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Short Quiz - Chapter 7 Class 12 Integrals

Chapter 7 Class 12 Integrals | 5 questions

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Question 1 of 5
Question 1 of 5
NCERT Exemplar
\(\quad \int_{-2}^2|x \cos \pi x| d x\) is equal to

Correct option: A

Answer: Solution (A) is the correct answer, since \(\mathrm{I}=\int_{-2}^2|x \cos \pi x| d x=2 \int_0^2|x \cos \pi x| d x\)

$$ =2\left\{\int_0^{\frac{1}{2}}|x \cos \pi x| d x+\int_{\frac{1}{2}}^{\frac{3}{2}}|x \cos \pi x| d x+\int_{\frac{3}{2}}^2|x \cos \pi x| d x\right\}=\frac{8}{\pi} . $$

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Question 2 of 5
NCERT Exemplar
\( \int_{a+c}^{b+c} f(x) d x\) is equal to

Correct option: B

Answer: Solution (B) is the correct answer, since by putting \(x=t+c\), we get

$$ \mathrm{I}=\int_a^b f(c+t) d t=\int_a^b f(x+c) d x $$

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Question 3 of 5
NCERT Exemplar
\(\int \frac{x+\sin x}{1+\cos x} d x\) is equal to

Correct option: D

Answer: $$ \int \frac{x+\sin x}{1+\cos x} d x $$


Split the fraction:

$$ \int\left(\frac{x}{1+\cos x}+\frac{\sin x}{1+\cos x}\right) d x $$


Use:

$$ 1+\cos x=2 \cos ^2 \frac{x}{2} $$


So,

$$ \frac{x}{1+\cos x}=\frac{x}{2 \cos ^2 \frac{x}{2}}=\frac{x}{2} \sec ^2 \frac{x}{2} $$


Also,

$$ \frac{\sin x}{1+\cos x}=\tan \frac{x}{2} $$


Therefore,

$$ \frac{x+\sin x}{1+\cos x}=\frac{x}{2} \sec ^2 \frac{x}{2}+\tan \frac{x}{2} $$


Now observe:

$$ \begin{gathered} \frac{d}{d x}\left(x \tan \frac{x}{2}\right)=\tan \frac{x}{2}+x \cdot \frac{1}{2} \sec ^2 \frac{x}{2} \\ =\tan \frac{x}{2}+\frac{x}{2} \sec ^2 \frac{x}{2} \end{gathered} $$


This is exactly the integrand.
Hence,

$$ x \tan \frac{x}{2}+C $$


Answer: (D)

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Question 4 of 5
NCERT Exemplar
\(\quad \int_0^{\frac{\pi}{2}} \sqrt{1-\sin 2 x} d x\) is equal to

Correct option: D

Answer: $$ \int_0^{\pi / 2} \sqrt{1-\sin 2 x} d x $$


Use:

$$ \sin 2 x=2 \sin x \cos x $$


So,

$$ 1-\sin 2 x=1-2 \sin x \cos x $$


But:

$$ 1=\sin ^2 x+\cos ^2 x $$


Therefore,

$$ \begin{aligned} 1-\sin 2 x= & \sin ^2 x+\cos ^2 x-2 \sin x \cos x \\ & =(\sin x-\cos x)^2 \end{aligned} $$


So,

$$ \sqrt{1-\sin 2 x}=|\sin x-\cos x| $$


Now, split the integral at:

$$ x=\frac{\pi}{4} $$

because at \(x=\frac{\pi}{4}\),

$$ \sin x=\cos x $$


From 0 to \(\frac{\pi}{4}\) :

$$ \cos x>\sin x $$


So,

$$ |\sin x-\cos x|=\cos x-\sin x $$


From \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\) :

$$ \sin x>\cos x $$


So,

$$ |\sin x-\cos x|=\sin x-\cos x $$


Therefore,

$$ \int_0^{\pi / 2} \sqrt{1-\sin 2 x} d x=\int_0^{\pi / 4}(\cos x-\sin x) d x+\int_{\pi / 4}^{\pi / 2}(\sin x-\cos x) d x $$


First integral:

$$ \begin{gathered} \int_0^{\pi / 4}(\cos x-\sin x) d x=[\sin x+\cos x]_0^{\pi / 4} \\ =\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-(0+1) \\ =\sqrt{2}-1 \end{gathered} $$


Second integral:

$$ \begin{gathered} \int_{\pi / 4}^{\pi / 2}(\sin x-\cos x) d x=[-\cos x-\sin x]_{\pi / 4}^{\pi / 2} \\ =(0-1)-\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right) \\ =-1+\sqrt{2} \\ =\sqrt{2}-1 \end{gathered} $$


Total:

$$ \begin{gathered} (\sqrt{2}-1)+(\sqrt{2}-1) \\ =2(\sqrt{2}-1) \\ 2(\sqrt{2}-1) \end{gathered} $$


Answer: (D)

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Question 5 of 5
NCERT Exemplar
\(\int \frac{x^3}{x+1}\) is equal to

Correct option: D

Answer: $$ \int \frac{x^3}{x+1} d x $$


Divide \(x^3\) by \(x+1\) :

$$ \frac{x^3}{x+1}=x^2-x+1-\frac{1}{x+1} $$


So,

$$ \begin{gathered} \int \frac{x^3}{x+1} d x=\int\left(x^2-x+1-\frac{1}{x+1}\right) d x \\ =\frac{x^3}{3}-\frac{x^2}{2}+x-\log |x+1|+C \end{gathered} $$


Rearrange:

$$ x-\frac{x^2}{2}+\frac{x^3}{3}-\log |1+x|+C $$


Answer: (D)

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