86594
The value of \(\int_{0}^{2 \pi} \frac{d x}{e^{\sin x}+1}\) is
1 \(\pi\)
2 0
3 \(3 \pi\)
4 \(\frac{\pi}{2}\)
Explanation:
(A) : Let \(I=\int_{0}^{2 \pi} \frac{d x}{e^{\sin x}+1}\) \(\mathrm{I}=\int_{0}^{2 \pi} \frac{\mathrm{dx}}{\mathrm{e}^{\sin (2 \pi-\mathrm{x})}+1}\) \(\mathrm{I}=\int_{0}^{2 \pi} \frac{\mathrm{dx}}{\mathrm{e}^{-\sin x}+1}\) \(I=\int_{0}^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+1} d x\) Adding equation (i) and (ii), we get - \(2 \mathrm{I}=\int_{0}^{2 \pi} 1 . \mathrm{dx}=2 \pi\) \(\therefore \mathrm{I}=\pi\)
COMEDK-2020
Integral Calculus
86595
If \(I_{1}=\int_{0}^{\pi / 2} x \sin x d x\) and \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\), then which one of the following is true?
1 \(\mathrm{I}_{1}+\mathrm{I}_{2}=\frac{\pi}{2}\)
2 \(\mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}\)
3 \(\mathrm{I}_{1}+\mathrm{I}_{2}=0\)
4 \(I_{1}=I_{2}\)
Explanation:
(B) : \(I_{1}=\int_{0}^{\pi / 2} x \sin x d x\) \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\) \(\mathrm{I}_{1}=\int_{0}^{\pi / 2} \mathrm{x} \sin \mathrm{x} d \mathrm{x}\) \(I_{1}=-[x \cos x]^{\frac{\pi}{2}}-\int_{0}^{\pi / 2}(-\cos x) d x\) \(=[0-0]+[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}\) \(I_{1}=-[x \cos x]_{0}^{\frac{\pi}{2}}-\int_{0}^{\pi / 2}(-\cos x) d x\) \(\mathrm{I}_{1}=1\) \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\) \(=[\mathrm{x} \sin \mathrm{x}]_{0}^{\pi / 2}-\int_{0}^{\pi / 2} \cos \mathrm{xd} \mathrm{x}\) \(=\left(\frac{\pi}{2}-0\right)+[\sin x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}+1\) Now, \(\mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}+1-1 \Rightarrow \mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}\)
Jamia Millia Islamia-2008
Integral Calculus
86596
Let \(f(x)\) be a positive function. Let \(I_{1}=\int_{1-k}^{k} x f\{x(1-x)\} d x\), \(I_{2}=\int_{1-k}^{k} f\{x(1-x)\} d x\), Where \(2 k-1>0\), then \(\frac{I_{1}}{I_{2}}\) is
86594
The value of \(\int_{0}^{2 \pi} \frac{d x}{e^{\sin x}+1}\) is
1 \(\pi\)
2 0
3 \(3 \pi\)
4 \(\frac{\pi}{2}\)
Explanation:
(A) : Let \(I=\int_{0}^{2 \pi} \frac{d x}{e^{\sin x}+1}\) \(\mathrm{I}=\int_{0}^{2 \pi} \frac{\mathrm{dx}}{\mathrm{e}^{\sin (2 \pi-\mathrm{x})}+1}\) \(\mathrm{I}=\int_{0}^{2 \pi} \frac{\mathrm{dx}}{\mathrm{e}^{-\sin x}+1}\) \(I=\int_{0}^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+1} d x\) Adding equation (i) and (ii), we get - \(2 \mathrm{I}=\int_{0}^{2 \pi} 1 . \mathrm{dx}=2 \pi\) \(\therefore \mathrm{I}=\pi\)
COMEDK-2020
Integral Calculus
86595
If \(I_{1}=\int_{0}^{\pi / 2} x \sin x d x\) and \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\), then which one of the following is true?
1 \(\mathrm{I}_{1}+\mathrm{I}_{2}=\frac{\pi}{2}\)
2 \(\mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}\)
3 \(\mathrm{I}_{1}+\mathrm{I}_{2}=0\)
4 \(I_{1}=I_{2}\)
Explanation:
(B) : \(I_{1}=\int_{0}^{\pi / 2} x \sin x d x\) \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\) \(\mathrm{I}_{1}=\int_{0}^{\pi / 2} \mathrm{x} \sin \mathrm{x} d \mathrm{x}\) \(I_{1}=-[x \cos x]^{\frac{\pi}{2}}-\int_{0}^{\pi / 2}(-\cos x) d x\) \(=[0-0]+[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}\) \(I_{1}=-[x \cos x]_{0}^{\frac{\pi}{2}}-\int_{0}^{\pi / 2}(-\cos x) d x\) \(\mathrm{I}_{1}=1\) \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\) \(=[\mathrm{x} \sin \mathrm{x}]_{0}^{\pi / 2}-\int_{0}^{\pi / 2} \cos \mathrm{xd} \mathrm{x}\) \(=\left(\frac{\pi}{2}-0\right)+[\sin x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}+1\) Now, \(\mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}+1-1 \Rightarrow \mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}\)
Jamia Millia Islamia-2008
Integral Calculus
86596
Let \(f(x)\) be a positive function. Let \(I_{1}=\int_{1-k}^{k} x f\{x(1-x)\} d x\), \(I_{2}=\int_{1-k}^{k} f\{x(1-x)\} d x\), Where \(2 k-1>0\), then \(\frac{I_{1}}{I_{2}}\) is
86594
The value of \(\int_{0}^{2 \pi} \frac{d x}{e^{\sin x}+1}\) is
1 \(\pi\)
2 0
3 \(3 \pi\)
4 \(\frac{\pi}{2}\)
Explanation:
(A) : Let \(I=\int_{0}^{2 \pi} \frac{d x}{e^{\sin x}+1}\) \(\mathrm{I}=\int_{0}^{2 \pi} \frac{\mathrm{dx}}{\mathrm{e}^{\sin (2 \pi-\mathrm{x})}+1}\) \(\mathrm{I}=\int_{0}^{2 \pi} \frac{\mathrm{dx}}{\mathrm{e}^{-\sin x}+1}\) \(I=\int_{0}^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+1} d x\) Adding equation (i) and (ii), we get - \(2 \mathrm{I}=\int_{0}^{2 \pi} 1 . \mathrm{dx}=2 \pi\) \(\therefore \mathrm{I}=\pi\)
COMEDK-2020
Integral Calculus
86595
If \(I_{1}=\int_{0}^{\pi / 2} x \sin x d x\) and \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\), then which one of the following is true?
1 \(\mathrm{I}_{1}+\mathrm{I}_{2}=\frac{\pi}{2}\)
2 \(\mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}\)
3 \(\mathrm{I}_{1}+\mathrm{I}_{2}=0\)
4 \(I_{1}=I_{2}\)
Explanation:
(B) : \(I_{1}=\int_{0}^{\pi / 2} x \sin x d x\) \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\) \(\mathrm{I}_{1}=\int_{0}^{\pi / 2} \mathrm{x} \sin \mathrm{x} d \mathrm{x}\) \(I_{1}=-[x \cos x]^{\frac{\pi}{2}}-\int_{0}^{\pi / 2}(-\cos x) d x\) \(=[0-0]+[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}\) \(I_{1}=-[x \cos x]_{0}^{\frac{\pi}{2}}-\int_{0}^{\pi / 2}(-\cos x) d x\) \(\mathrm{I}_{1}=1\) \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\) \(=[\mathrm{x} \sin \mathrm{x}]_{0}^{\pi / 2}-\int_{0}^{\pi / 2} \cos \mathrm{xd} \mathrm{x}\) \(=\left(\frac{\pi}{2}-0\right)+[\sin x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}+1\) Now, \(\mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}+1-1 \Rightarrow \mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}\)
Jamia Millia Islamia-2008
Integral Calculus
86596
Let \(f(x)\) be a positive function. Let \(I_{1}=\int_{1-k}^{k} x f\{x(1-x)\} d x\), \(I_{2}=\int_{1-k}^{k} f\{x(1-x)\} d x\), Where \(2 k-1>0\), then \(\frac{I_{1}}{I_{2}}\) is
86594
The value of \(\int_{0}^{2 \pi} \frac{d x}{e^{\sin x}+1}\) is
1 \(\pi\)
2 0
3 \(3 \pi\)
4 \(\frac{\pi}{2}\)
Explanation:
(A) : Let \(I=\int_{0}^{2 \pi} \frac{d x}{e^{\sin x}+1}\) \(\mathrm{I}=\int_{0}^{2 \pi} \frac{\mathrm{dx}}{\mathrm{e}^{\sin (2 \pi-\mathrm{x})}+1}\) \(\mathrm{I}=\int_{0}^{2 \pi} \frac{\mathrm{dx}}{\mathrm{e}^{-\sin x}+1}\) \(I=\int_{0}^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+1} d x\) Adding equation (i) and (ii), we get - \(2 \mathrm{I}=\int_{0}^{2 \pi} 1 . \mathrm{dx}=2 \pi\) \(\therefore \mathrm{I}=\pi\)
COMEDK-2020
Integral Calculus
86595
If \(I_{1}=\int_{0}^{\pi / 2} x \sin x d x\) and \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\), then which one of the following is true?
1 \(\mathrm{I}_{1}+\mathrm{I}_{2}=\frac{\pi}{2}\)
2 \(\mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}\)
3 \(\mathrm{I}_{1}+\mathrm{I}_{2}=0\)
4 \(I_{1}=I_{2}\)
Explanation:
(B) : \(I_{1}=\int_{0}^{\pi / 2} x \sin x d x\) \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\) \(\mathrm{I}_{1}=\int_{0}^{\pi / 2} \mathrm{x} \sin \mathrm{x} d \mathrm{x}\) \(I_{1}=-[x \cos x]^{\frac{\pi}{2}}-\int_{0}^{\pi / 2}(-\cos x) d x\) \(=[0-0]+[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}\) \(I_{1}=-[x \cos x]_{0}^{\frac{\pi}{2}}-\int_{0}^{\pi / 2}(-\cos x) d x\) \(\mathrm{I}_{1}=1\) \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\) \(=[\mathrm{x} \sin \mathrm{x}]_{0}^{\pi / 2}-\int_{0}^{\pi / 2} \cos \mathrm{xd} \mathrm{x}\) \(=\left(\frac{\pi}{2}-0\right)+[\sin x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}+1\) Now, \(\mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}+1-1 \Rightarrow \mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}\)
Jamia Millia Islamia-2008
Integral Calculus
86596
Let \(f(x)\) be a positive function. Let \(I_{1}=\int_{1-k}^{k} x f\{x(1-x)\} d x\), \(I_{2}=\int_{1-k}^{k} f\{x(1-x)\} d x\), Where \(2 k-1>0\), then \(\frac{I_{1}}{I_{2}}\) is
86594
The value of \(\int_{0}^{2 \pi} \frac{d x}{e^{\sin x}+1}\) is
1 \(\pi\)
2 0
3 \(3 \pi\)
4 \(\frac{\pi}{2}\)
Explanation:
(A) : Let \(I=\int_{0}^{2 \pi} \frac{d x}{e^{\sin x}+1}\) \(\mathrm{I}=\int_{0}^{2 \pi} \frac{\mathrm{dx}}{\mathrm{e}^{\sin (2 \pi-\mathrm{x})}+1}\) \(\mathrm{I}=\int_{0}^{2 \pi} \frac{\mathrm{dx}}{\mathrm{e}^{-\sin x}+1}\) \(I=\int_{0}^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+1} d x\) Adding equation (i) and (ii), we get - \(2 \mathrm{I}=\int_{0}^{2 \pi} 1 . \mathrm{dx}=2 \pi\) \(\therefore \mathrm{I}=\pi\)
COMEDK-2020
Integral Calculus
86595
If \(I_{1}=\int_{0}^{\pi / 2} x \sin x d x\) and \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\), then which one of the following is true?
1 \(\mathrm{I}_{1}+\mathrm{I}_{2}=\frac{\pi}{2}\)
2 \(\mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}\)
3 \(\mathrm{I}_{1}+\mathrm{I}_{2}=0\)
4 \(I_{1}=I_{2}\)
Explanation:
(B) : \(I_{1}=\int_{0}^{\pi / 2} x \sin x d x\) \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\) \(\mathrm{I}_{1}=\int_{0}^{\pi / 2} \mathrm{x} \sin \mathrm{x} d \mathrm{x}\) \(I_{1}=-[x \cos x]^{\frac{\pi}{2}}-\int_{0}^{\pi / 2}(-\cos x) d x\) \(=[0-0]+[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}\) \(I_{1}=-[x \cos x]_{0}^{\frac{\pi}{2}}-\int_{0}^{\pi / 2}(-\cos x) d x\) \(\mathrm{I}_{1}=1\) \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\) \(=[\mathrm{x} \sin \mathrm{x}]_{0}^{\pi / 2}-\int_{0}^{\pi / 2} \cos \mathrm{xd} \mathrm{x}\) \(=\left(\frac{\pi}{2}-0\right)+[\sin x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}+1\) Now, \(\mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}+1-1 \Rightarrow \mathrm{I}_{2}-\mathrm{I}_{1}=\frac{\pi}{2}\)
Jamia Millia Islamia-2008
Integral Calculus
86596
Let \(f(x)\) be a positive function. Let \(I_{1}=\int_{1-k}^{k} x f\{x(1-x)\} d x\), \(I_{2}=\int_{1-k}^{k} f\{x(1-x)\} d x\), Where \(2 k-1>0\), then \(\frac{I_{1}}{I_{2}}\) is
