad\left(\frac{1}{2} x \sqrt{1+x^2}+\frac{1}{2} \ln \left|\sqrt{1+x^2}+x\right|\right)\right]_0^1 \end{aligned}

\begin{aligned} & =2\left[\left(1+\frac{3}{2} \ln 3-\frac{3}{2} \ln \sqrt{3}\right)-\left(\frac{\sqrt{2}}{2}+\frac{1}{2} \ln (\sqrt{2}+1)\right)\right] \\ & =2\left(1+\frac{3}{4} \ln 3-\frac{1}{\sqrt{2}}-\frac{1}{2} \ln (\sqrt{2}+1)\right) \\ & =3 \ln \sqrt{3}+2-\sqrt{2}-\ln (\sqrt{2}+1) \\ & I-3 \ln \sqrt{3}=2-\sqrt{2}-\ln (\sqrt{2}+1) \end{aligned}

Q195

Let the domain of the function

f(x)=\log _2 \log _4 \log _6\left(3+4 x-x^2\right)

be

(a, b)

. If

\int_0^{b-a}\left[x^2\right] d x=p-\sqrt{q}-\sqrt{r}, p, q, r \in \mathbb{N}, \operatorname{gcd}(p, q, r)=1

, where

[\cdot]

is the greatest integer function, then

p+q+r

is equal to

A 10

B 11

C 9

D 8

Correct Answer

Option A

Solution

Step 1: Ensure the innermost function is greater than zero: $\log_6(3 + 4x - x^2) > 1$ Step 2: Simplify the inequality from Step 1: $3 + 4x - x^2 > 6 \implies x^2 - 4x + 3 Step 3: Solve the quadratic inequality:$ (x - 1)(x - 3) This inequality indicates that $x$ must lie between the roots, giving the interval $x \in (1, 3)$ .

With the domain of $f(x)$ identified as $(a, b) = (1, 3)$ , we calculate the definite integral over $[0, b-a]$ : Calculate the Integral: Given: $\int_0^{b-a} [x^2] \, dx$ For $b - a = 2$ , we compute: $\int_1^2 [x^2] \, dx = \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^2 3 \, dx$ Computation of Each Integral Segment: $\int_1^{\sqrt{2}} 1 \, dx = \sqrt{2} - 1$ $\int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx = 2(\sqrt{3} - \sqrt{2})$ $\int_{\sqrt{3}}^2 3 \, dx = 3(2 - \sqrt{3})$ Summing these, we have: $5 - \sqrt{2} - \sqrt{3}$ Conclusion: The values for $p, q,$ and $r$ are $p = 5$ , $q = 2$ , and $r = 3$ , with the greatest common divisor of these numbers being 1.

Therefore, adding them together gives: $p + q + r = 5 + 2 + 3 = 10$

Q196

Let

f(x)+2 f\left(\dfrac{1}{x}\right)=x^2+5

and

2 g(x)-3 g\left(\dfrac{1}{2}\right)=x, x>0

. If

\alpha=\int_1^2 f(x) \mathrm{d} x

, and

\beta=\int_1^2 g(x) \mathrm{d} x

, then the value of

9 \alpha+\beta

is :

A 0

B 10

C 1

D 11

Correct Answer

Option D

Solution

\begin{aligned} & f(x)+2 f\left(\frac{1}{x}\right)=x^2+5 \\ & 2 f\left(\frac{1}{x}\right)+4 f(x)=2\left(\frac{1}{x^2}+5\right) \\ & 3 f(x)=\frac{2}{x^2}-x^2+5 \\ & f(x)=\frac{1}{3}\left(\frac{2}{x^2}-x^2+5\right) \\ & 2 g(x)-3 g\left(\frac{1}{x}\right)=x \\ & 2 g\left(\frac{1}{x}\right)-3 g(x)=\frac{1}{x} \\ & \text{ Or } 4 g(x)-6 g\left(\frac{1}{x}\right)=2 x \\ & 6 g\left(\frac{1}{x}\right)-9 g(x)=\frac{3}{x} \\ & -5 g(x)=2 x+\frac{3}{x} \end{aligned}

\begin{aligned} & \text{ Or } g(x)=-\frac{1}{5}\left(2 x+\frac{3}{x}\right) \\ & \int_1^2 f(x) d x=\int_1^2 \frac{1}{3}\left(\frac{2}{x^2}-x^2+5\right) d x \end{aligned}

\begin{aligned} &\begin{aligned} & =\frac{1}{3}\left[-\frac{2}{x}-\frac{x^3}{3}+5 x\right]_1^2 \\ & =\frac{1}{3}\left[\left(-\frac{2}{2}-\frac{8}{3}+10\right)-\left(-2-\frac{1}{3}+5\right)\right] \\ & =\frac{1}{3}\left[-1-\frac{8}{3}+10+2+\frac{1}{3}-5\right] \\ & \alpha=\frac{11}{9} \end{aligned}\\ &\text{ Now, } 2 g(x)=x+3 g\left(\frac{1}{2}\right)\\ &\begin{aligned} & 2 g\left(\frac{1}{2}\right)=\frac{1}{2}+3 g\left(\frac{1}{2}\right) \\ & g\left(\frac{1}{2}\right)=-\frac{1}{2} \end{aligned} \end{aligned}

\begin{aligned} & \therefore \beta=\int_1^2 g(x) d x \\ & =\frac{1}{2} \int_1^2\left(x+3 g\left(\frac{1}{2}\right)\right) d x \\ & =\frac{1}{2}\left[\frac{x^2}{2}+3 g\left(\frac{1}{2}\right) x\right]_1^2 \\ & =0 \\ & \therefore 9 \alpha+\beta=11 \end{aligned}

Q197

The value of

\int\limits_{-1}^1 \dfrac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} d x

is equal to

A

1+\dfrac{2 \sqrt{2}}{3}

B

1-\dfrac{2 \sqrt{2}}{3}

C

2+\dfrac{2 \sqrt{2}}{3}

D

3-\dfrac{2 \sqrt{2}}{3}

Correct Answer

Option A

Solution

\begin{aligned} & I=\int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} d x \\ & =\int_0^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} \end{aligned}

\begin{aligned} & +\left(\frac{(1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^{-x}}{e^{-x}+e^x}\right) d x \\ = & \int_0^1 \frac{(1+\sqrt{|x|-x}+\sqrt{|x|+x})\left(e^x+e^{-x}\right)}{e^x+e^{-x}} d x \\ = & \int_0^1(1+\sqrt{|x|-x}+\sqrt{|x|+x}) d x \end{aligned}

\begin{aligned} & =\int_0^1(1+\sqrt{2 x}) d x=\left.x\right|_0 ^1+\left.\frac{\sqrt{2} x^{\frac{3}{2}}}{\frac{3}{2}}\right|_0 ^1 \\ & =1+\frac{2 \sqrt{2}}{3} \end{aligned}

Q198

The integral

\int_0^\pi \dfrac{8 x d x}{4 \cos ^2 x+\sin ^2 x}

is equal to

A

2 \pi^2

B

4 \pi^2

C

\pi^2

D

\dfrac{3 \pi^2}{2}

Correct Answer

Option A

Solution

\begin{aligned} & I=\int_0^\pi \frac{8 x}{4 \cos ^2 x+\sin ^2 x} d x \ldots(1) \\ & I=\int_0^\pi \frac{8(\pi-x)}{4 \cos ^2(\pi-x)+\sin ^2(\pi-x)} d x \\ & I=\int_0^\pi \frac{8(\pi-x)}{4 \cos ^2 x+\sin ^2 x} d x \ldots(2) \end{aligned}

Adding (1) and (2)

\begin{aligned} & 2 I=8 \pi \int_0^\pi \frac{1}{4 \cos ^2 x+\sin ^2 x} d x \\ & I=4 \pi \times 2 \int_0^\pi \frac{\sec ^2 x}{4 \tan ^2 x} d x \end{aligned}

Put $\tan x=t$

\sec ^2 x d x=d t

I=8 \pi \int_0^{\infty} \frac{d t}{4+t^2}

\begin{aligned} & I=8 \pi \frac{1}{2}\left(\tan ^{-1} \frac{t}{2}\right)_0^{\infty} \\ & I=4 \pi\left(\frac{\pi}{2}\right) \\ & I=2 \pi^2 \end{aligned}

Q199

For x

\in

R, x

\ne

0, if y(x) is a differentiable function such that x

\int\limits_1^x y

(t) dt = (x + 1)

\int\limits_1^x ty

(t) dt, then y (x) equals : (where C is a constant.)

A

{C \over x}{e^{ - {1 \over x}}}

B

{C \over {{x^2}}}{e^{ - {1 \over x}}}

C

{C \over {{x^3}}}{e^{ - {1 \over x}}}

D

C{x^3}\,{1 \over {{e^x}}}

Correct Answer

Option C

Solution

x\int\limits_1^x {y\left( t \right)dt} = x\int\limits_1^x {ty} \left( t \right)dt + \int\limits_1^x {ty\left( t \right)} \,\,dt

Differentiate w.r. to x.

\int\limits_1^x {y\left( t \right)dt + x\left[ {y\left( x \right) - y\left( 1 \right)} \right]}

= \int\limits_1^x {ty\left( t \right)dt + x\left[ {xy\left( x \right) - y\left( 1 \right)} \right] + xy\left( x \right) - y\left( 1 \right)}

\int\limits_1^x {y\left( t \right)dt = \int\limits_1^x {ty\left( t \right)dt + {x^2}y\left( x \right) - y\left( 1 \right)} }

Diff. again w.r.t to x

y\left( x \right) - y\left( a \right) = xy\left( x \right) - y\left( a \right) + 2x\,y\left( x \right) + {x^2}{y^3}\left( x \right)

\left( {1 - 3x} \right)y\left( x \right) = {x^2}{y^3}\,\left( x \right)

{{{y^3}\left( x \right)} \over {y\left( x \right)}} = {{1 - 3x} \over {{x^2}}}

{{1dy} \over {ydx}} = {{1 - 3x} \over {{x^2}}} \Rightarrow \ln \,y = - {1 \over x} - 3\ln x

\ln \left( {y{x^3}} \right) = - {1 \over x}

y{x^3} = - {e^{ - 1/x}}

y = {{{e^{ - {1 \over x}}}} \over {{x^3}}}

or

y = {{c{e^{ - {1 \over x}}}} \over {{x^3}}}

Q200

The integral

\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}}

is equal to

A 2

B 4

C

-

1

D

-

2

Correct Answer

Option A

Solution

\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}}

=

\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {2{{\cos }^2}{x \over 2}}}}

=

{1 \over 2}\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{\sec }^2}} {x \over 2}\,dx

{1 \over 2}\left[ {{{\tan {x \over 2}} \over {{1 \over 2}}}} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}

=

\left[ {\tan {x \over 2}} \right]_{{\pi \over 4}}^{{3 \over 4}}

= tan

{{3\pi } \over 8}

$-$ tan

{\pi \over 8}

=

\left( {\sqrt 2 + 1} \right) - \left( {\sqrt 2 - 1} \right)

= 2