Integrals Test

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Integrals Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

The value of the definite integral $$\displaystyle \int_{0}^{\pi/2}sinx\space sin2x\space sin3xdx$$ is equal to

A
$$\dfrac{1}{3}$$

B
$$-\dfrac{2}{3}$$

C
$$-\dfrac{1}{3}$$

D
$$\dfrac{1}{6}$$

Solution

$$=\displaystyle \int_{0}^{\pi/2}\sin x\sin2x\sin3xdx$$
$$=\displaystyle \int_{0}^{\pi/2}sin\left(\dfrac{\pi}{2}-x\right) sin2\left(\dfrac{\pi}{2}-x\right) sin3\left(\dfrac{\pi}{2}-x\right)dx$$
$$\Rightarrow I=\displaystyle \int_{0}^{\pi/2}-cosxsin2xcos3xdx$$
$$\therefore 2I=\displaystyle \int_{0}^{\pi/2}-sin2x(cosxcos3x-sin3xsinx)dx$$
$$=\displaystyle \int_{0}^{\pi/2}-sin2xcos(4x)dx$$
$$=-\displaystyle \int_{0}^{\pi/2}sin2x(2cos^22x-1)dx$$
Put $$cos2x=t \Rightarrow -sin2x\times 2dx=dt$$
$$\therefore 2I=\displaystyle \int_{1}^{-1}\dfrac{1}{2}(2t^2-1)dt=\dfrac{1}{2}\left[ \dfrac{2t^3}{3}-t \right]^{-1}$$
$$=\dfrac{1}{2}\left[ \dfrac{2}{3}(-1)^3-(-1)-\left( \dfrac{2}{3}(1)^3 -1 \right)\right]$$
$$=\dfrac{1}{2}\left[ -\dfrac{2}{3}+1-\dfrac{2}{3}+1\right]=\dfrac{1}{2}\left[\dfrac{2}{3}\right]\qquad I\Rightarrow \dfrac{1}{6}$$
Question 2
1 / -0

If $$\displaystyle \int f(x) dx = F(x)$$, then $$\displaystyle \int x^3 f(x^2) dx$$ is equal to

A
$$\frac{1}{2} [x^2$$ {$$F(x)$$}$$^2 - \displaystyle \int$$ {$$F(x)$$}$$^2 dx$$]

B
$$\frac{1}{2} [x^2 F(x^2) - \displaystyle \int F(x^2) d (x^2)]$$

C
$$\frac{1}{2} [x^2 F(x^2) - \frac{1}{2} \displaystyle \int$$ {$$F(x)$$}$$^2 dx$$]

D
None of the above

Solution

We have, $$\displaystyle \int f(x) dx = F(x)$$
$$\therefore \displaystyle \int x^3 f(x^2) dx = \frac{1}{2} \displaystyle \int \underbrace{x^2} \underbrace{f(x^2) d(x^2)}$$
I II
$$ = \frac{1}{2} [x^2 F(x^2) - \displaystyle \int F(x^2) d (x^2)]$$
Question 3
1 / -0

Directions For Questions

$$f(x)=\displaystyle \int_{0}^{x}(4t^4-at^3)dt $$ and $$g(x)$$ is quadratic satisfying $$g(0)+6=g'(0)-c=g''(c)+2b=0, y=h(x) \space and y=g(x)$$ intersect in 4 distinct points with abscissae $$x_i;i=1,2,3,4 $$ such that $$\displaystyle \sum \dfrac{i}{x_i}=8,a,b,c \in R^+ and h(x) = f'(x).$$

...view full instructions

Abscissae of point of intersection are in

A
AP

B
GP

C
HP

D
None of these

Solution

We have $$g(x)=g(0)+xg'(0)+\dfrac{x^2}{2}g''(0)=-bx^2+cx-6$$
$$h(x)=g(x)=4x^4-ax^3+bx^2-cs+6=0 $$ has 4 distinct real roots. Using Descarte's rule of signs.
Given biquadratic equation has 4 distinct positive roots.
Let the roots be $$x_1,x_2,x_3$$ and $$x_4$$
Now, $$\dfrac{\dfrac{1}{x_1}+\dfrac{2}{x_2}|+\dfrac{3}{x_3}+\dfrac{4}{x_4}}{4}\geq \sqrt{\dfrac{24}{x_1x_2x_3x_4}}$$
$$\Rightarrow 2\geq2 \Rightarrow \dfrac{1}{x_1}=\dfrac{2}{x_2}=\dfrac{3}{x_3}=\dfrac{4}{x_4}=k $$
$$\Rightarrow \dfrac{1}{x_1}.\dfrac{2}{x_2}.\dfrac{3}{x_3}.\dfrac{4}{x_4}=k^4$$
$$\Rightarrow \dfrac{24}{3/2}=k^4 \Rightarrow k=2$$
$$\therefore$$ Roots are $$\dfrac{1}{2},1,\dfrac{3}{2}$$ and $$2.$$ Also,a=20 and c=25
Question 4
1 / -0

The value of
$$\int_{2}^{5} \dfrac {\sqrt x}{\sqrt {x} + \sqrt {7-x}} dx$$
is :

A
3

B
2

C
$$\dfrac {3}{2}$$

D
$$\dfrac {1}{2}$$

Solution

Let
$$I = \int_{2}^{5} \dfrac {\sqrt x}{\sqrt {x} + \sqrt {7-x}} dx$$ .................(1)
$$\Rightarrow \int_{2}^{5} \dfrac {\sqrt 2+5-x}{\sqrt {2+5-x} + \sqrt {7-2-5+x}} dx$$
$$\Rightarrow \int_{2}^{5} \dfrac {\sqrt 7 -x}{\sqrt {7-x} + \sqrt {x}} dx$$ ............(2)

Adding equation (1) and (2)
$$2I = \int_{2}^{5} \dfrac {\sqrt x + \sqrt{7-x}}{\sqrt {7-x} + \sqrt {x}} dx$$
= $$\int_{2}^{5} 1 dx = []_{2}^{5} = 5-2 = 3$$
$$I = \dfrac {3}{2}$$

hence, $$\int_{2}^{5} \dfrac {\sqrt x}{\sqrt {x} + \sqrt {7-x}} dx=\dfrac{3}{2}$$
option C is correct.
Question 5
1 / -0

If $$A(x) = \int_{0}^{x} \theta ^2 d\theta$$
then value of $$A(3)$$ will be :

A
9

B
27

C
3

D
81

Solution

$$A(x) = \int_{0}^{x} \theta ^2 d\theta$$
= $$[\dfrac {\theta ^3}{3}]_{0}^{x}$$
= $$\dfrac {1}{3} x^3$$
$$\Rightarrow A(3) = \dfrac {1}{3} 3^3$$
= $$\dfrac {1}{3} \times 3 \times 3 \times 3$$
= $$3 \times 3$$
= $$9$$

Thus, option A is correct
Question 6
1 / -0

$$\int_{a-c}^{b-c} f(x+c) dx $$ is:

A
$$\int_{a}^{b} f(x+c) dx $$

B
$$\int_{a}^{b} f(x) dx $$

C
$$\int_{a-2c}^{b-2c} f(x) dx $$

D
$$\int_{a}^{b} f(x+2c) dx $$
Question 7
1 / -0

$$\int_{0}^{log 5} \displaystyle \frac{e^{x}\sqrt{e^{x}-1}}{e^{x}+3} dx =$$

A
$$3+ 2 \pi$$

B
$$4 - \pi$$

C
$$2 + \pi$$

D
$$\pi -4$$

Solution

Let $$I=\int { \cfrac { { e }^{ x }\sqrt { { e }^{ x }-1 } }{ { e }^{ x }+3 } } $$
Substitute $$t=e^{ x }\Rightarrow dt=e^{ x }dx$$, we get
$$I=\int { \cfrac { \sqrt { t-1 } }{ t+3 } dt } $$
Substitute $$u=\sqrt { t-1 } \Rightarrow du=\cfrac { 1 }{ 2\sqrt { t-1 } } dt$$, we get
$$I=2\int { \cfrac { { u }^{ 2 } }{ { u }^{ 2 }+4 } du } =2\int { \left( 1-\cfrac { 4 }{ { u }^{ 2 }+4 } \right) du } =2\int { du } -2\int { \cfrac { 4 }{ { u }^{ 2 }+4 } } du\\ =2u-4{ tan }^{ -1 }\left( \cfrac { u }{ 2 } \right) =2\sqrt { t-1 } -4{ tan }^{ -1 }\left( \cfrac { \sqrt { t-1 } }{ 2 } \right) \\ =2\sqrt { e^{ x }-1 } -4{ tan }^{ -1 }\left( \cfrac { \sqrt { e^{ x }-1 } }{ 2 } \right) $$
Therefore,
$$\int _{ 0 }^{ \log { 5 } }{ Idx } ={ \left[ 2\sqrt { e^{ x }-1 } -4{ tan }^{ -1 }\left( \cfrac { \sqrt { e^{ x }-1 } }{ 2 } \right) \right] }_{ 0 }^{ \log { 5 } }=4-\pi $$
Question 8
1 / -0

If $$\displaystyle I_n=\int_{0}^{\tfrac{\pi}{4}} \tan^nx\sec^2xdx,$$ then $$I_1, I_2, I_3..$$ are in

A
A.P.

B
G.P.

C
H.P.

D
A.G.P.

Solution

$$\displaystyle I_n=\int_{0}^{\tfrac{\pi}{4}} \tan^nx\sec^2xdx,$$
Substituting $$t=\tan x \Rightarrow dt =\sec^2xdx$$
$$I_n=\int_{0}^{1}t^ndt$$ =$$\dfrac{1}{n\;+\;1}$$

$$\Rightarrow I_n\;=\;\dfrac{1}{n\;+\;1}$$

Reciprocal of the terms $$I_1,I_2,I_3..$$ are in A.P, therefore $$\dfrac{1}{n+1}, \dfrac{1}{n+2}, \dfrac{1}{n+3}$$ are in H.P.
Hence, option 'C' is correct.
Question 9
1 / -0

Evaluate the integral
$$\displaystyle \int_{0}^{\pi/4}\frac{ {s}i {n}\theta+ {c} {o} {s}\theta}{9+16 {s}i {n}2\theta} \ {d}\theta $$

A
$$\displaystyle \frac{1}{20} \log 2$$

B
$$\displaystyle \frac{1}{20} \log 3$$

C
$$\log 3$$

D
$$\log 2$$

Solution

$$\displaystyle \int_{0}^{\pi/ 4} \dfrac {\sin \theta + \cos \theta}{9 + 16\sin 2\theta} d\theta$$

Let $$\sin \theta - \cos \theta = t$$

$$\therefore (\cos \theta + \sin \theta) d\theta = dt$$

$$= \displaystyle \int_{-1}^{0} \dfrac {dt}{9 + 16(1 – t^{2})} (\because t^{2} = 1 – 2\cdot \sin \theta \cdot \cos \theta)$$

$$= \displaystyle \int_{-1}^{0} \dfrac {dt}{25 – 16t^{2}}$$

$$= \dfrac {1}{16} \displaystyle \int_{-1}^{0} \dfrac {dt}{\left (\dfrac {5}{4}\right )^{2} – t^{2}}$$

$$= \dfrac {1}{16}\times \dfrac {1}{2\times \dfrac {5}{4}}\left [ln \left |\dfrac {\dfrac {5}{4} +t}{\dfrac {5}{4} – t}\right |\right ]_{-1}^{0}$$

$$\left (\because \displaystyle \int \dfrac {1}{a^{2} – x^{2}} dx = \dfrac {1}{2a} ln \left |\dfrac {a + x}{a – x}\right | + c\right )$$

$$= \dfrac {1}{40} \left [ln |1| - ln \left |\dfrac {\dfrac {1}{4}}{\dfrac {9}{4}}\right |\right ]$$

$$= \dfrac {1}{40} ln 9 = \dfrac {ln 3}{20}$$

$$\therefore \displaystyle \int_{0}^{\pi/ 4} \dfrac {\sin \theta + \cos \theta}{9 + 16 \sin 2\theta} d\theta = \dfrac {ln 3}{20}$$.
Question 10
1 / -0

Evaluate: $$\displaystyle \int_{0}^{\pi}\frac{dx}{5+4\cos x}$$

A
$$\dfrac{\pi}{2}$$

B
$$\dfrac{\pi}{6}$$

C
$$\dfrac{\pi}{3}$$

D
$$-\pi$$

Solution

Putting $$\cos{x}=\dfrac{1-\tan^2{(x/2)}}{1+\tan^2{(x/2)}}$$,

$$\displaystyle\int_{0}^{\pi}\dfrac{dx}{5+4\dfrac{1-\tan^2{(x/2)}}{1+\tan^2{(x/2)}}}$$

$$=\displaystyle\int_{0}^{\pi} \dfrac{\{1+\tan^2{(x/2)\}}dx}{9+\tan^2{(x/2)}}$$

$$=\displaystyle\int_{0}^{\pi} \dfrac{\sec^2{(x/2)}dx}{\tan^2{(x/2)}+3^2}$$

Let , $$z=\tan{(x/2)}\implies dz=\dfrac{1}{2}\sec^2{(x/2)}dx\implies \sec^2{(x/2)}dx=2dz$$

And when, $$x=0, z=0$$ and when $$x=\pi, z=\infty$$

Hence, integration becomes:-

$$\displaystyle\int_{0}^{\infty} \dfrac{2dz}{z^2+3^2}$$

$$=\dfrac{2}{3}\left[\tan^{-1} {\dfrac{z}{3}}\right]_{0}^{\infty}$$

$$=\dfrac{2}{3}\left[\tan^{-1} {\infty}-\tan^{-1} {0}\right]$$

$$=\dfrac{2}{3}\left(\dfrac{\pi}{2}-0\right)$$

$$=\dfrac{\pi}{3}$$

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Re-Attempt Weekly Quiz Competition

Integrals Test - 45

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Integrals Test - 45

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SHARING IS CARING

Question 1

$$\dfrac{1}{3}$$

$$-\dfrac{2}{3}$$

$$-\dfrac{1}{3}$$

$$\dfrac{1}{6}$$

Solution

Question 2

$$\frac{1}{2} [x^2$$ {$$F(x)$$}$$^2 - \displaystyle \int$$ {$$F(x)$$}$$^2 dx$$]

$$\frac{1}{2} [x^2 F(x^2) - \displaystyle \int F(x^2) d (x^2)]$$

$$\frac{1}{2} [x^2 F(x^2) - \frac{1}{2} \displaystyle \int$$ {$$F(x)$$}$$^2 dx$$]

None of the above

Solution

Question 3

AP

GP

HP

None of these

Solution

Question 4

3

2

$$\dfrac {3}{2}$$

$$\dfrac {1}{2}$$

Solution

Question 5

9

27

3

81

Solution

Question 6

$$\int_{a}^{b} f(x+c) dx $$

$$\int_{a}^{b} f(x) dx $$

$$\int_{a-2c}^{b-2c} f(x) dx $$

$$\int_{a}^{b} f(x+2c) dx $$

Question 7

$$3+ 2 \pi$$

$$4 - \pi$$

$$2 + \pi$$

$$\pi -4$$

Solution

Question 8

A.P.

G.P.

H.P.

A.G.P.

Solution

Question 9

$$\displaystyle \frac{1}{20} \log 2$$

$$\displaystyle \frac{1}{20} \log 3$$

$$\log 3$$

$$\log 2$$

Solution

Question 10

$$\dfrac{\pi}{2}$$

$$\dfrac{\pi}{6}$$

$$\dfrac{\pi}{3}$$

$$-\pi$$

Solution

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