86404
Let \(f\) be the function on \([-\pi, \pi]\) given by \(f(0)=\) 9 and \(f(x)=\sin \left(\frac{9 x}{2}\right) / \sin \left(\frac{x}{2}\right)\) for \(x \neq 0\). The value of \(\frac{2}{\pi} \int_{-\pi}^{\pi} f(x) d x\) is
1 0
2 4
3 8
4 None of these
Explanation:
(B) : Given, \(\mathrm{f}(\mathrm{x})=\frac{\sin \left(\frac{9 \mathrm{x}}{2}\right)}{\sin \left(\frac{\mathrm{x}}{2}\right)}\) Since, \(\mathrm{f}(\mathrm{x})\) is even function \(\mathrm{f}(-\mathrm{x})=\mathrm{f}(\mathrm{x})\) is \(\int_{-\pi}^{\pi} f(x) d x=2 \int_{0}^{\pi} f(x) d x\) Now, \(\quad \frac{2}{\pi} \int_{-\pi}^{\pi} \frac{\sin \left(\frac{9 x}{2}\right)}{\sin \left(\frac{x}{2}\right)} d x\) \(=\frac{2 \times 2}{\pi} \int_{0}^{\pi} \frac{\sin \left(\frac{9 x}{2}\right)}{\sin \left(\frac{x}{2}\right)} d x\) \(I=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin \left(\frac{9(\pi-x)}{2}\right)}{\sin \left(\frac{\pi-x}{2}\right)} d x=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin 9\left(\frac{\pi}{2}-\frac{x}{2}\right)}{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)} d x\) \(=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin \left[4 \pi+\left(\frac{\pi}{2}-\frac{x}{2}\right)\right]}{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)} d x\) \(=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)}{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)} d x=\frac{4}{\pi} \int_{0}^{\pi} 1 d x\) \(=\frac{4}{\pi} \times \pi\) \(=4\)
AMU-2014
Integral Calculus
86396
The value of the integral \(\int_{1}^{2}\left(\frac{t^{4}+1}{\mathbf{t}^{6}+1}\right) d t\) is
(C) : Given, \(I=\int_{1}^{2}\left(\frac{t^{4}+1}{t^{6}+1}\right) d t \Rightarrow I=\int_{1}^{2}\left(\frac{t^{4}+1}{t^{6}+1}\right)\left(\frac{t^{2}+1}{t^{2}+1}\right) d t\) \(\int_{1}^{2} \frac{\mathrm{t}^{6}+\mathrm{t}^{4}+\mathrm{t}^{2}+1}{\left(\mathrm{t}^{6}+1\right)\left(\mathrm{t}^{2}+1\right)} \mathrm{dt}\) \(=\int_{1}^{2} \frac{\mathrm{t}^{6}+1}{\left(\mathrm{t}^{6}+1\right)\left(\mathrm{t}^{2}+1\right)} \mathrm{dt}+\int_{1}^{2} \frac{\mathrm{t}^{4}+\mathrm{t}^{2}}{\left(\mathrm{t}^{6}+1\right)\left(\mathrm{t}^{2}+1\right)} d \mathrm{t}\) \(\int_{1}^{2} \frac{1}{\mathrm{t}^{2}+1} \mathrm{dt}+\int_{1}^{2} \frac{\mathrm{t}^{2}}{\mathrm{t}^{6}+1} \mathrm{dt}\) \(\mathrm{I}=\left[\tan ^{-1} \mathrm{t}\right]_{1}^{2}+\mathrm{I}_{1}\) \(I_{1}=\int_{1}^{2} \frac{t^{2}}{t^{6}+1} d t\) Put, \(\mathrm{t}^{3}=\theta \Rightarrow 3 \mathrm{t}^{2} \mathrm{dt}=\mathrm{d} \theta\) When, \(\mathrm{t}=1\) then \(\theta=1\) And, when \(\quad t=2\) then \(\theta=8\) Then, \(\mathrm{I}_{1}=\frac{1}{3} \int_{1}^{8} \frac{\mathrm{d} \theta}{\theta^{2}+1}\) \(=\frac{1}{3}\left[\tan ^{-1}(\theta)\right]_{1}^{8}=\frac{1}{3}\left[\tan ^{-1} 8-\frac{\pi}{4}\right]\) Now, \(\left[\tan ^{-1} \mathrm{t}\right]_{1}^{2}+\mathrm{I}_{1}\) \(=\tan ^{-1} 2-\frac{\pi}{4}+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{12}\) \(=\tan ^{-1} 2+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{3}\)
JEE Main-2023-29.01.2023
Integral Calculus
86397
The value of the integral \(\int_{-2}^{2} \frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)} d x\) is equal to:
1 \(5 \mathrm{e}^{2}\)
2 \(3 \mathrm{e}^{-2}\)
3 4
4 6
Explanation:
(D) : \(f(x)=\frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)} d x\) \(\int_{-2}^{2} f(x) d x=\int_{0}^{2}(f(x)+f(-x)) d x\) \(=\int_{0}^{2}\left(\frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)}+\frac{\left|-x^{3}-x\right|}{\left(e^{-x|-x|}+1\right)}\right) d x\) \(=\int_{0}^{2}\left(\frac{\left|\mathrm{x}^{3}+\mathrm{x}\right|}{\left(\mathrm{e}^{\mathrm{x}|\mathrm{x}|}+1\right)}+\frac{\left|\mathrm{x}^{3}+\mathrm{x}\right|}{\left(\mathrm{e}^{-\mathrm{x}|x|}+1\right)}\right) \mathrm{dx}\) \(=\int_{0}^{2}\left(\frac{x^{3}+x}{\left(e^{x^{2}}+1\right)}+\frac{x^{3}+x}{\left(e^{-x^{2}}+1\right)}\right) d x\) \(=\int_{0}^{2}\left(\frac{\mathrm{x}^{3}+\mathrm{x}}{1+\mathrm{e}^{\mathrm{x}^{2}}}+\frac{\mathrm{e}^{\mathrm{x}^{2}}\left(\mathrm{x}^{3}+\mathrm{x}\right)}{1+\mathrm{e}^{\mathrm{x}^{2}}}\right) \mathrm{dx}\) \(=\int_{0}^{2}\left(\mathrm{x}^{3}+\mathrm{x}\right) \mathrm{dx}\) \(=\left[\frac{\mathrm{x}^{4}}{4}+\frac{\mathrm{x}^{2}}{2}\right]_{0}^{2}=4+2=6\)
JEE Main-2022-27.06.2022
Integral Calculus
86398
If \(\int_{1}^{4} x \sqrt{x^{2}-1} d x=\alpha(k)^{\beta}\), then \(\alpha \beta=\)
1 \(\frac{9}{2}\)
2 \(\frac{1}{2}\)
3 \(\frac{1}{3}\)
4 \(\frac{3}{2}\)
Explanation:
(B): Given that:\(\int_{1}^{4} \mathrm{x} \sqrt{\mathrm{x}^{2}-1} \mathrm{dx}=\alpha(\mathrm{k})^{\beta}\) \(I=\int_{1}^{4} x \sqrt{x^{2}-1} d x\) Let, \(x^{2}-1=t\) \(2 x d x=d t\) upper limit \(=t=4^{2}-1=15\) lower limit \(=\mathrm{t}=1^{2}-1=0\) \(\mathrm{I}=\int_{0}^{15} \sqrt{\mathrm{t}} \frac{\mathrm{dt}}{2}\) \(\mathrm{I}=\frac{1}{2} \times \frac{2}{3}\left[\mathrm{t}^{3 / 2}\right]_{0}^{15} \Rightarrow \mathrm{I}=\frac{1}{3}\left[(15)^{3 / 2}\right]\) Compare with \(\alpha(\mathrm{k})^{\beta}\), We have, \(\alpha=\frac{1}{3}, \mathrm{k}=15, \beta=\frac{3}{2}\) \(\therefore \alpha \beta=\frac{1}{3} \times \frac{3}{2}=\frac{1}{2}\)
86404
Let \(f\) be the function on \([-\pi, \pi]\) given by \(f(0)=\) 9 and \(f(x)=\sin \left(\frac{9 x}{2}\right) / \sin \left(\frac{x}{2}\right)\) for \(x \neq 0\). The value of \(\frac{2}{\pi} \int_{-\pi}^{\pi} f(x) d x\) is
1 0
2 4
3 8
4 None of these
Explanation:
(B) : Given, \(\mathrm{f}(\mathrm{x})=\frac{\sin \left(\frac{9 \mathrm{x}}{2}\right)}{\sin \left(\frac{\mathrm{x}}{2}\right)}\) Since, \(\mathrm{f}(\mathrm{x})\) is even function \(\mathrm{f}(-\mathrm{x})=\mathrm{f}(\mathrm{x})\) is \(\int_{-\pi}^{\pi} f(x) d x=2 \int_{0}^{\pi} f(x) d x\) Now, \(\quad \frac{2}{\pi} \int_{-\pi}^{\pi} \frac{\sin \left(\frac{9 x}{2}\right)}{\sin \left(\frac{x}{2}\right)} d x\) \(=\frac{2 \times 2}{\pi} \int_{0}^{\pi} \frac{\sin \left(\frac{9 x}{2}\right)}{\sin \left(\frac{x}{2}\right)} d x\) \(I=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin \left(\frac{9(\pi-x)}{2}\right)}{\sin \left(\frac{\pi-x}{2}\right)} d x=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin 9\left(\frac{\pi}{2}-\frac{x}{2}\right)}{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)} d x\) \(=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin \left[4 \pi+\left(\frac{\pi}{2}-\frac{x}{2}\right)\right]}{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)} d x\) \(=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)}{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)} d x=\frac{4}{\pi} \int_{0}^{\pi} 1 d x\) \(=\frac{4}{\pi} \times \pi\) \(=4\)
AMU-2014
Integral Calculus
86396
The value of the integral \(\int_{1}^{2}\left(\frac{t^{4}+1}{\mathbf{t}^{6}+1}\right) d t\) is
(C) : Given, \(I=\int_{1}^{2}\left(\frac{t^{4}+1}{t^{6}+1}\right) d t \Rightarrow I=\int_{1}^{2}\left(\frac{t^{4}+1}{t^{6}+1}\right)\left(\frac{t^{2}+1}{t^{2}+1}\right) d t\) \(\int_{1}^{2} \frac{\mathrm{t}^{6}+\mathrm{t}^{4}+\mathrm{t}^{2}+1}{\left(\mathrm{t}^{6}+1\right)\left(\mathrm{t}^{2}+1\right)} \mathrm{dt}\) \(=\int_{1}^{2} \frac{\mathrm{t}^{6}+1}{\left(\mathrm{t}^{6}+1\right)\left(\mathrm{t}^{2}+1\right)} \mathrm{dt}+\int_{1}^{2} \frac{\mathrm{t}^{4}+\mathrm{t}^{2}}{\left(\mathrm{t}^{6}+1\right)\left(\mathrm{t}^{2}+1\right)} d \mathrm{t}\) \(\int_{1}^{2} \frac{1}{\mathrm{t}^{2}+1} \mathrm{dt}+\int_{1}^{2} \frac{\mathrm{t}^{2}}{\mathrm{t}^{6}+1} \mathrm{dt}\) \(\mathrm{I}=\left[\tan ^{-1} \mathrm{t}\right]_{1}^{2}+\mathrm{I}_{1}\) \(I_{1}=\int_{1}^{2} \frac{t^{2}}{t^{6}+1} d t\) Put, \(\mathrm{t}^{3}=\theta \Rightarrow 3 \mathrm{t}^{2} \mathrm{dt}=\mathrm{d} \theta\) When, \(\mathrm{t}=1\) then \(\theta=1\) And, when \(\quad t=2\) then \(\theta=8\) Then, \(\mathrm{I}_{1}=\frac{1}{3} \int_{1}^{8} \frac{\mathrm{d} \theta}{\theta^{2}+1}\) \(=\frac{1}{3}\left[\tan ^{-1}(\theta)\right]_{1}^{8}=\frac{1}{3}\left[\tan ^{-1} 8-\frac{\pi}{4}\right]\) Now, \(\left[\tan ^{-1} \mathrm{t}\right]_{1}^{2}+\mathrm{I}_{1}\) \(=\tan ^{-1} 2-\frac{\pi}{4}+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{12}\) \(=\tan ^{-1} 2+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{3}\)
JEE Main-2023-29.01.2023
Integral Calculus
86397
The value of the integral \(\int_{-2}^{2} \frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)} d x\) is equal to:
1 \(5 \mathrm{e}^{2}\)
2 \(3 \mathrm{e}^{-2}\)
3 4
4 6
Explanation:
(D) : \(f(x)=\frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)} d x\) \(\int_{-2}^{2} f(x) d x=\int_{0}^{2}(f(x)+f(-x)) d x\) \(=\int_{0}^{2}\left(\frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)}+\frac{\left|-x^{3}-x\right|}{\left(e^{-x|-x|}+1\right)}\right) d x\) \(=\int_{0}^{2}\left(\frac{\left|\mathrm{x}^{3}+\mathrm{x}\right|}{\left(\mathrm{e}^{\mathrm{x}|\mathrm{x}|}+1\right)}+\frac{\left|\mathrm{x}^{3}+\mathrm{x}\right|}{\left(\mathrm{e}^{-\mathrm{x}|x|}+1\right)}\right) \mathrm{dx}\) \(=\int_{0}^{2}\left(\frac{x^{3}+x}{\left(e^{x^{2}}+1\right)}+\frac{x^{3}+x}{\left(e^{-x^{2}}+1\right)}\right) d x\) \(=\int_{0}^{2}\left(\frac{\mathrm{x}^{3}+\mathrm{x}}{1+\mathrm{e}^{\mathrm{x}^{2}}}+\frac{\mathrm{e}^{\mathrm{x}^{2}}\left(\mathrm{x}^{3}+\mathrm{x}\right)}{1+\mathrm{e}^{\mathrm{x}^{2}}}\right) \mathrm{dx}\) \(=\int_{0}^{2}\left(\mathrm{x}^{3}+\mathrm{x}\right) \mathrm{dx}\) \(=\left[\frac{\mathrm{x}^{4}}{4}+\frac{\mathrm{x}^{2}}{2}\right]_{0}^{2}=4+2=6\)
JEE Main-2022-27.06.2022
Integral Calculus
86398
If \(\int_{1}^{4} x \sqrt{x^{2}-1} d x=\alpha(k)^{\beta}\), then \(\alpha \beta=\)
1 \(\frac{9}{2}\)
2 \(\frac{1}{2}\)
3 \(\frac{1}{3}\)
4 \(\frac{3}{2}\)
Explanation:
(B): Given that:\(\int_{1}^{4} \mathrm{x} \sqrt{\mathrm{x}^{2}-1} \mathrm{dx}=\alpha(\mathrm{k})^{\beta}\) \(I=\int_{1}^{4} x \sqrt{x^{2}-1} d x\) Let, \(x^{2}-1=t\) \(2 x d x=d t\) upper limit \(=t=4^{2}-1=15\) lower limit \(=\mathrm{t}=1^{2}-1=0\) \(\mathrm{I}=\int_{0}^{15} \sqrt{\mathrm{t}} \frac{\mathrm{dt}}{2}\) \(\mathrm{I}=\frac{1}{2} \times \frac{2}{3}\left[\mathrm{t}^{3 / 2}\right]_{0}^{15} \Rightarrow \mathrm{I}=\frac{1}{3}\left[(15)^{3 / 2}\right]\) Compare with \(\alpha(\mathrm{k})^{\beta}\), We have, \(\alpha=\frac{1}{3}, \mathrm{k}=15, \beta=\frac{3}{2}\) \(\therefore \alpha \beta=\frac{1}{3} \times \frac{3}{2}=\frac{1}{2}\)
86404
Let \(f\) be the function on \([-\pi, \pi]\) given by \(f(0)=\) 9 and \(f(x)=\sin \left(\frac{9 x}{2}\right) / \sin \left(\frac{x}{2}\right)\) for \(x \neq 0\). The value of \(\frac{2}{\pi} \int_{-\pi}^{\pi} f(x) d x\) is
1 0
2 4
3 8
4 None of these
Explanation:
(B) : Given, \(\mathrm{f}(\mathrm{x})=\frac{\sin \left(\frac{9 \mathrm{x}}{2}\right)}{\sin \left(\frac{\mathrm{x}}{2}\right)}\) Since, \(\mathrm{f}(\mathrm{x})\) is even function \(\mathrm{f}(-\mathrm{x})=\mathrm{f}(\mathrm{x})\) is \(\int_{-\pi}^{\pi} f(x) d x=2 \int_{0}^{\pi} f(x) d x\) Now, \(\quad \frac{2}{\pi} \int_{-\pi}^{\pi} \frac{\sin \left(\frac{9 x}{2}\right)}{\sin \left(\frac{x}{2}\right)} d x\) \(=\frac{2 \times 2}{\pi} \int_{0}^{\pi} \frac{\sin \left(\frac{9 x}{2}\right)}{\sin \left(\frac{x}{2}\right)} d x\) \(I=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin \left(\frac{9(\pi-x)}{2}\right)}{\sin \left(\frac{\pi-x}{2}\right)} d x=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin 9\left(\frac{\pi}{2}-\frac{x}{2}\right)}{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)} d x\) \(=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin \left[4 \pi+\left(\frac{\pi}{2}-\frac{x}{2}\right)\right]}{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)} d x\) \(=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)}{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)} d x=\frac{4}{\pi} \int_{0}^{\pi} 1 d x\) \(=\frac{4}{\pi} \times \pi\) \(=4\)
AMU-2014
Integral Calculus
86396
The value of the integral \(\int_{1}^{2}\left(\frac{t^{4}+1}{\mathbf{t}^{6}+1}\right) d t\) is
(C) : Given, \(I=\int_{1}^{2}\left(\frac{t^{4}+1}{t^{6}+1}\right) d t \Rightarrow I=\int_{1}^{2}\left(\frac{t^{4}+1}{t^{6}+1}\right)\left(\frac{t^{2}+1}{t^{2}+1}\right) d t\) \(\int_{1}^{2} \frac{\mathrm{t}^{6}+\mathrm{t}^{4}+\mathrm{t}^{2}+1}{\left(\mathrm{t}^{6}+1\right)\left(\mathrm{t}^{2}+1\right)} \mathrm{dt}\) \(=\int_{1}^{2} \frac{\mathrm{t}^{6}+1}{\left(\mathrm{t}^{6}+1\right)\left(\mathrm{t}^{2}+1\right)} \mathrm{dt}+\int_{1}^{2} \frac{\mathrm{t}^{4}+\mathrm{t}^{2}}{\left(\mathrm{t}^{6}+1\right)\left(\mathrm{t}^{2}+1\right)} d \mathrm{t}\) \(\int_{1}^{2} \frac{1}{\mathrm{t}^{2}+1} \mathrm{dt}+\int_{1}^{2} \frac{\mathrm{t}^{2}}{\mathrm{t}^{6}+1} \mathrm{dt}\) \(\mathrm{I}=\left[\tan ^{-1} \mathrm{t}\right]_{1}^{2}+\mathrm{I}_{1}\) \(I_{1}=\int_{1}^{2} \frac{t^{2}}{t^{6}+1} d t\) Put, \(\mathrm{t}^{3}=\theta \Rightarrow 3 \mathrm{t}^{2} \mathrm{dt}=\mathrm{d} \theta\) When, \(\mathrm{t}=1\) then \(\theta=1\) And, when \(\quad t=2\) then \(\theta=8\) Then, \(\mathrm{I}_{1}=\frac{1}{3} \int_{1}^{8} \frac{\mathrm{d} \theta}{\theta^{2}+1}\) \(=\frac{1}{3}\left[\tan ^{-1}(\theta)\right]_{1}^{8}=\frac{1}{3}\left[\tan ^{-1} 8-\frac{\pi}{4}\right]\) Now, \(\left[\tan ^{-1} \mathrm{t}\right]_{1}^{2}+\mathrm{I}_{1}\) \(=\tan ^{-1} 2-\frac{\pi}{4}+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{12}\) \(=\tan ^{-1} 2+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{3}\)
JEE Main-2023-29.01.2023
Integral Calculus
86397
The value of the integral \(\int_{-2}^{2} \frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)} d x\) is equal to:
1 \(5 \mathrm{e}^{2}\)
2 \(3 \mathrm{e}^{-2}\)
3 4
4 6
Explanation:
(D) : \(f(x)=\frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)} d x\) \(\int_{-2}^{2} f(x) d x=\int_{0}^{2}(f(x)+f(-x)) d x\) \(=\int_{0}^{2}\left(\frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)}+\frac{\left|-x^{3}-x\right|}{\left(e^{-x|-x|}+1\right)}\right) d x\) \(=\int_{0}^{2}\left(\frac{\left|\mathrm{x}^{3}+\mathrm{x}\right|}{\left(\mathrm{e}^{\mathrm{x}|\mathrm{x}|}+1\right)}+\frac{\left|\mathrm{x}^{3}+\mathrm{x}\right|}{\left(\mathrm{e}^{-\mathrm{x}|x|}+1\right)}\right) \mathrm{dx}\) \(=\int_{0}^{2}\left(\frac{x^{3}+x}{\left(e^{x^{2}}+1\right)}+\frac{x^{3}+x}{\left(e^{-x^{2}}+1\right)}\right) d x\) \(=\int_{0}^{2}\left(\frac{\mathrm{x}^{3}+\mathrm{x}}{1+\mathrm{e}^{\mathrm{x}^{2}}}+\frac{\mathrm{e}^{\mathrm{x}^{2}}\left(\mathrm{x}^{3}+\mathrm{x}\right)}{1+\mathrm{e}^{\mathrm{x}^{2}}}\right) \mathrm{dx}\) \(=\int_{0}^{2}\left(\mathrm{x}^{3}+\mathrm{x}\right) \mathrm{dx}\) \(=\left[\frac{\mathrm{x}^{4}}{4}+\frac{\mathrm{x}^{2}}{2}\right]_{0}^{2}=4+2=6\)
JEE Main-2022-27.06.2022
Integral Calculus
86398
If \(\int_{1}^{4} x \sqrt{x^{2}-1} d x=\alpha(k)^{\beta}\), then \(\alpha \beta=\)
1 \(\frac{9}{2}\)
2 \(\frac{1}{2}\)
3 \(\frac{1}{3}\)
4 \(\frac{3}{2}\)
Explanation:
(B): Given that:\(\int_{1}^{4} \mathrm{x} \sqrt{\mathrm{x}^{2}-1} \mathrm{dx}=\alpha(\mathrm{k})^{\beta}\) \(I=\int_{1}^{4} x \sqrt{x^{2}-1} d x\) Let, \(x^{2}-1=t\) \(2 x d x=d t\) upper limit \(=t=4^{2}-1=15\) lower limit \(=\mathrm{t}=1^{2}-1=0\) \(\mathrm{I}=\int_{0}^{15} \sqrt{\mathrm{t}} \frac{\mathrm{dt}}{2}\) \(\mathrm{I}=\frac{1}{2} \times \frac{2}{3}\left[\mathrm{t}^{3 / 2}\right]_{0}^{15} \Rightarrow \mathrm{I}=\frac{1}{3}\left[(15)^{3 / 2}\right]\) Compare with \(\alpha(\mathrm{k})^{\beta}\), We have, \(\alpha=\frac{1}{3}, \mathrm{k}=15, \beta=\frac{3}{2}\) \(\therefore \alpha \beta=\frac{1}{3} \times \frac{3}{2}=\frac{1}{2}\)
86404
Let \(f\) be the function on \([-\pi, \pi]\) given by \(f(0)=\) 9 and \(f(x)=\sin \left(\frac{9 x}{2}\right) / \sin \left(\frac{x}{2}\right)\) for \(x \neq 0\). The value of \(\frac{2}{\pi} \int_{-\pi}^{\pi} f(x) d x\) is
1 0
2 4
3 8
4 None of these
Explanation:
(B) : Given, \(\mathrm{f}(\mathrm{x})=\frac{\sin \left(\frac{9 \mathrm{x}}{2}\right)}{\sin \left(\frac{\mathrm{x}}{2}\right)}\) Since, \(\mathrm{f}(\mathrm{x})\) is even function \(\mathrm{f}(-\mathrm{x})=\mathrm{f}(\mathrm{x})\) is \(\int_{-\pi}^{\pi} f(x) d x=2 \int_{0}^{\pi} f(x) d x\) Now, \(\quad \frac{2}{\pi} \int_{-\pi}^{\pi} \frac{\sin \left(\frac{9 x}{2}\right)}{\sin \left(\frac{x}{2}\right)} d x\) \(=\frac{2 \times 2}{\pi} \int_{0}^{\pi} \frac{\sin \left(\frac{9 x}{2}\right)}{\sin \left(\frac{x}{2}\right)} d x\) \(I=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin \left(\frac{9(\pi-x)}{2}\right)}{\sin \left(\frac{\pi-x}{2}\right)} d x=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin 9\left(\frac{\pi}{2}-\frac{x}{2}\right)}{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)} d x\) \(=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin \left[4 \pi+\left(\frac{\pi}{2}-\frac{x}{2}\right)\right]}{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)} d x\) \(=\frac{4}{\pi} \int_{0}^{\pi} \frac{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)}{\sin \left(\frac{\pi}{2}-\frac{x}{2}\right)} d x=\frac{4}{\pi} \int_{0}^{\pi} 1 d x\) \(=\frac{4}{\pi} \times \pi\) \(=4\)
AMU-2014
Integral Calculus
86396
The value of the integral \(\int_{1}^{2}\left(\frac{t^{4}+1}{\mathbf{t}^{6}+1}\right) d t\) is
(C) : Given, \(I=\int_{1}^{2}\left(\frac{t^{4}+1}{t^{6}+1}\right) d t \Rightarrow I=\int_{1}^{2}\left(\frac{t^{4}+1}{t^{6}+1}\right)\left(\frac{t^{2}+1}{t^{2}+1}\right) d t\) \(\int_{1}^{2} \frac{\mathrm{t}^{6}+\mathrm{t}^{4}+\mathrm{t}^{2}+1}{\left(\mathrm{t}^{6}+1\right)\left(\mathrm{t}^{2}+1\right)} \mathrm{dt}\) \(=\int_{1}^{2} \frac{\mathrm{t}^{6}+1}{\left(\mathrm{t}^{6}+1\right)\left(\mathrm{t}^{2}+1\right)} \mathrm{dt}+\int_{1}^{2} \frac{\mathrm{t}^{4}+\mathrm{t}^{2}}{\left(\mathrm{t}^{6}+1\right)\left(\mathrm{t}^{2}+1\right)} d \mathrm{t}\) \(\int_{1}^{2} \frac{1}{\mathrm{t}^{2}+1} \mathrm{dt}+\int_{1}^{2} \frac{\mathrm{t}^{2}}{\mathrm{t}^{6}+1} \mathrm{dt}\) \(\mathrm{I}=\left[\tan ^{-1} \mathrm{t}\right]_{1}^{2}+\mathrm{I}_{1}\) \(I_{1}=\int_{1}^{2} \frac{t^{2}}{t^{6}+1} d t\) Put, \(\mathrm{t}^{3}=\theta \Rightarrow 3 \mathrm{t}^{2} \mathrm{dt}=\mathrm{d} \theta\) When, \(\mathrm{t}=1\) then \(\theta=1\) And, when \(\quad t=2\) then \(\theta=8\) Then, \(\mathrm{I}_{1}=\frac{1}{3} \int_{1}^{8} \frac{\mathrm{d} \theta}{\theta^{2}+1}\) \(=\frac{1}{3}\left[\tan ^{-1}(\theta)\right]_{1}^{8}=\frac{1}{3}\left[\tan ^{-1} 8-\frac{\pi}{4}\right]\) Now, \(\left[\tan ^{-1} \mathrm{t}\right]_{1}^{2}+\mathrm{I}_{1}\) \(=\tan ^{-1} 2-\frac{\pi}{4}+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{12}\) \(=\tan ^{-1} 2+\frac{1}{3} \tan ^{-1} 8-\frac{\pi}{3}\)
JEE Main-2023-29.01.2023
Integral Calculus
86397
The value of the integral \(\int_{-2}^{2} \frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)} d x\) is equal to:
1 \(5 \mathrm{e}^{2}\)
2 \(3 \mathrm{e}^{-2}\)
3 4
4 6
Explanation:
(D) : \(f(x)=\frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)} d x\) \(\int_{-2}^{2} f(x) d x=\int_{0}^{2}(f(x)+f(-x)) d x\) \(=\int_{0}^{2}\left(\frac{\left|x^{3}+x\right|}{\left(e^{x|x|}+1\right)}+\frac{\left|-x^{3}-x\right|}{\left(e^{-x|-x|}+1\right)}\right) d x\) \(=\int_{0}^{2}\left(\frac{\left|\mathrm{x}^{3}+\mathrm{x}\right|}{\left(\mathrm{e}^{\mathrm{x}|\mathrm{x}|}+1\right)}+\frac{\left|\mathrm{x}^{3}+\mathrm{x}\right|}{\left(\mathrm{e}^{-\mathrm{x}|x|}+1\right)}\right) \mathrm{dx}\) \(=\int_{0}^{2}\left(\frac{x^{3}+x}{\left(e^{x^{2}}+1\right)}+\frac{x^{3}+x}{\left(e^{-x^{2}}+1\right)}\right) d x\) \(=\int_{0}^{2}\left(\frac{\mathrm{x}^{3}+\mathrm{x}}{1+\mathrm{e}^{\mathrm{x}^{2}}}+\frac{\mathrm{e}^{\mathrm{x}^{2}}\left(\mathrm{x}^{3}+\mathrm{x}\right)}{1+\mathrm{e}^{\mathrm{x}^{2}}}\right) \mathrm{dx}\) \(=\int_{0}^{2}\left(\mathrm{x}^{3}+\mathrm{x}\right) \mathrm{dx}\) \(=\left[\frac{\mathrm{x}^{4}}{4}+\frac{\mathrm{x}^{2}}{2}\right]_{0}^{2}=4+2=6\)
JEE Main-2022-27.06.2022
Integral Calculus
86398
If \(\int_{1}^{4} x \sqrt{x^{2}-1} d x=\alpha(k)^{\beta}\), then \(\alpha \beta=\)
1 \(\frac{9}{2}\)
2 \(\frac{1}{2}\)
3 \(\frac{1}{3}\)
4 \(\frac{3}{2}\)
Explanation:
(B): Given that:\(\int_{1}^{4} \mathrm{x} \sqrt{\mathrm{x}^{2}-1} \mathrm{dx}=\alpha(\mathrm{k})^{\beta}\) \(I=\int_{1}^{4} x \sqrt{x^{2}-1} d x\) Let, \(x^{2}-1=t\) \(2 x d x=d t\) upper limit \(=t=4^{2}-1=15\) lower limit \(=\mathrm{t}=1^{2}-1=0\) \(\mathrm{I}=\int_{0}^{15} \sqrt{\mathrm{t}} \frac{\mathrm{dt}}{2}\) \(\mathrm{I}=\frac{1}{2} \times \frac{2}{3}\left[\mathrm{t}^{3 / 2}\right]_{0}^{15} \Rightarrow \mathrm{I}=\frac{1}{3}\left[(15)^{3 / 2}\right]\) Compare with \(\alpha(\mathrm{k})^{\beta}\), We have, \(\alpha=\frac{1}{3}, \mathrm{k}=15, \beta=\frac{3}{2}\) \(\therefore \alpha \beta=\frac{1}{3} \times \frac{3}{2}=\frac{1}{2}\)
