Integrals Test

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Integrals Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

The equation $$\displaystyle\int^{\pi/4}_{-\pi/4}\left(a|\sin x|+\dfrac{b\sin x}{1+\cos x}+c\right)dx=0$$, where a, b, c are constants, gives a relation between.

A
a, b and c

B
a and c

C
a and b

D
b and c

Solution
Question 2
1 / -0

The value of $$\displaystyle\int^{199\pi/2}_{-\pi/2}\sqrt{(1+\cos 2x)}dx$$ is?

A
$$50\sqrt{2}$$

B
$$100\sqrt{2}$$

C
$$150\sqrt{2}$$

D
$$200\sqrt{2}$$

Solution
Question 3
1 / -0

Evaluate : $$\displaystyle\int^2_1|x^2-3x+2|dx$$

A
$$\dfrac{-1}{6}$$

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

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

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

Solution
Question 4
1 / -0

The value of the integral $$\displaystyle\int^1_0\dfrac{x^{\alpha}-1}{log \alpha}dx$$, is?

A
$$log \alpha$$

B
$$2 log(\alpha +1)$$

C
$$3 log\alpha$$

D
None of these

Solution
Question 5
1 / -0

Value of $$\displaystyle\int^3_2\dfrac{dx}{\sqrt{(1+x^3)}}$$ is?

A
Less than $$1$$

B
Greater than $$2$$

C
Lies between $$3$$ and $$4$$

D
None of these

Solution

Given, $$ \int_{2}^{3} \dfrac{dx}{\sqrt{1+x^3}} $$
Let us consider the function $$ f(x) = \int_{2}^{3} \dfrac{dx}{\sqrt{1+x^3}} $$
according to the question putting the lower and upper limit we get,
$$ f(2) = \dfrac{1}{\sqrt{1+2^3}} = \dfrac{1}{3}$$
$$ f(3) = \dfrac{1}{\sqrt{1+3^3}} = \dfrac{1}{\sqrt{28}}$$
$$ \therefore \dfrac{1}{3} > \dfrac{1}{\sqrt{28}} $$ we conclude that the Larger integrating value lies in the point 2
$$ \therefore \dfrac{1}{\sqrt{28}} < f(x) < \dfrac{1}{3} $$
Since, m < f(x) < M $$ \Rightarrow \int m dx < \int f(x) dx < \int M dx \, \, \, [ \, \forall \, m , M \in \mathbb{R} $$ and be the lower and upper limit of the f(n) ]
we can conclude that,
$$ \int_2^3 f(x) dx < \dfrac{1}{3} $$
Therefore the value is less than 1. (Ans)
Question 6
1 / -0

The value of $$\int_{1}^{e} \dfrac{1+x^{2} \ln x}{x+x^{2} \ln x} d x$$ is

A
$$e$$

B
$$\ln (1+e)$$

C
$$e+\ln (1+e)$$

D
$$e-\ln (1+e)$$

Solution

$$\int_{1}^{e} \dfrac{1+x^{2} \ln x}{x+x^{2} \ln x} d x$$

$$=\int_{1}^{e} \dfrac{1+x-x+x^{2} \ln x}{x(1+x \ln x)} d x$$

Splitting them we get,

$$I =\int_{1}^{e}\left(\dfrac{1}{x}+1\right) d x-\int_{1}^{e} \dfrac{1+\ln x}{1+x \ln x} d x \\$$
$$=[\ln x+x]_{1}^{e}-[\ln (1+x \ln x)]_{1}^{e} \\$$
$$=e-\ln (1+e)$$
Question 7
1 / -0

$$I_{1}=\int_{0}^{\frac{\pi}{2}} \dfrac{\sin x-\cos x}{1+\sin x \cos x} d x, I_{2}=\int_{0}^{2 \pi} \cos ^{6} x d x$$$$I_{3}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{3} x d x, I_{4}=\int_{0}^{1} \ln \left(\dfrac{1}{x}-1\right) d x,$$ then

A
$$I_{2}=I_{3}=I_{4}=0, I_{1} \neq 0$$

B
$$I_{1}=I_{2}=I_{3}=0, I_{4} \neq 0$$

C
$$I_{1}=I_{3}=I_{4}=0, I_{2} \neq 0$$

D
$$I_{1}=I_{2}=I_{3}=0, I_{4} \neq 0$$

Solution

$$\text { } I_{1} =\int_{0}^{\pi / 2} \dfrac{\sin x-\cos x}{1+\sin x \cos x} d x \\ $$

$$=\int_{0}^{\pi / 2} \dfrac{\sin \left(\dfrac{\pi}{2}-x\right)-\cos \left(\dfrac{\pi}{2}-x\right)}{1+\sin \left(\dfrac{\pi}{2}-x\right) \cos \left(\dfrac{\pi}{2}-x\right)} d x\\$$

$$=\int_{0}^{\pi / 2} \dfrac{\cos x-\sin x}{1+\sin x \cos x} d x=-I_{1}\\$$

$$\Rightarrow I_{1}=0\\$$

$$I_{3}=0$$ as $$\sin ^{3} x\\$$ is odd

$$I_{4}=\int_{0}^{1} \ln \left(\dfrac{1-x}{x}\right) d x\\$$

$$=\int_{0}^{1} \ln \left(\dfrac{1-(1-x)}{1-x}\right) d x\\$$

$$=\int_{0}^{1} \ln \dfrac{x}{1-x} d x=-I_{4}\\$$

$$\Rightarrow I_{4}=0\\$$

$$I_{2}=\int_{0}^{2 \pi} \cos ^{6} x d x=2 \int_{0}^{\pi} \cos ^{6} x d x \neq 0$$
Question 8
1 / -0

If $$\int_{1}^{2} e^{x^{2}} d x=a,$$ then $$\int_{e}^{e^{4}} \sqrt{\ln x} d x$$ is equal to

A
$$2 e^{4}-2 e-a$$

B
$$2 e^{4}-e-a$$

C
$$2 e^{4}-e-2 a$$

D
$$e^{4}-e-a$$

Solution

$$I_{1}=\int_{e}^{e^{4}} \sqrt{\ln x} d x,$$ putting $$t=\sqrt{\ln x},$$ i.e., $$d t=\dfrac{d x}{2 x \sqrt{\ln x}}\\$$

$$\Rightarrow d x=2 t e^{t^{2}} d t \\$$

$$\Rightarrow \int_{e}^{e^{4} \sqrt{\ln x} d x} \\$$

$$=\int_{1}^{2} 2 t^{2} e^{t^{2}} d t \\$$

using $$uv$$ integration formula , taking $$2t^2$$ as first function and $$e^{t^2}$$ as second function, we get

$$=\left.t \cdot e^{t^{2}}\right|_{1} ^{2}-\int_{1}^{2} e^{t^{2}} d t=2 e^{4}-e-a $$
Question 9
1 / -0

The value of the definite integral $$\displaystyle \int_{0}^{\pi / 2} \dfrac{\sin 5 x}{\sin x} d x$$ is

A
0

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

C
$$\pi$$

D
$$2 \pi$$

Solution

$$\sin n x-\sin (n-2) x=2 \cos (n-1) x \sin x\\$$
$$\Rightarrow \displaystyle \int \dfrac{\sin n x}{\sin x} d x=\int 2 \cos (n-1) d x+\int \dfrac{\sin (n-2) x}{\sin x} d x\\$$
$$\therefore \displaystyle \int_{0}^{\pi / 2} \dfrac{\sin 5 x}{\sin x} d x=\int_{0}^{\pi / 2} 2 \cos 4 x d x+\int_{0}^{\pi / 2} \dfrac{\sin 3 x}{\sin x} d x\\$$
$$=0+\displaystyle \int_{0}^{\pi / 2} \dfrac{\sin 3 x}{\sin x} d x\\=\displaystyle\int_{0}^{\pi / 2} 3-4sin^2x\ \ d x=\displaystyle\int_{0}^{\pi / 2} 1+2cos2x\ \ d x=\dfrac{\pi}{2}$$
Question 10
1 / -0

If $$\displaystyle \int_{0}^{t} \dfrac{b x \cos 4 x-a \sin 4 x}{x^{2}} d x=\dfrac{a \sin 4 t}{t}-1,$$ where $$0<t<\dfrac{\pi}{4}$$then the values of $$a, b$$ are equal to

A
$$\dfrac{1}{4}, 1$$

B
$$-1,4$$

C
$$2,2$$

D
$$2,4$$

Solution

$$\displaystyle I =b \int_{0}^{t} \dfrac{1}{x} \cos 4 x d x-a \int_{0}^{t} \dfrac{1}{x^{2}} \sin 4 x dx$$

$$=b I_{1}-a I_{2}$$

$$\displaystyle I_{2} =\int_{0}^{t} \dfrac{1}{x^{2}} \sin 4 x dx$$

$$\displaystyle =\left\{\left[-\dfrac{1}{x} \sin 4 x\right]_{0}^{t}+4 \int_{0}^{t} \dfrac{\cos 4 x}{x} d x\right\}$$

$$=\left[-\dfrac{\sin 4 t}{t}+4+4 I_{1}\right],\left\{\lim _{x \rightarrow 0} \dfrac{\sin 4 x}{x}=4\right\}$$

$$\therefore I=b I_{1}-a\left\{-\dfrac{\sin 4 t}{t}+4+4 I_{1}\right\}$$

$$\displaystyle =(b-4 a) \int_{0}^{t} \dfrac{1}{x} \cos 4 x d x+\dfrac{a \sin 4 t}{t}-4 a$$

$$=\dfrac{a \sin 4 t}{t}-1 \\$$
$$\displaystyle \therefore (b-4 a) \int_{0}^{t} \dfrac{1}{x} \cos 4 x d x=4 a-1\\$$
L.H.S. is a function of $$t,$$ whereas R.H:S. is a constant. Hence, we must have $$b-4 a=0$$ and $$4 a-1=0\\$$
$$\therefore a=\dfrac{1}{4}, b=1$$