Definite Integration — Page 19 | JEE Mathematics

Q181

Let

f(x)=\int\limits_0^x \mathrm{t}\left(\mathrm{t}^2-9 \mathrm{t}+20\right) \mathrm{dt}, 1 \leq x \leq 5

. If the range of

f

is

[\alpha, \beta]

, then

4(\alpha+\beta)

equals :

A 253

B 157

C 154

D 125

Correct Answer

Option B

Solution

$f^{\prime}(x)=x^3-9 x^2+20 x=x(x-4)(x-5)$

\begin{aligned} & \therefore f(x)=\frac{x^4}{4}-\frac{9 x^3}{3}+\frac{20 x^2}{2} \\ & f(1)=\frac{1}{4}-3+10=\frac{29}{4}=\alpha \\ & \left.f(4)=\frac{256}{4}-3(64) \right\rvert\,+10(16)=32=\beta \\ & 4(\alpha+\beta)=4\left(\frac{29}{4}+32\right)=157 \end{aligned}

Q182

The integral

80 \int\limits_0^{\dfrac{\pi}{4}}\left(\dfrac{\sin \theta+\cos \theta}{9+16 \sin 2 \theta}\right) d \theta

is equal to :

A 3

\log 4

B 4

\log 3

C 6

\log \dfrac{4}{3}

D 2

\log 3

Correct Answer

Option B

Solution

\begin{aligned} & I=80 \int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9+16(2 \sin \theta \cdot \cos \theta)}\right) d \theta \\ & =80 \int_0^{\frac{\pi}{4}} \frac{\sin \theta+\cos \theta}{9-16(1-2 \sin \theta \cdot \cos \theta-1)} d \theta \end{aligned}

\begin{aligned} &=80 \int_0^{\frac{\pi}{4}} \frac{\sin \theta+\cos \theta}{9+16-16(\sin \theta-\cos \theta)^2} \mathrm{~d} \theta\\ &\text{ Let } \sin \theta-\cos \theta=\mathrm{t}\\ &(\cos \theta+\sin \theta) \mathrm{d} \theta=\mathrm{dt}\\ &=80 \int_{-1}^0 \frac{\mathrm{dt}}{25-16 \mathrm{t}^2}\\ &=\frac{80}{16} \int_{-1}^0 \frac{\mathrm{dt}}{\left(\frac{5}{4}\right)^2-\mathrm{t}^2}\\ &\left.=\frac{5}{2\left(\frac{5}{4}\right)} \ln \left|\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right|\right]_{-1}^0\\ &=2 \ln (1)+4 \ln 3\\ &=4 \ln 3 \end{aligned}

Q183

Let for

f(x)=7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x, \quad \mathrm{I}_1=\int_0^{\pi / 4} f(x) \mathrm{d} x

and

\mathrm{I}_2=\int_0^{\pi / 4} x f(x) \mathrm{d} x

. Then

7 \mathrm{I}_1+12 \mathrm{I}_2

is equal to :

A 2

\pi

B 1

C

\pi

D 2

Correct Answer

Option B

Solution

$\begin{aligned} f(x) & =7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x \\ & =7 \tan ^6 x\left(1+\tan ^2 x\right)-3 \tan ^2 x\left(1+\tan ^2 x\right) \\ & =\left(7 \tan ^6 x-3 \tan ^2 x\right)\left(1+\tan ^2 x\right) \\ & =\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x\end{aligned}$ $\begin{aligned} I_1= & \int_0^{\dfrac{\pi}{4}} f(x) d x=\int_0^{\dfrac{\pi}{4}}\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x d x \\ & =\left.\left(\dfrac{7 \tan ^7 x}{7}-\dfrac{3 \tan ^3 x}{3}\right)\right|_0 ^{\dfrac{\pi}{4}}=1-1=0\end{aligned}$ $\begin{aligned} I_2= & \int_0^{\dfrac{\pi}{4}} x f(x) d x=\int_0^{\dfrac{\pi}{4}} x\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x d x \\ & =\left.x\left(\tan ^7 x-\tan ^3 x\right)\right|_0 ^{\dfrac{\pi}{4}}-\int_0^{\dfrac{\pi}{4}} 1 \cdot\left(\tan ^7 x-\tan ^3 x\right) d x \\ & =0-\int_0^{\dfrac{\pi}{4}} \tan ^3 x\left(\tan ^2 x-1\right)\left(\tan ^2 x+1\right) d x\end{aligned}$

\begin{aligned} & =\int_0^{\frac{\pi}{4}}\left(\tan ^3 x-\tan ^5 x\right) \sec ^2 x d x=\frac{\tan ^4 x}{4}-\left.\frac{\tan ^6 x}{6}\right|_0 ^{\frac{\pi}{4}} \\ & =\frac{1}{12} \end{aligned}

Hence $7 I_1+12 I_2=1$

Q184

Let

\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}

be a twice differentiable function such that

f(2)=1

. If

\mathrm{F}(\mathrm{x})=\mathrm{x} f(\mathrm{x})

for all

\mathrm{x} \in \mathrm{R}

,

\int\limits_0^2 x F^{\prime}(x) d x=6

and

\int\limits_0^2 x^2 F^{\prime \prime}(x) d x=40

, then

F^{\prime}(2)+\int\limits_0^2 F(x) d x

is equal to :

A 13

B 11

C 9

D 15

Correct Answer

Option B

Solution

\begin{aligned} &\begin{aligned} & \int_0^2 \mathrm{xF}^{\prime}(\mathrm{x}) \mathrm{dx}=6 \\ & =\left.\mathrm{xF}(\mathrm{x})\right|_0 ^2-\int_0^2 \mathrm{f}(\mathrm{x}) \mathrm{dx}=6 \\ & =2 \mathrm{~F}(2)-\int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=6[\therefore \mathrm{f}(2)=2 \mathrm{~F}(2)=2] \\ & \int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=-2 \quad \ldots(1) \\ & \Rightarrow \int_0^2 \mathrm{~F}(\mathrm{x}) \mathrm{dx}=-2\quad \ldots(2) \end{aligned}\\ &\text{ Also }\\ &\begin{aligned} & \int_0^2 x^2 F^{\prime \prime}(x) d x=\left.x^2 F^{\prime}(x)\right|_0 ^2-2 \int_0^2 x F^{\prime}(x) d x=40 \\ & =4 F^{\prime}(2)-2 \times 6=40 \\ & F^{\prime}(2)=13 \\ & \therefore F^{\prime}(2)+\int_0^2 F(x)=13-2=11 \end{aligned} \end{aligned}

Q185

Let

f

be a real valued continuous function defined on the positive real axis such that

g(x)=\int\limits_0^x t f(t) d t

. If

g\left(x^3\right)=x^6+x^7

, then value of

\sum\limits_{r=1}^{15} f\left(r^3\right)

is :

A 270

B 340

C 310

D 320

Correct Answer

Option C

Solution

\begin{aligned} & g(x)=x 2+x^{\frac{7}{3}} \\ & g^{\prime}(x)=2 x+\frac{7}{3} x^{\frac{4}{3}} \\ & f(x)=\frac{g^{\prime}(x)}{x} \\ & f(x)=2+\frac{7}{3} x^{\frac{1}{3}} \\ & f\left(r^3\right)=2+\frac{7 r}{3} \\ & \sum_{r=1}^{15}\left(1+\frac{7}{3} r\right)=310 \end{aligned}

Q186

The value of

\int_{e^2}^{e^4} \dfrac{1}{x}\left(\dfrac{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}}{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}+e^{\left(\left(6-\log _e x\right)^2+1\right)^{-1}}}\right) d x

is

A 1

B

\log_e2

C

e^2

D 2

Correct Answer

Option A

Solution

\begin{aligned} &\text{ Let } \ln \mathrm{x}=\mathrm{t} \Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{dt}\\ &\begin{aligned} & I=\int_2^4 \frac{e^{\frac{1}{1+t^2}}}{e^{\frac{1}{1+t^2}}+e^{\frac{1}{1+(6-t)^2}}} d t \\ & I=\int_2^4 \frac{\frac{1}{e^{1+(6-t)^2}}}{\frac{1}{e^{1+(6-t)^2}}+e^{\frac{1}{1+t^2}}} d t \\ & 2 I=\int_2^4 d t=(t)_2^4=4-2=2 \\ & I=1 \end{aligned} \end{aligned}

Q187

If

\mathrm{I}=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin ^{\dfrac{3}{2}} x}{\sin ^{\dfrac{3}{2}} x+\cos ^{\dfrac{3}{2}} x} \mathrm{~d} x

, then

\int_0^{2I} \dfrac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} \mathrm{~d} x

equals :

A

\dfrac{\pi^2}{12}

B

\dfrac{\pi^2}{4}

C

\dfrac{\pi^2}{16}

D

\dfrac{\pi^2}{8}

Correct Answer

Option C

Solution

For I Apply king ( $\mathrm{P}-5$ ) and add

\begin{aligned} & 2 I=\int_0^{\pi / 2} d x=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4} \\ & I_2=\int_0^{\pi / 2} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \end{aligned}

Apply king and add

\begin{aligned} & \mathrm{I}_2=\frac{\pi}{4} \int_0^{\pi / 2} \frac{\tan \mathrm{xsec}{ }^2 \mathrm{xdx}}{\tan ^4 \mathrm{x}+1} \\ & \text{ put } \tan ^2 \mathrm{x}=\mathrm{t} \\ & \frac{\pi}{8} \int_0^{\infty} \frac{\mathrm{dt}}{\mathrm{t}^2+1} \\ & =\frac{\pi}{8} \cdot \frac{\pi}{2}=\frac{\pi^2}{16} \end{aligned}

Q188

If

I(m, n)=\int_0^1 x^{m-1}(1-x)^{n-1} d x, m, n>0

, then

I(9,14)+I(10,13)

is

A

I(9,1)

B

I(1,13)

C

\mathrm{I}(19,27)

D

\mathrm{I}(9,13)

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{I}(\mathrm{~m}, \mathrm{~m})=\int_0^1 \mathrm{x}^{\mathrm{m}-1}(1-\mathrm{x})^{\mathrm{n}-1} \mathrm{dx} \\ & \text{ Let } \mathrm{x}=\sin ^2 \theta \quad \mathrm{dx}=2 \sin \theta \cos \theta \mathrm{~d} \theta \\ & \mathrm{I}(\mathrm{~m}, \mathrm{n})=2 \int_0^{\pi / 2}(\sin \theta)^{2 \mathrm{~m}-1}(\cos \theta)^{2 \mathrm{n}-1} \mathrm{~d} \theta \\ & \mathrm{I}(9,14)+\mathrm{I}(10,13)=2 \int_0^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{27} \mathrm{~d} \theta \\ & +2 \int_0^{\pi / 2}(\sin \theta)^{19}(\cos \theta)^{25} \mathrm{~d} \theta \\ & =2 \int_0^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{25} \mathrm{~d} \theta \\ & =\mathrm{I}(9,13) \end{aligned}

Q189

The integral

\int\limits_{-1}^{\dfrac{3}{2}} \left(| \pi^2 x \sin(\pi x) \right|) dx

is equal to:

A

2 + 3\pi

B

4 + \pi

C

1 + 3\pi

D

3 + 2\pi

Correct Answer

Option C

Solution

\begin{aligned} &\text{ Let, } \mathrm{I}=\pi^2 \int_{-1}^{3 / 2}|\mathrm{x} \sin \pi \mathrm{x}| \mathrm{dx}\\ &\begin{aligned} & =\pi^2\left\{\int_{-1}^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_1^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\} \\ & =\pi^2\left\{2 \int_0^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_{-1}^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\} \end{aligned} \end{aligned}

\begin{aligned} &\text{ Consider }\\ &\begin{aligned} & \int \mathrm{x} \sin \pi \mathrm{xdx} \\ & -\mathrm{x} \cdot \frac{1}{\pi} \cos \pi \mathrm{x}+\int 1 \cdot \frac{1}{\pi} \cos \pi \mathrm{xdx} \\ & =--\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2} \\ & \mathrm{I}=\pi^2\left\{2\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_0^1-\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_1^{3 / 2}\right\} \\ & =\pi^2\left\{\frac{2}{\pi}-\left(-\frac{1}{\pi^2}-\frac{1}{\pi}\right)\right\} \\ & =\pi^2\left\{\frac{3}{\pi}+\frac{1}{\pi^2}\right\} \\ & =3 \pi+1 \end{aligned} \end{aligned}

Q190

Let f(x) be a positive function and

I_{1} = \int\limits_{-\dfrac{1}{2}}^{1} 2x \, f(2x(1-2x)) \, dx

and

I_{2} = \int\limits_{-1}^{2} f(x(1-x)) \, dx

. Then the value of

\dfrac{I_{2}}{I_{1}}

is equal to ________

A 12

B 9

C 6

D 4

Correct Answer

Option D

Solution

\begin{aligned} & I_1=\int_{-\frac{1}{2}}^1 2 x f(2 x(1-2 x)) d x \\ & \Rightarrow 2 x=t \Rightarrow 2 d x=d t \quad \Rightarrow I_1=\frac{1}{2} \int_{-1}^2 t f(t(1-t)) d t \\ & \Rightarrow 2 I_1=\int_{-1}^2(1-t) f(1-t)(1-(1-t)) d t \end{aligned}

\Rightarrow 2{I_1} = \int\limits_{ - 1}^2 {f(t(1 - t)dt - \int\limits_{ - 1}^2 {tf(t(1 - t)dt} }

\begin{aligned} & \Rightarrow 2 \mathrm{I}_1=\mathrm{I}_2-2 \mathrm{I}_1 \\ & \Rightarrow 4 \mathrm{I}_1=\mathrm{I}_2 \\ & \Rightarrow \frac{\mathrm{I}_2}{\mathrm{I}_1}=4 \end{aligned}