Integrals Test

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Integrals Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

$$\frac { 1 }{ \pi } \int _{ -2 }^{ 2 }{ \frac { 1 }{ 4+{ x }^{ 2 } } dx= } $$

A
$$0$$

B
$$\frac { 1 }{ 4 } $$

C
$$\frac { \pi }{ 4 } $$

D
$$\frac { 1 }{ 2 } $$

Solution
Question 2
1 / -0

$$\overset { -5 }{ \underset { -4 }{ \int } } e^(x + 5)^2 dx + 3 \overset { 2/3}{ \underset { 1/3 }{ \int } } e^9(9(x-2/3)^2$$ dx is equal toi

A
$$e^5$$

B
$$e^4$$

C
$$3e^2$$

D
0

Solution
Question 3
1 / -0

For $$\displaystyle x\in \,R$$ let $$f(x)=|\sin\,x|$$ and $$g(x)=\int^x_0\,f(t)dt.$$ Let $$p(x)=g(x)-\dfrac{2}{\pi}x$$. Then

A
$$p(x+\pi)=p(x)$$ for all $$x$$

B
$$p(x+\pi)\neq p(x)$$ for all least one but finitely many $$x$$

C
$$p(x+\pi)\neq p(x)$$ for infinitely many $$x$$

D
$$p$$ is a one-one function

Solution

Given:$$f\left(x\right)=\left|\sin{x}\right|$$
$$g\left(x\right)=\displaystyle\int_{0}^{x}{f\left(t\right).dt}$$
$$p\left(x\right)=g\left(x\right)-\dfrac{2}{\pi}x$$

Now, $$p\left(x+\pi\right)=g\left(x+\pi\right)-\dfrac{2}{\pi}\left(x+\pi\right)$$
$$=\displaystyle\int_{0}^{\pi+x}{f\left(t\right)\,dt}-\dfrac{2}{\pi}x-2$$
$$=\displaystyle\int_{0}^{\pi}{f\left(t\right)\,dt}+\displaystyle\int_{\pi}^{\pi+x}{f\left(t\right)\,dt}-\dfrac{2}{\pi}x-2$$
$$f\left(x\right)$$ is periodic function with period $$\pi\,\,\,\therefore\,\,\displaystyle\int_{\pi}^{\pi+x}{f\left(t\right)dt}=\displaystyle\int_{0}^{x}{f\left(t\right)dt}$$
$$\Rightarrow\,p\left(x+\pi\right)=\displaystyle\int_{0}^{\pi}{\left|\sin{x}\right|dx}+g\left(x\right)-\dfrac{2}{\pi}x-2$$
$$=2+g\left(x\right)-\dfrac{2}{\pi}x-2$$
$$\Rightarrow\,p\left(x+\pi\right)=p\left(x\right)$$ for all $$x$$
Question 4
1 / -0

The area of the region bounded by the lines $$x = 1, x = 2$$, and the curves $$x(y - e^x) = \sin x$$ and $$2xy = 2 \sin x + x^3$$ is

A
$$e^2 - e - \dfrac{1}{6}$$

B
$$e^2 - e - \dfrac{7}{6}$$

C
$$e^2 - e + \dfrac{1}{6}$$

D
$$e^2 - e + \dfrac{7}{6}$$

Solution

Given:$$x\left(y-{e}^{x}\right)=\sin{x}$$
$$\Rightarrow\,xy-x{e}^{x}=\sin{x}$$
$$\Rightarrow\,xy=\sin{x}+x{e}^{x}$$
$$\Rightarrow\,y=\dfrac{\sin{x}}{x}+{e}^{x}$$
Given:$$2xy=2\sin{x}+{x}^{3}$$
$$\Rightarrow\,y=\dfrac{\sin{x}}{x}+\dfrac{{x}^{2}}{2}$$
$$A=\displaystyle\int_{1}^{2}{\left\{\dfrac{\sin{x}}{x}+{e}^{x}-\left(\dfrac{\sin{x}}{x}+\dfrac{{x}^{2}}{2}\right)\right\}dx}$$
$$=\displaystyle\int_{1}^{2}{\left({e}^{x}-\dfrac{{x}^{2}}{2}\right)dx}$$
$$=\left[{e}^{x}-\dfrac{{x}^{3}}{6}\right]_{1}^{2}$$
$$={e}^{2}-\dfrac{8}{6}-\left(e-\dfrac{1}{6}\right)$$
$$={e}^{2}-e-\dfrac{7}{6}$$
Question 5
1 / -0

The value of the definite integral $$\displaystyle\int^1_0\dfrac{xdx}{(x^2+16)}$$ lies in the interval $$[a, b]$$. Then smallest such interval is?

A
$$\left[0, \dfrac{1}{17}\right]$$

B
$$[0, 1]$$

C
$$\left[0, \dfrac{1}{27}\right]$$

D
None of these

Solution
Question 6
1 / -0

Let $$I=\overset{1}{\underset{0}{\int}}\dfrac{\sin x}{\sqrt x}dx$$ and $$J=\overset{1}{\underset{0}{\int}}\dfrac{\cos x}{\sqrt x}dx$$. Then which one of the following is true?

A
$$1>\dfrac{2}{3}$$ and $$J>2$$

B
$$1<\dfrac{2}{3}$$ and $$J<2$$

C
$$1<\dfrac{2}{3}$$ and $$J>2$$

D
$$1>\dfrac{2}{3}$$ and $$J<2$$

Solution
Question 7
1 / -0

$$\overset { 1 }{ \underset { -1 }{ \int } } \dfrac{x^3+|x|+1}{x^2+2|x|+1}$$dx is equal to

A
$$In 3$$

B
$$2In 3$$

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

D
none of these
Question 8
1 / -0

Let $$I_{1}=\int_{-2}^{2} \dfrac{x^{6}+3 x^{5}+7 x^{4}}{x^{4}+2} d x$$ and$$I_{2}=\int_{-3}^{1} \dfrac{2(x+1)^{2}+11(x+1)+14}{(x+1)^{4}+2} d x,$$ then the value of$$I_{1}+I_{2}$$ is

A
8

B
$$200 / 3$$

C
$$100 / 3$$

D
None of these

Solution

In $$I_{2},$$ put $$x+1=t,$$
then $$I_{2}=\int_{-2}^{2} \dfrac{2 t^{2}+11 t+14}{t^{4}+2} d t=\int_{-2}^{2} \dfrac{2 x^{2}+11 x+14}{x^{4}+2} d x\\$$
$$\therefore I_{1}+I_{2} = \\$$
$$=\int_{-2}^{2} \dfrac{\left(x^{2}+3 x+7\right)\left(x^{4}+2\right)+5 x}{x^{4}+2} d x \\$$
$$=\int_{-2}^{2}\left(x^{2}+3 x+7\right) d x+5 \int_{-2}^{2} \dfrac{x}{x^{4}+2} d x \\$$
$$=2 \int_{0}^{2}\left(x^{2}+7\right) d x=\dfrac{100}{3}$$
(The other integrals are zero, being integrals of odd functions.)
Question 9
1 / -0

If $$z=x+3i$$ then value of $$\displaystyle\int^4_2\left[arg\left|\dfrac{z-i}{z+i}\right|\right]dx$$, where $$[.]$$ denotes the greatest integer function, is?

A
$$3\sqrt{2}$$

B
$$6\sqrt{3}$$

C
$$\sqrt{6}$$

D
None

Solution
Question 10
1 / -0

If $$I=\displaystyle\int^1_0\cos\left(2\cot^{-1}\sqrt{\left(\dfrac{1-x}{1+x}\right)}\right)dx$$ then?

A
$$I > \dfrac{1}{2}$$

B
$$I=-\dfrac{1}{2}$$

C
$$0 < I < \dfrac{1}{2}$$

D
None of these

Solution